Rd Sharma XII Vol 1 2020 for Class 12 Science Maths Chapter 18 - Indefinite Integrals

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Rd Sharma XII Vol 1 2020 Solutions for Class 12 Science Maths Chapter 18 Indefinite Integrals are provided here with simple step-by-step explanations. These solutions for Indefinite Integrals are extremely popular among Class 12 Science students for Maths Indefinite Integrals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2020 Book of Class 12 Science Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2020 Solutions. All Rd Sharma XII Vol 1 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 18.104:

Question 1:

$\int \frac{x}{x^{2} + 3 x + 2} d x$

Answer:

$\int \frac{x}{x^{2} + 3 x + 2} d x x = A \frac{d}{d x} (x^{2} + 3 x + 2) + B x = A (2 x + 3) + B x = (2 A x) + 3 A + B$

Comparing the Coefficients of like powers of x we get
$2 A = 1 \Rightarrow A = \frac{1}{2} 3 A + B = 0 \frac{3}{2} + B = 0 B = - \frac{3}{2} x = \frac{1}{2} (2 x + 3) - \frac{3}{2}$

$Now, \int \frac{x}{x^{2} + 3 x + 2} d x = \int [\frac{\frac{1}{2} (2 x + 3) - \frac{3}{2}}{x^{2} + 3 x + 2}] d x = \frac{1}{2} \int \frac{(2 x + 3) d x}{x^{2} + 3 x + 2} - \frac{3}{2} \int \frac{d x}{x^{2} + 3 x + 2} = \frac{1}{2} \int \frac{(2 x + 3) d x}{x^{2} + 3 x + 2} - \frac{3}{2} \int \frac{d x}{x^{2} + 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2} = \frac{1}{2} \int \frac{(2 x + 3) d x}{x^{2} + 3 x + 2} - \frac{3}{2} \int \frac{d x}{{(x + \frac{3}{2})}^{2} - \frac{9}{4} + 2} = \frac{1}{2} \int \frac{(2 x + 3) d x}{x^{2} + 3 x + 2} - \frac{3}{2} \int \frac{d x}{{(x + \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}} = \frac{1}{2} \log |x^{2} + 3 x + 2| - \frac{3}{2} \times \frac{1}{2 \times \frac{1}{2}} \log |\frac{x + \frac{3}{2} - \frac{1}{2}}{x + \frac{3}{2} + \frac{1}{2}}| + C = \frac{1}{2} \log |x^{2} + 3 x + 2| - \frac{3}{2} \log |\frac{x + 1}{x + 2}| + C$

Page No 18.104:

Question 2:

$\int \frac{x + 1}{x^{2} + x + 3} d x$

Answer:

$\int \frac{(x + 1) d x}{x^{2} + x + 3} x + 1 = \frac{A d}{d x} (x^{2} + x + 3) + B x + 1 = A (2 x + 1) + B x + 1 = 2 A x + A + B$

Comparing Coefficients of like powers of x
$2 A = 1 A = \frac{1}{2} A + B = 1 \frac{1}{2} + B = 1 B = \frac{1}{2} (x + 1) = \frac{1}{2} (2 x + 1) + \frac{1}{2}$

$Now, \int \frac{(x + 1) d x}{x^{2} + x + 3} = \int \frac{\frac{1}{2} (2 x + 1) d x}{x^{2} + x + 3} + \frac{1}{2} \int \frac{d x}{x^{2} + x + 3} = \frac{1}{2} \int \frac{(2 x + 1) d x}{x^{2} + x + 3} + \frac{1}{2} \int \frac{d x}{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 3} = \frac{1}{2} \int \frac{(2 x + 1) d x}{x^{2} + x + 3} + \frac{1}{2} \int \frac{d x}{{(x + \frac{1}{2})}^{2} + 3 - \frac{1}{4}} = \frac{1}{2} \int \frac{(2 x + 1) d x}{x^{2} + x + 3} + \frac{1}{2} \int \frac{d x}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{11}}{2})}^{2}} = \frac{1}{2} \log |x^{2} + x + 3| + \frac{1}{2} \times \frac{2}{\sqrt{11}} \tan^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{11}}{2}}) + C = \frac{1}{2} \log |x^{2} + x + 3| + \frac{1}{\sqrt{11}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{11}}) + C$

Page No 18.104:

Question 3:

$\int \frac{x - 3}{x^{2} + 2 x - 4} d x$

Answer:

$\int (\frac{x - 3}{x^{2} + 2 x - 4}) d x x - 3 = A \frac{d}{d x} (x^{2} + 2 x - 4) + B x - 3 = A (2 x + 2) + B x - 3 = (2 A) x + 2 A + B$

Comparing Coefficients of like powers of x

$2 A = 1 A = \frac{1}{2} 2 A + B = - 3 2 \times \frac{1}{2} + B = - 3 B = - 4$

$Now, \int (\frac{x - 3}{x^{2} + 2 x - 4}) d x = \int (\frac{\frac{1}{2} (2 x + 2) - 4}{x^{2} + 2 x - 4}) d x = \frac{1}{2} \int \frac{(2 x + 2) d x}{(x^{2} + 2 x - 4)} - 4 \int \frac{d x}{x^{2} + 2 x + 1 - 1 - 4} = \frac{1}{2} \int \frac{(2 x + 2) d x}{(x^{2} + 2 x - 4)} - 4 \int \frac{d x}{{(x + 1)}^{2} - {(\sqrt{5})}^{2}} = \frac{1}{2} \log |x^{2} + 2 x - 4| - \frac{4}{2 \sqrt{5}} \log |\frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}}| + C = \frac{1}{2} \log |x^{2} + 2 x - 4| - \frac{2}{\sqrt{5}} \log |\frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}}| + C$

Page No 18.104:

Question 4:

$\int \frac{2 x - 3}{x^{2} + 6 x + 13} d x$

Answer:

$\int \frac{(2 x - 3) d x}{x^{2} + 6 x + 13} 2 x - 3 = A \frac{d}{d x} (x^{2} + 6 x + 13) + B 2 x - 3 = A (2 x + 6) + B 2 x - 3 = (2 A) x + 6 A + B$

Comparing Coefficients of like powers of x
$2 A = 2 A = 1 6 A + B = - 3 6 + B = - 3 B = - 9 ∴ 2 x - 3 = 1 (2 x + 6) - 9$

$∴ \int \frac{(2 x - 3)}{x^{2} + 6 x + 13} d x = \int (\frac{2 x + 6 - 9}{x^{2} + 6 x + 13}) d x = \int (\frac{2 x + 6}{x^{2} + 6 x + 13}) d x - \int \frac{9 d x}{x^{2} + 6 x + 13} = \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 13} - 9 \int \frac{d x}{x^{2} + 6 x + 3^{2} - 3^{2} + 13} = \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 13} - 9 \int \frac{d x}{{(x + 3)}^{2} + 2^{2}} = \log |x^{2} + 6 x + 13| - 9 \times \frac{1}{2} \tan^{- 1} (\frac{x + 3}{2}) + C = \log |x^{2} + 6 x + 13| - \frac{9}{2} \tan^{- 1} (\frac{x + 3}{2}) + C$

Page No 18.104:

Question 5:

$\int \frac{x - 1}{3 x^{2} - 4 x + 3} d x$

Answer:

$\int (\frac{x - 1}{3 x^{2} - 4 x + 3}) d x x - 1 = A \frac{d}{d x} (3 x^{2} - 4 x + 3) + B x - 1 = A (6 x - 4) + B x - 1 = (6 A) x + B - 4 A$

Comparing the Coefficients of like powers of x
$6 A = 1 A = \frac{1}{6} B - 4 A = - 1 B - 4 \times \frac{1}{6} = - 1 B = - 1 + \frac{2}{3} B = \frac{1}{3}$

$Now, \int \frac{(x - 1) d x}{3 x^{2} - 4 x + 3} = \int [\frac{\frac{1}{6} (6 x - 4) + \frac{1}{3}}{3 x^{2} - 4 x + 3}] d x = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 3} + \frac{1}{3} \int \frac{d x}{3 x^{2} - 4 x + 3} = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 3} + \frac{1}{9} \int \frac{d x}{x^{2} - \frac{4}{3} x + 1} = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 3} + \frac{1}{9} \int \frac{d x}{x^{2} - \frac{4}{3} x + {(\frac{2}{3})}^{2} {(\frac{2}{3})}^{2} + 1} = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 3} + \frac{1}{9} \int \frac{d x}{{(x - \frac{2}{3})}^{2} - \frac{4}{9} + 1} = \frac{1}{6} \int \frac{(6 x - 4) d x}{3 x^{2} - 4 x + 13} + \frac{1}{9} \int \frac{d x}{{(x - \frac{2}{3})}^{2} + {(\frac{\sqrt{5}}{3})}^{2}} = \frac{1}{6} \log |3 x^{2} - 4 x + 3| + \frac{1}{9} \times \frac{3}{\sqrt{5}} \tan^{- 1} (\frac{x^{- \frac{2}{3}}}{\frac{\sqrt{5}}{3}}) + C = \frac{1}{6} \log |3 x^{2} - 4 x + 3| + \frac{1}{3 \sqrt{5}} \tan^{- 1} (\frac{3 x - 2}{\sqrt{5}}) + C = \frac{1}{6} \log |3 x^{2} - 4 x + 3| + \frac{\sqrt{5}}{15} \tan^{- 1} (\frac{3 x - 2}{\sqrt{5}}) + C$

Page No 18.104:

Question 6:

$\int \frac{2 x}{2 + x - x^{2}} d x$

Answer:

$\int \frac{2 x d x}{(2 + x - x^{2})} 2 x = A \frac{d}{d x} (2 + x - x^{2}) + B 2 x = A (0 + 1 - 2 x) + B 2 x = (- 2 A) x + A + B$

Comparing the Coefficients of like powers of x
$- 2 A = 2 A = - 1 A + B = 0 - 1 + B = 0 B = 1$

$Now, \int \frac{2 x d x}{(2 + x - x^{2})} = \int (\frac{- 1 (1 - 2 x) + 1}{- x^{2} + x + 2}) d x = - \int (\frac{1 - 2 x}{- x^{2} + x + 2}) d x + \int \frac{d x}{- x^{2} + x + 2} = - I_{1} + I_{2} . . . (1) (say) where I_{1} = \int (\frac{1 - 2 x}{- x^{2} + x + 2}) d x I_{2} = \int \frac{d x}{- x^{2} + x + 2} I_{1} = \int (\frac{1 - 2 x}{- x^{2} + x + 2}) d x let - x^{2} + x + 2 = t \Rightarrow (1 - 2 x) d x = d t I_{1} = \int \frac{d t}{t} I_{1} = \log |t| + C_{1} = \log |2 + x - x^{2}| + C_{1} . . . (2) I_{2} = \int \frac{d x}{- x^{2} + x + 2} I_{2} = \int \frac{- d x}{x^{2} - x - 2} I_{2} = \int \frac{- d x}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 2} I_{2} = \int \frac{- d x}{{(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}} I_{2} = - \frac{1}{2 \times \frac{3}{2}} \log |\frac{x - \frac{1}{2} - \frac{3}{2}}{x - \frac{1}{2} + \frac{3}{2}}| + C_{2} I_{2} = - \frac{1}{3} \log |\frac{x - 2}{x + 1}| + C_{2} . . . (3) from (1) (2) and (3) \int (\frac{2 x}{2 + x - x^{2}}) d x = - \log |2 + x - x^{2}| - \frac{1}{3} \log |\frac{x - 2}{x + 1}| + C_{1} + C_{2} = - \log |2 + x - x^{2}| + \frac{1}{3} \log |\frac{1 + x}{x - 2}| + C where C = C_{1} + C_{2}$

Page No 18.104:

Question 7:

$\int \frac{1 - 3 x}{3 x^{2} + 4 x + 2} d x$

Answer:

$\int \frac{(1 - 3 x) d x}{3 x^{2} + 4 x + 2} 1 - 3 x = A \frac{d}{d x} (3 x^{2} + 4 x + 2) + B 1 - 3 x = A (6 x + 4) + B 1 - 3 x = (6 A) x + 4 A + B$

Comparing the Coefficients of like powers of x

$6 A = - 3 A = \frac{- 1}{2} 4 A + B = 1 4 \times \frac{- 1}{2} + B = 1 B = 3$

$1 - 3 x = - \frac{1}{2} (6 x + 4) + 3 Now, \int \frac{(1 - 3 x) d x}{3 x^{2} + 4 x + 2} = \int (\frac{\frac{- 1}{2} (6 x + 4) + 3}{3 x^{2} + 4 x + 2}) d x = - \frac{1}{2} \int \frac{(6 x + 4) d x}{3 x^{2} + 4 x + 2} + 3 \int \frac{d x}{3 x^{2} + 4 x + 2} = - \frac{1}{2} I_{1} + 3 I_{2} (say) . . . (1) where I_{1} = \int \frac{6 x + 4}{3 x^{2} + 4 x + 2} and I_{2} = \int \frac{d x}{3 x^{2} + 4 x + 2} I_{1} = \int (\frac{6 x + 4}{3 x^{2} + 4 x + 2}) d x let 3 x^{2} + 4 x + 2 = t \Rightarrow (6 x + 4) d x = d t I_{1} = \int \frac{d t}{t} = \log |t| + C_{1} = \log |3 x^{2} + 4 x + 2| + C_{1} . . . (2) I_{2} = \int \frac{d x}{3 x^{2} + 4 x + 2} I_{2} = \frac{1}{3} \int \frac{d x}{x^{2} + \frac{4}{3} x + \frac{2}{3}} I_{2} = \frac{1}{3} \int \frac{d x}{x^{2} + \frac{4 x}{x} + {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{2} + \frac{2}{3}} I_{2} = \frac{1}{3} \int \frac{d x}{{(x - \frac{2}{3})}^{2} - \frac{4}{9} + \frac{2}{3}} I_{2} = \frac{1}{3} \int \frac{d x}{{(x + \frac{2}{3})}^{2} + \frac{2}{9}} I_{2} = \frac{1}{3} \int \frac{d x}{{(x + \frac{2}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} I_{2} = \frac{1}{3} \times \frac{3}{\sqrt{2}} \tan^{- 1} (\frac{x + \frac{2}{3}}{\frac{\sqrt{2}}{3}}) + C_{2} I_{2} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 2}{\sqrt{2}}) + C_{2} . . . (3) from (1), (2) and (3) \int \frac{(1 - 3 x) d x}{3 x^{2} + 4 x + 2} = - \frac{1}{2} \log |3 x^{2} + 4 x + 2| + \frac{3 \times 1}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 2}{\sqrt{2}}) + C_{1} + C_{2} = - \frac{1}{2} \log |3 x^{2} + 4 x + 2| + \frac{3}{\sqrt{2}} \tan^{- 1} (\frac{3 x + 2}{\sqrt{2}}) + C (Where C = C_{1} + C_{2})$

Page No 18.104:

Question 8:

$\int \frac{2 x + 5}{x^{2} - x - 2} d x$

Answer:

$\int \frac{(2 x + 5) d x}{x^{2} - x - 2} 2 x + 5 = A \frac{d}{d x} (x^{2} - x - 2) + B 2 x + 5 = A (2 x - 1) + B 2 x + 5 = (2 A) x + B - A$

Comparing the Coefficients of like powers of x

$2 A = 2 A = 1 B - A = 5 B - 1 = 5 B = 6$

$∴ 2 x + 5 = 1 \cdot (2 x - 1) + 6 ∴ \int (\frac{2 x + 5}{x^{2} - x - 2}) d x \Rightarrow \int (\frac{(2 x - 1) + 6}{x^{2} - x - 2}) d x \Rightarrow \int (\frac{2 x - 1}{x^{2} - x - 2}) d x + 6 \int \frac{d x}{x^{2} - x - 2} = I_{1} + 6 I_{2} (say) . . . (1) where I_{1} = \int (\frac{2 x - 1}{x^{2} - x - 2}) d x I_{2} = \int \frac{d x}{x^{2} - x - 2} I_{1} = \int (\frac{2 x - 1}{x^{2} - x - 2}) d x let x^{2} - x - 2 = t \Rightarrow (2 x - 1) d x = d t I_{1} = \int \frac{d t}{t} I_{1} = \log (t) I_{1} = \log |x^{2} - x - 2| + C_{1} . . . (2) I_{2} = \int \frac{d x}{x^{2} - x - 2} I_{2} = \int \frac{d x}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 2} I_{2} = \int \frac{d x}{{(x - \frac{1}{2})}^{2} - \frac{1}{4} - 2} I_{2} = \int \frac{d x}{{(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}} I_{2} = \frac{1}{2 \times \frac{3}{2}} \log |\frac{x - \frac{1}{2} - \frac{3}{2}}{x - \frac{1}{2} + \frac{3}{2}}| I_{2} = \frac{1}{3} \log |\frac{x - 2}{x + 1}| + C_{2} . . . (3) \int \frac{(2 x + 5) d x}{x^{2} - x - 2} = \log |x^{2} - x - 2| + \frac{6}{3} \log |\frac{x - 2}{x + 1}| + C_{1} + C_{2} = \log |x^{2} - x - 2| + 2 \log |\frac{x - 2}{x + 1}| + C (Where C = C_{1} + C_{2})$

Page No 18.104:

Question 9:

$\int \frac{a x^{3} + b x}{x^{4} + c^{2}} d x$

Answer:

$\int \frac{(a x^{3} + b x)}{x^{4} + c^{2}} d x = \int \frac{a x^{3}}{x^{4} + c^{2}} d x + \int \frac{b x}{{(x^{2})}^{2} + c^{2}} d x = I_{1} + I_{2} (say) Where I_{1} = \int \frac{a x^{3}}{x^{4} + c^{2}} d x & I_{2} = \int \frac{b x}{{(x^{2})}^{2} + c^{2}} d x Now, I_{1} = \int \frac{a x^{3}}{x^{4} + c^{2}} d x let x^{4} + c^{2} = t \Rightarrow 4 x^{3} d x = d t \Rightarrow x^{3} d x = \frac{d t}{4} I_{1} = \frac{a}{4} \int \frac{d t}{t} = \frac{a}{4} \log |t| + C_{1} = \frac{a}{4} \log |x^{4} + c^{2}| + C_{1} Now, I_{2} = \int \frac{b x}{{(x^{2})}^{2} + c^{2}} d x let x^{2} = p \Rightarrow 2 x d x = d p \Rightarrow x d x = \frac{d p}{2}$

$I_{2} = \frac{b}{2} \int \frac{d p}{p^{2} + c^{2}} = \frac{b}{2} \times \frac{1}{c} \tan^{- 1} (\frac{p}{c}) + C_{2} = \frac{b}{2 c} \tan^{- 1} (\frac{x^{2}}{c}) + C_{2} \int \frac{(a x^{3} + b x)}{x^{4} + c^{2}} d x = \frac{a}{4} \log |x^{4} + c^{2}| + \frac{b}{2 c} \tan^{- 1} (\frac{x^{2}}{c}) + C_{1} + C_{2} = \frac{a}{4} \log |x^{4} + c^{2}| + \frac{b}{2 c} \tan^{- 1} (\frac{x^{2}}{c}) + C (Where C = C_{1} + C_{2})$

Page No 18.104:

Question 10:

$\int \frac{(3 \sin x - 2) \cos x}{5 - \cos^{2} x - 4 \sin x} d x$

Answer:

$\int \frac{(3 \sin x - 2) \cos x d x}{5 - \cos^{2} x - 4 \sin x} = \int \frac{(3 \sin x - 2) \cos x d x}{5 - (1 - \sin^{2} x) - 4 \sin x} = \int \frac{(3 \sin x - 2) \cos x d x}{\sin^{2} x - 4 \sin x + 4} Let \sin x = t \Rightarrow \cos x d x = d t \int \frac{(3 t - 2) d t}{t^{2} - 4 t + 4} 3 t - 2 = A \frac{d}{d x} (t^{2} - 4 t + 4) + B 3 t - 2 = A (2 t - 4) + B 3 t - 2 = (2 A) t + B - 4 A$

Comparing the Coefficients of like powers of t

$2 A = 3 A = \frac{3}{2} B - 4 A = - 2 B - 4 \times \frac{3}{2} = - 2 B = - 2 + 6 B = 4$

$3 t - 2 = \frac{3}{2} (2 t - 4) + 4 ∴ \int \frac{(3 t - 2) d t}{t^{2} - 4 t + 4} = \int (\frac{\frac{3}{2} (2 t - 4) + 4}{t^{2} - 4 t + 4}) d t = \frac{3}{2} \int (\frac{2 t - 4}{t^{2} - 4 t + 4}) d t + 4 \int \frac{d t}{t^{2} - 4 t + 4} = \frac{3}{2} I_{1} + 4 I_{2} . . . (1) where I_{1} = \int \frac{(2 t - 4) d t}{t^{2} - 4 t + 4}, I_{2} = \int \frac{d t}{t^{2} - 4 t + 4} I_{1} = \int \frac{(2 t - 4) d t}{t^{2} - 4 t + 4} Let t^{2} - 4 t + 4 = p \Rightarrow (2 t - 4) d t = d p I_{1} = \int \frac{(2 t - 4) d t}{t^{2} - 4 t + 4} = \int \frac{d p}{p} = \log |p| + C_{1} = \log |t^{2} - 4 t + 4| + C_{1} . . . (2) I_{2} = \int \frac{d t}{t^{2} - 4 t + 4} I_{2} = \int \frac{d t}{{(t - 2)}^{2}} I_{2} = \int {(t - 2)}^{- 2} d t I_{2} = \frac{{(t - 2)}^{- 2 + 1}}{- 2 + 1} + C_{2} I_{2} = \frac{- 1}{t - 2} + C_{2} . . . (3) from (1), (2) and (3) \int \frac{(3 \sin x - 2) \cos x d x}{5 - \cos^{2} x - 4 \sin x} = \frac{3}{2} \log |t^{2} - 4 t + 4| + 4 \times \frac{- 1}{t - 2} + C_{1} + C_{2} = \frac{3}{2} \log |\sin^{2} x - 4 \sin x + 4| + \frac{4}{2 - t} + C (Where C = C_{1} + C_{2}) = \frac{3}{2} \log |{(\sin x - 2)}^{2}| + \frac{4}{2 - \sin x} + C = \frac{3}{2} \times 2 \log |\sin x - 2| + \frac{4}{2 - \sin x} + C = 3 \log |2 - \sin x| + \frac{4}{2 - \sin x} + C$

Page No 18.104:

Question 11:

$\int \frac{x + 2}{2 x^{2} + 6 x + 5} d x$

Answer:

$\int (\frac{x + 2}{2 x^{2} + 6 x + 5}) d x x + 2 = A \frac{d}{d x} (2 x^{2} + 6 x + 5) + B x + 2 = A (4 x + 6) + B x + 2 = (4 A) x + 6 A + B$

Comparing the Coefficients of like powers of x

$4 A = 1 A = \frac{1}{4} 6 A + B = 2 6 \times \frac{1}{4} + B = 2 B = \frac{1}{2}$

$∴ \int (\frac{x + 2}{2 x^{2} + 6 x + 5}) d x = \int [\frac{\frac{1}{4} (4 x + 6) + \frac{1}{2}}{2 x^{2} + 6 x + 5}] d x = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{2} \int \frac{1}{2 x^{2} + 6 x + 5} d x = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{x^{2} + 3 x + \frac{5}{2}} = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{x^{2} + 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{5}{2}} = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{{(x + \frac{3}{2})}^{2} - \frac{9}{4} + \frac{5}{2}} = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{{(x + \frac{3}{2})}^{2} + \frac{1}{4}} = \frac{1}{4} \int \frac{(4 x + 6)}{2 x^{2} + 6 x + 5} d x + \frac{1}{4} \int \frac{d x}{{(x + \frac{3}{2})}^{2} + {(\frac{1}{2})}^{2}} = \frac{1}{4} \log |2 x^{2} + 6 x + 5| + \frac{1}{4} \times 2 \tan^{- 1} (\frac{x + \frac{3}{2}}{\frac{1}{2}}) + C = \frac{1}{4} \log |2 x^{2} + 6 x + 5| + \frac{1}{2} \tan^{- 1} (2 x + 3) + C$

Page No 18.104:

Question 12:

Evaluate the following integrals:

$\int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x$

Answer:

$Let I = \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x = \int \frac{5 x - 2}{3 x^{2} + 2 x + 1} d x We express 5 x - 2 = A (\frac{d}{d x} (3 x^{2} + 2 x + 1)) + B 5 x - 2 = A (6 x + 2) + B Equating the coefficients of x and constants, we get 5 = 6 A and - 2 = 2 A + B or A = \frac{5}{6} and B = - \frac{11}{3} ∴ I = \int \frac{\frac{5}{6} (6 x + 2) - \frac{11}{3}}{3 x^{2} + 2 x + 1} d x = \frac{5}{6} \int \frac{(6 x + 2)}{3 x^{2} + 2 x + 1} d x - \frac{11}{3} \int \frac{1}{3 x^{2} + 2 x + 1} d x = \frac{5}{6} I_{1} - \frac{11}{3} I_{2} . . . (1) Now, I_{1} = \int \frac{(6 x + 2)}{3 x^{2} + 2 x + 1} d x Let 3 x^{2} + 2 x + 1 = t On differentiating both sides, we get (6 x + 2) d x = d t ∴ I_{1} = \int \frac{1}{t} d t = \log |t| + c_{1} = \log |3 x^{2} + 2 x + 1| + c_{1} . . . (2) And, I_{2} = \int \frac{1}{3 x^{2} + 2 x + 1} d x = \frac{1}{3} \int \frac{1}{x^{2} + \frac{2}{3} x + \frac{1}{3}} d x = \frac{1}{3} \int \frac{1}{x^{2} + \frac{2}{3} x + \frac{1}{9} - \frac{1}{9} + \frac{1}{3}} d x = \frac{1}{3} \int \frac{1}{{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} d x Let x + \frac{1}{3} = t On differentiating both sides, we get d x = d t ∴ I_{2} = \frac{1}{3} \int \frac{1}{{(t)}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} d t = \frac{1}{3} \times \frac{1}{\frac{\sqrt{2}}{3}} \tan^{- 1} \frac{3 t}{\sqrt{2}} + c_{2} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 (x + \frac{1}{3})}{\sqrt{2}} + c_{2} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}} + c_{2} . . . (3) From (1), (2) and (3), we get ∴ I = \frac{5}{6} (\log |3 x^{2} + 2 x + 1| + c_{1}) - \frac{11}{3} (\frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}} + c_{2}) = \frac{5}{6} (\log |3 x^{2} + 2 x + 1|) - \frac{11}{3} (\frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}}) + c Hence, \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x = \frac{5}{6} (\log |3 x^{2} + 2 x + 1|) - \frac{11}{3} (\frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}}) + c$

Page No 18.104:

Question 13:

$\int \frac{x + 5}{3 x^{2} + 13 x - 10} d x$

Answer:

$I = \int \frac{x + 5}{3 x^{2} + 13 x - 10} d x = \int \frac{x + 5}{3 x^{2} + 15 x - 2 x - 10} d x = \int \frac{x + 5}{3 x (x + 5) - 2 (x + 5)} d x = \int \frac{x + 5}{(3 x - 2) (x + 5)} d x$

$= \int \frac{x + 5}{(3 x - 2) (x + 5)} d x = \int \frac{1}{3 x - 2} d x ∴ I = \frac{1}{3} \ln |3 x - 2| + c$

Page No 18.104:

Question 14:

$\int \frac{(3 \sin x - 2) \cos x}{13 - \cos^{2} x - 7 \sin x} d x$

Answer:

$I = \int \frac{(3 \sin x - 2) \cos x}{13 - \cos^{2} x - 7 \sin x} d x$

$= \int \frac{(3 \sin x - 2) \cos x}{13 - (1 - \sin^{2} x) - 7 \sin x} d x (∵ \cos^{2} x = 1 - \sin^{2} x) = \int \frac{(3 \sin x - 2) \cos x}{\sin^{2} x - 7 \sin x + 12} d x = \int \frac{(3 \sin x - 2) \cos x}{\sin^{2} x - 4 \sin x - 3 \sin x + 12} d x = \int \frac{(3 \sin x - 2) \cos x}{\sin x (\sin x - 4) - 3 (\sin x - 4)} d x = \int \frac{(3 \sin x - 2) \cos x}{(\sin x - 3) (\sin x - 4)} d x$

$Let \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \frac{(3 t - 2)}{(t - 3) (t - 4)} d t$

Using partial fraction, we get

$\frac{(3 t - 2)}{(t - 3) (t - 4)} = \frac{A}{(t - 3)} + \frac{B}{(t - 4)} = \frac{A (t - 4) + B (t - 3)}{(t - 3) (t - 4)} \Rightarrow 3 t - 2 = (A + B) t - 4 A - 3 B$

Comparing coefficients, we get

A = $-$ 7 and B = 10

So, $I = - 7 \int \frac{1}{(t - 3)} d t + 10 \int \frac{1}{(t - 4)} d t$

$\Rightarrow I = - 7 \ln |t - 3| + 10 \ln |t - 4| + c ∴ I = - 7 \ln |\sin x - 3| + 10 \ln |\sin x - 4| + c$

Page No 18.104:

Question 15:

$\int \frac{x + 7}{3 x^{2} + 25 x + 28} d x$

Answer:

$I = \int \frac{x + 7}{3 x^{2} + 25 x + 28} d x = \int \frac{x + 7}{3 x^{2} + 21 x + 4 x + 28} d x = \int \frac{x + 7}{3 x (x + 7) + 4 (x + 7)} d x = \int \frac{x + 7}{(3 x + 4) (x + 7)} d x$

$= \int \frac{1}{(3 x + 4)} d x = \frac{1}{3} \ln |3 x + 4| + c$

Page No 18.104:

Question 16:

$\int \frac{3 x + 5}{x^{2} + 3 x - 18} d x$

Answer:

$I = \int \frac{3 x + 5}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = 3 \int \frac{x + \frac{5}{3}}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = \frac{3}{2} \int \frac{2 x + \frac{10}{3}}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = \frac{3}{2} \int \frac{2 x + 3 + \frac{1}{3}}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = \frac{3}{2} \int \frac{2 x + 3}{x^{2} + 3 x - 18} d x + \frac{1}{2} \int \frac{1}{x^{2} + 3 x - 18} d x$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{2} \int \frac{1}{x^{2} + 3 x + \frac{9}{4} - 18 - \frac{9}{4}} d x$ $(\begin{matrix} \int \frac{f' (x) d x}{f (x)} = \log |f (x)| + C \\ f (x) = x^{2} + 3 x - 18 \Rightarrow f' (x) = 2 x = 3 \end{matrix})$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{2} \int \frac{1}{{(x + \frac{3}{2})}^{2} - \frac{81}{4}} d x$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{2} \int \frac{1}{{(x + \frac{3}{2})}^{2} - {(\frac{9}{2})}^{2}} d x$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{2} \times \frac{1}{2 \times \frac{9}{2}} \log |\frac{(x + \frac{3}{2}) - \frac{9}{2}}{(x + \frac{3}{2}) + \frac{9}{2}}| + C$ $(\int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} \log |\frac{x - a}{x + a}| + C)$

$\Rightarrow I = \frac{3}{2} \log |x^{2} + 3 x - 18| + \frac{1}{18} \log |\frac{x - 3}{x + 6}| + C$

Page No 18.104:

Question 17:

Evaluate the following integrals:
$\int \frac{x^{3}}{x^{4} + x^{2} + 1} d x$

Answer:

$I = \int \frac{x^{3}}{x^{4} + x^{2} + 1} d x = \int \frac{x^{2} \cdot x}{{(x^{2})}^{2} + x^{2} + 1} d x Let x^{2} = t or 2 x d x = d t \Rightarrow I = \frac{1}{2} \int \frac{t}{t^{2} + t + 1} d t = \frac{1}{4} \int \frac{2 t}{t^{2} + t + 1} d t = \frac{1}{4} \int \frac{2 t + 1 - 1}{t^{2} + t + 1} d t$

$= \frac{1}{4} \int [\frac{(2 t + 1)}{(t^{2} + t + 1)} - \frac{1}{(t^{2} + t + 1)}] d t = \frac{1}{4} [\log |t^{2} + t + 1| - \int \frac{1}{(t^{2} + t + \frac{1}{4} + \frac{3}{4})} d t] = \frac{1}{4} [\log |t^{2} + t + 1| - \int \frac{1}{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d t] = \frac{1}{4} [\log |t^{2} + t + 1| - \frac{2}{\sqrt{3}} \tan \frac{(t + \frac{1}{2})}{(\frac{\sqrt{3}}{2})}] + c = \frac{1}{4} [\log |t^{2} + t + 1| - \frac{2}{\sqrt{3}} \tan (\frac{2 t + 1}{\sqrt{3}})] + c$

$= \frac{1}{4} [\log |x^{4} + x^{2} + 1| - \frac{2}{\sqrt{3}} \tan (\frac{2 x^{2} + 1}{\sqrt{3}})] + c$

Page No 18.104:

Question 18:

Evaluate the following integrals:
$\int \frac{x^{3} - 3 x}{x^{4} + 2 x^{2} - 4} d x$

Answer:

$I = \int \frac{x^{3} - 3 x}{x^{4} + 2 x^{2} - 4} d x$

$= \int \frac{x (x^{2} - 3)}{x^{4} + 2 x^{2} - 4} d x$

Let $x^{2} = t$ , or, $2 x d x = d t$

$\Rightarrow I = \frac{1}{2} \int \frac{(t - 3)}{t^{2} + 2 t - 4} d t = \frac{1}{4} \int \frac{2 t - 6}{t^{2} + 2 t - 4} d t = \frac{1}{4} \int \frac{2 t + 2 - 8}{t^{2} + 2 t - 4} d t = \frac{1}{4} \int (\frac{2 t + 2}{t^{2} + 2 t - 4} - \frac{8}{t^{2} + 2 t - 4}) d t = \frac{1}{4} (\int \frac{2 t + 2}{t^{2} + 2 t - 4} d t - \int \frac{8}{t^{2} + 2 t - 4} d t)$

$\Rightarrow I = \frac{1}{4} (I_{1} + I_{2}) . . . (i)$

Now,

$I_{1} = \int \frac{2 t + 2}{t^{2} + 2 t - 4} d t$

Let $t^{2} + 2 t - 4 = u$

$or, (2 t + 2) d t = d u \Rightarrow I_{1} = \int \frac{1}{u} d u = \ln |u| + c_{1} \Rightarrow I_{1} = \ln |t^{2} + 2 t - 4| + c_{1} ∴ I_{1} = \ln |x^{4} + 2 x^{2} - 4| + c_{1}$

Now,

$I_{2} = \int \frac{- 8}{(t + 1)^{2} - 5} d t \Rightarrow I_{2} = \int \frac{8}{(\sqrt{5})^{2} - (t + 1)^{2}} d t ∴ I_{2} = \frac{8}{2 \sqrt{5}} \ln |\frac{\sqrt{5} + x^{2} + 1}{\sqrt{5} - x^{2} - 1}| + c_{2}$

$So, from (i), we get I = \frac{1}{4} [\ln |x^{4} + 2 x^{2} - 4| + \frac{4}{\sqrt{5}} \ln |\frac{\sqrt{5} + x^{2} + 1}{\sqrt{5} - x^{2} - 1}|] + C ∴ I = \frac{1}{4} \ln |x^{4} + 2 x^{2} - 4| + \frac{1}{\sqrt{5}} \ln |\frac{\sqrt{5} + x^{2} + 1}{\sqrt{5} - x^{2} - 1}| + C$

Page No 18.106:

Question 1:

$\int \frac{x^{2} + x + 1}{x^{2} - x} d x$

Answer:

$\int (\frac{x^{2} + x + 1}{x^{2} - x}) d x \frac{x^{2} + x + 1}{x^{2} - x} = 1 + \frac{2 x + 1}{x^{2} - x} ∴ \int (\frac{x^{2} + x + 1}{x^{2} - x}) d x = \int (1 + \frac{2 x + 1}{x^{2} - x}) d x = \int 1 + (\frac{2 x - 1 + 2}{x^{2} - x}) d x = \int d x + \int \frac{(2 x - 1) d x}{x^{2} - x} + \int \frac{2 d x}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} = \int d x + \int \frac{(2 x - 1) d x}{x^{2} - x} + 2 \int \frac{d x}{{(x - \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} = x + \log |x^{2} - x| + 2 \times \frac{1}{2 \times \frac{1}{2}} \log |\frac{x - \frac{1}{2} - \frac{1}{2}}{x - \frac{1}{2} + \frac{1}{2}}| = x + \log |x^{2} - x| + 2 \log |\frac{x - 1}{x}| + C$

Page No 18.106:

Question 2:

$\int \frac{x^{2} + x - 1}{x^{2} + x - 6} d x$

Answer:

$\int (\frac{x^{2} + x - 1}{x^{2} + x - 6}) d x \frac{x^{2} + x - 1}{x^{2} + x - 6} = 1 + \frac{5}{x^{2} + x - 6} \int (\frac{x^{2} + x - 1}{x^{2} + x - 6}) d x = \int d x + 5 \int \frac{d x}{x^{2} + x - 6} = \int d x + 5 \int \frac{d x}{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 6} = \int d x + 5 \int \frac{d x}{{(x + \frac{1}{2})}^{2} - \frac{1}{4} - 6} = \int d x + 5 \int \frac{d x}{{(x + \frac{1}{2})}^{2} - {(\frac{5}{2})}^{2}} = x + 5 \times \frac{1}{2 \times \frac{5}{2}} \log |\frac{x + \frac{1}{2} - \frac{5}{2}}{x + \frac{1}{2} + \frac{5}{2}}| + C = x + \log |\frac{x - 2}{x + 3}| + C$

Page No 18.106:

Question 3:

$\int \frac{(1 - x^{2})}{x (1 - 2 x)} d x$

Answer:

$We have, I = \int \frac{(1 - x^{2})}{x (1 - 2 x)} d x = \int (\frac{- x^{2} + 1}{- 2 x^{2} + x}) d x = \int \frac{1}{2} d x + \int (\frac{1 - \frac{x}{2}}{- 2 x^{2} + x}) d x = \frac{1}{2} \int d x + \frac{1}{2} \int \frac{2 - x}{- 2 x^{2} + x} d x = \frac{1}{2} [\int d x + \int \frac{2 - x}{- 2 x^{2} + x} d x] = \frac{1}{2} [I_{1} + I_{2}] (say) where I_{1} = \int d x & I_{2} = \int \frac{2 - x}{- 2 x^{2} + x} d x Now, I_{1} = \int d x = x + C_{1} I_{2} = \int \frac{2 - x}{- 2 x^{2} + x} d x Let 2 - x = A \frac{d}{d x} (- 2 x^{2} + x) + B \Rightarrow 2 - x = A (- 4 x + 1) + B \Rightarrow 2 - x = - 4 A x + A + B$

Comparing coefficients of like terms

$- 1 = - 4 A \Rightarrow A = \frac{1}{4} & A + B = 2 \Rightarrow \frac{1}{4} + B = 2 \Rightarrow B = 2 - \frac{1}{4} = \frac{8 - 1}{4} = \frac{7}{4}$

$∴ \int \frac{2 - x}{- 2 x^{2} + x} d x = \int \frac{\frac{1}{4} (- 4 x + 1) + \frac{7}{4}}{- 2 x^{2} + x} d x = \int \frac{\frac{1}{4} (- 4 x + 1)}{- 2 x^{2} + x} d x + \int \frac{\frac{7}{4}}{- 2 x^{2} + x} d x = \frac{1}{4} \log |- 2 x^{2} + x| + \frac{7}{4} \int \frac{1}{- 2 (x^{2} - \frac{x}{2} + \frac{1}{16} - \frac{1}{16})} d x = \frac{1}{4} \log |- 2 x^{2} + x| - \frac{7}{8} \int \frac{1}{\{{(x - \frac{1}{4})}^{2} - {(\frac{1}{4})}^{2}\}} d x = \frac{1}{4} \log |x (- 2 x + 1)| - \frac{7}{8} \times \frac{1}{2 \times \frac{1}{4}} \log |\frac{x - \frac{1}{4} - \frac{1}{4}}{x - \frac{1}{4} + \frac{1}{4}}| + C_{2} = \frac{1}{4} \log |x| + \frac{1}{4} \log |- 2 x + 1| - \frac{7}{4} \log |\frac{x - \frac{1}{2}}{x}| + C_{2} = \frac{1}{4} \log |x| + \frac{1}{4} \log |- 2 x + 1| - \frac{7}{4} \log |x - \frac{1}{2}| + \frac{7}{4} \log |x| + C_{2} = 2 \log |x| + \frac{1}{4} \log |- 2 x + 1| - \frac{7}{4} \log |2 x - 1| + C_{3}, where C_{3} = C_{2} + \frac{7}{4} \log 2 = 2 \log |x| + \frac{1}{4} \log |- 2 x + 1| - \frac{7}{4} \log |1 - 2 x| + C_{3} = 2 \log |x| - \frac{3}{2} \log |1 - 2 x| + C_{3} Thus, I = \frac{1}{2} [x + C_{1} + 2 \log |x| - \frac{3}{2} \log |1 - 2 x| + C_{3}] = \frac{1}{2} x + \log |x| - \frac{3}{4} \log |1 - 2 x| + C, where C = \frac{1}{2} [C_{1} + C_{3}]$

Page No 18.106:

Question 4:

$\int \frac{x^{2} + 1}{x^{2} - 5 x + 6} d x$

Answer:

$Let I \int (\frac{x^{2} + 1}{x^{2} - 5 x + 6}) d x Dividing Numerator by Denominator x^{2} - 5 x + 6 \overset{1}{x^{2} + 1} x^{2} - 5 x + 6 - + - 5 x - 5 \frac{x^{2} + 1}{x^{2} - 5 x + 6} = 1 + (\frac{5 x - 5}{x^{2} - 5 x + 6}) . . . . . (1) Also \frac{5 x - 5}{x^{2} - 5 x + 6} = \frac{5 x - 5}{(x - 2) (x - 3)} Let \frac{5 x - 5}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3} \Rightarrow \frac{5 x - 5}{(x - 2) (x - 3)} = \frac{A (x - 3) + B (x - 2)}{(x - 2) (x - 3)} \Rightarrow 5 x - 5 = A (x - 3) + B (x - 2) let x = 3 5 \times 3 - 5 = A \times 0 + B (3 - 2) 10 = B let x = 2 5 \times 2 - 5 = A (2 - 3) + B \times 0 A = - 5 \frac{5 x - 5}{(x - 2) (x - 3)} = \frac{- 5}{x - 2} + \frac{10}{x - 3} . . . . . (2) from (1) and (2) I = \int d x - 5 \int \frac{d x}{x - 2} + 10 \int \frac{d x}{x - 3} = x - 5 \log |x - 2| + 10 \log |x - 3| + C$

Page No 18.106:

Question 5:

$\int \frac{x^{2}}{x^{2} + 7 x + 10} d x$

Answer:

$Let I = \int (\frac{x^{2}}{x^{2} + 7 x + 10}) d x Now, x^{2} + 7 x + 10 \overset{1}{x^{2}} x^{2} + 7 x + 10 - - - - 7 x - 10 ∴ \frac{x^{2}}{x^{2} + 7 x + 10} = 1 - \frac{(7 x + 10)}{x^{2} + 7 x + 10} \Rightarrow \frac{x^{2}}{x^{2} + 7 x + 10} = 1 - (\frac{7 x + 10}{x^{2} + 2 x + 5 x + 10}) \frac{x^{2}}{x^{2} + 7 x + 10} = 1 - (\frac{7 x + 10}{x (x + 2) + 5 (x + 2)}) \frac{x^{2}}{x^{2} + 7 x + 10} = 1 - [\frac{7 x + 10}{(x + 2) (x + 5)}] . . . . . (1) Consider, \frac{7 x + 10}{(x + 2) (x + 5)} = \frac{A}{(x + 2)} + \frac{B}{x + 5} 7 x + 10 = A (x + 5) + B (x + 2) let x + 5 = 0 x = - 5 \Rightarrow 7 (- 5) + 10 = A \times 0 + B (- 5 + 2) - 25 = B (- 3) \Rightarrow B = \frac{25}{3} let x + 2 = 0 x = - 2 7 (- 2) + 10 = A (- 2 + 5) \Rightarrow - 4 = A (3) \Rightarrow A = - \frac{4}{3} \frac{7 x + 10}{(x + 2) (x + 5)} = \frac{- 4}{3 (x + 2)} + \frac{25}{3 (x + 5)} . . . . . (2) from (1) and (2) \frac{x^{2}}{x^{2} + 7 x + 10} = 1 + \frac{4}{3 (x + 2)} - \frac{25}{3 (x + 5)} \Rightarrow \int \frac{x^{2} d x}{x^{2} + 7 x + 10} = \int d x + \frac{4}{3} \int \frac{d x}{x + 2} - \frac{25}{3} \int \frac{d x}{x + 5} = x + \frac{4}{3} \log |x + 2| - \frac{25}{3} \log |x + 5| + C$

Page No 18.106:

Question 6:

$\int \frac{x^{2} + x + 1}{x^{2} - x + 1} d x$

Answer:

$Let I = \int (\frac{x^{2} + x + 1}{x^{2} - x + 1}) d x Now, x^{2} - x + 1 \overset{1}{x^{2} + x + 1} x^{2} - x + 1 - + - 2 x Therefore, \frac{x^{2} + x + 1}{x^{2} - x + 1} = 1 + \frac{2 x}{x^{2} - x + 1} \Rightarrow \int (\frac{x^{2} + x + 1}{x^{2} - x + 1}) d x = \int d x + \int (\frac{2 x - 1 + 1}{x^{2} - x + 1}) d x = \int d x + \int (\frac{2 x - 1}{x^{2} - x + 1}) d x + \int \frac{d x}{x^{2} - x + 1} = \int d x + \int \frac{(2 x - 1) d x}{x^{2} - x + 1} + \int \frac{d x}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} = \int d x + \int \frac{(2 x - 1) d x}{x^{2} - x + 1} + \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = x + \log |x^{2} - x + 1| + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + C$

Page No 18.106:

Question 7:

$\int \frac{{(x - 1)}^{2}}{x^{2} + 2 x + 2} d x$

Answer:

$Let I = \int \frac{{(x - 1)}^{2}}{x^{2} + 2 x + 2} d x = \int (\frac{x^{2} - 2 x + 1}{x^{2} + 2 x + 2}) d x Here, x^{2} + 2 x + 2 \overset{1}{x^{2} - 2 x + 1} x^{2} + 2 x + 2 - - - - 4 x - 1 Therefore, \frac{x^{2} - 2 x + 1}{x^{2} + 2 x + 2} = 1 - \frac{(4 x + 1)}{x^{2} + 2 x + 2} . . . . . (1) Let 4 x + 1 = A \frac{d}{d x} (x^{2} + 2 x + 2) + B 4 x + 1 = A (2 x + 2) + B 4 x + 1 = (2 A) x + 2 A + B Equating Coefficients of like terms 2 A = 4 A = 2 2 A + B = 1 2 \times 2 + B = 1 B = - 3 \int (\frac{x^{2} - 2 x + 1}{x^{2} + 2 x + 2}) d x = \int d x - 2 \int \frac{(2 x + 2)}{x^{2} + 2 x + 2} d x + 3 \int \frac{d x}{x^{2} + 2 x + 2} = \int d x - 2 \int \frac{(2 x + 2)}{x^{2} + 2 x + 2} d x + 3 \int \frac{d x}{{(x + 1)}^{2} + 1^{2}} = x - 2 \log |x^{2} + 2 x + 2| + \frac{3}{1} \tan^{- 1} (\frac{x + 1}{1}) + C = x - 2 \log |x^{2} + 2 x + 2| + 3 \tan^{- 1} (x + 1) + C$

Page No 18.106:

Question 8:

$\int \frac{x^{3} + x^{2} + 2 x + 1}{x^{2} - x + 1} d x$

Answer:

$Let I = \int (\frac{x^{3} + x^{2} + 2 x + 1}{x^{2} - x + 1}) d x x^{2} - x + 1 \overset{x + 2}{x^{3} + x^{2} + 2 x + 1} x^{3} - x^{2} + x - + - 2 x^{2} + x + 1 2 x^{2} - 2 x + 2 - + - 3 x - 1 Therefore, \frac{x^{3} + x^{2} + 2 x + 1}{x^{2} - x + 1} = x + 2 + \frac{3 x - 1}{x^{2} - x + 1} . . . . . (1) Let 3 x - 1 = A \frac{d}{d x} (x^{2} - x + 1) + B 3 x - 1 = A (2 x - 1) + B 3 x - 1 = (2 A) x + B - A Equating Coefficients of like terms 2 A = 3 A = \frac{3}{2} B - A = - 1 B - \frac{3}{2} = - 1 B = \frac{1}{2} \int (\frac{x^{3} + x^{2} + 2 x + 1}{x^{2} - x + 1}) d x = \int (x + 2) d x + \int (\frac{\frac{3}{2} (2 x - 1) + \frac{1}{2}}{x^{2} - x + 1}) d x = \int (x + 2) d x + \frac{3}{2} \int (\frac{2 x - 1}{x^{2} - x + 1}) d x + \frac{1}{2} \int \frac{d x}{x^{2} - x + 1} = \int (x + 2) d x + \frac{3}{2} \int \frac{(2 x - 1) d x}{x^{2} - x + 1} + \frac{1}{2} \int \frac{d x}{x^{2} - x + \frac{1}{4} - \frac{1}{4} + 1} = \int (x + 2) d x + \frac{3}{2} \int \frac{(2 x - 1) d x}{x^{2} - x + 1} + \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{x^{2}}{2} + 2 x + \frac{3}{2} \log |x^{2} - x + 1| + \frac{1}{2} \times \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C = \frac{x^{2}}{2} + 2 x + \frac{3}{2} \log |x^{2} - x + 1| + \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) + C$

Page No 18.106:

Question 9:

$\int \frac{x^{2} (x^{4} + 4)}{x^{2} + 4} d x$

Answer:

$Let I = \int \frac{x^{2} (x^{4} + 4)}{(x^{2} + 4)} d x = \int (\frac{x^{6} + 4 x^{2}}{x^{2} + 4}) d x Now, x^{2} + 4 \overset{x^{4} - 4 x^{2} + 20}{x^{6} + 4 x^{2}} x^{6} + 4 x^{4} - - - 4 x^{4} + 4 x^{2} - 4 x^{4} - 16 x^{2} + + 20 x^{2} 20 x^{2} + 80 - - - 80 Therefore, \frac{x^{2} (x^{4} + 4)}{(x^{2} + 4)} = (x^{4} - 4 x^{2} + 20) - \frac{80}{x^{2} + 4} I = \int \frac{x^{2} (x^{4} + 4)}{(x^{2} + 4)} d x = \int (x^{4} - 4 x^{2} + 20) d x - 80 \int \frac{d x}{x^{2} + 2^{2}} = \int x^{4} d x - 4 \int x^{2} d x + 20 \int d x - 80 \int \frac{d x}{x^{2} + 2^{2}} = \frac{x^{4 + 1}}{4 + 1} - 4 [\frac{x^{3}}{3}] + 20 (x) - 80 \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + C = \frac{x^{5}}{5} - \frac{4}{3} x^{3} + 20 x - 40 \tan^{- 1} (\frac{x}{2}) + C$

Page No 18.106:

Question 10:

$\int \frac{x^{2}}{x^{2} + 6 x + 12} d x$

Answer:

$Let I = \int \frac{x^{2} d x}{x^{2} + 6 x + 12} Now, x^{2} + 6 x + 12 \overset{1}{x^{2}} x^{2} + 6 x + 12 - - - - 6 x - 12 Therefore, \frac{x^{2}}{x^{2} + 6 x + 12} = 1 - \frac{(6 x + 12)}{x^{2} + 6 x + 12} . . . . . (1) Let 6 x + 12 = A \frac{d}{d x} (x^{2} + 6 x + 12) + B \Rightarrow 6 x + 12 = A (2 x + 6) + B \Rightarrow 6 x + 12 = (2 A) x + 6 A + B Equating Coefficients of like terms 2 A = 6 A = 3 6 A + B = 12 18 + B = 12 B = - 6 ∴ \frac{x^{2}}{x^{2} + 6 x + 12} = 1 - \frac{3 (2 x + 6) - 6}{x^{2} + 6 x + 12} I = \int \frac{x^{2} d x}{x^{2} + 6 x + 12} = \int d x - 3 \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 12} + 6 \int \frac{d x}{x^{2} + 6 x + 12} = \int d x - 3 \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 12} + 6 \int \frac{d x}{x^{2} + 6 x + 9 + 3} = \int d x - 3 \int \frac{(2 x + 6) d x}{x^{2} + 6 x + 12} + 6 \int \frac{d x}{{(x + 3)}^{2} + {(\sqrt{3})}^{2}} = x - 3 \log |x^{2} + 6 x + 12| + \frac{6}{\sqrt{3}} \tan^{- 1} (\frac{x + 3}{\sqrt{3}}) + C = x - 3 \log |x^{2} + 6 x + 12| + 2 \sqrt{3} \tan^{- 1} (\frac{x + 3}{\sqrt{3}}) + C$

Page No 18.110:

Question 1:

$\int \frac{x}{\sqrt{x^{2} + 6 x + 10}} d x$

Answer:

$Let I = \int \frac{x d x}{\sqrt{x^{2} + 6 x + 10}} x = A \frac{d}{d x} (x^{2} + 6 x + 10) + B x = A (2 x + 6) + B x = (2 A) x + 6 A + B Equating Coefficients of like terms 2 A = 1 A = \frac{1}{2} 6 A + B = 0 6 \times \frac{1}{2} + B = 0 B = - 3 I = \int \frac{x d x}{\sqrt{x^{2} + 6 x + 10}} = \int (\frac{\frac{1}{2} (2 x + 6) - 3}{\sqrt{x^{2} + 6 x + 10}}) d x = \frac{1}{2} \int \frac{(2 x + 6) d x}{\sqrt{x^{2} + 6 x + 10}} - 3 \int \frac{d x}{\sqrt{x^{2} + 6 x + 3^{2} - 3^{2} + 10}} = \frac{1}{2} \int \frac{(2 x + 6) d x}{\sqrt{x^{2} + 6 x + 10}} - 3 \int \frac{d x}{\sqrt{{(x + 3)}^{2} + 1^{2}}} let x^{2} + 6 x + 10 = t \Rightarrow (2 x + 6) d x = d t I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} - 3 \int \frac{d x}{\sqrt{{(x + 3)}^{2} + 1}} = \frac{1}{2} \times 2 \sqrt{t} - 3 \log |x + 3 + \sqrt{{(x + 3)}^{2} + 1}| + C = \sqrt{t} - 3 \log |x + 3 + \sqrt{x^{2} + 6 x + 10}| + C = \sqrt{x^{2} + 6 x + 10} - 3 \log |x + 3 + \sqrt{x^{2} + 6 x + 10}| + C$

Page No 18.110:

Question 2:

$\int \frac{2 x + 1}{\sqrt{x^{2} + 2 x - 1}} d x$

Answer:

$Let I = \int \frac{(2 x + 1) d x}{\sqrt{x^{2} + 2 x - 1}} = \int \frac{(2 x + 2 - 1) d x}{\sqrt{x^{2} + 2 x - 1}} = \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} - \int \frac{d x}{\sqrt{x^{2} + 2 x - 1}} = \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} - \int \frac{d x}{\sqrt{x^{2} + 2 x + 1 - 1 - 1}} = \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} - \int \frac{d x}{\sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}} let x^{2} + 2 x - 1 = t \Rightarrow (2 x + 2) d x = d t I = \int \frac{d t}{\sqrt{t}} - \int \frac{d x}{\sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}} = 2 \sqrt{t} - \log |x + 1 + \sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}| + C = 2 \sqrt{x^{2} + 2 x - 1} - \log |x + 1 + \sqrt{x^{2} + 2 x - 1}| + C$

Page No 18.110:

Question 3:

$\int \frac{x + 1}{\sqrt{4 + 5 x - x^{2}}} d x$

Answer:

$Let I = \int \frac{(x + 1) d x}{\sqrt{4 + 5 x - x^{2}}} Also, x + 1 = A \frac{d}{d x} (4 + 5 x - x^{2}) + B x + 1 = A (5 - 2 x) + B x + 1 = (- 2 A) x + 5 A + B Equating Coefficients of like terms - 2 A = 1 \Rightarrow A = - \frac{1}{2} And 5 A + B = 1 \Rightarrow - \frac{5}{2} + B = 1 B = \frac{7}{2} I = \int \frac{(x + 1) d x}{\sqrt{4 + 5 x - x^{2}}} = \int (\frac{- \frac{1}{2} (5 - 2 x) + \frac{7}{2}}{\sqrt{4 + 5 x - x^{2}}}) d x = - \frac{1}{2} \int \frac{(5 - 2 x) d x}{\sqrt{4 + 5 x - x^{2}}} + \frac{7}{2} \int \frac{d x}{\sqrt{4 - (x^{2} - 5 x)}} = - \frac{1}{2} \int \frac{(5 - 2 x) d x}{\sqrt{4 + 5 x - x^{2}}} + \frac{7}{2} \int \frac{d x}{\sqrt{4 - [x^{2} - 5 x + {(\frac{5}{2})}^{2} - {(\frac{5}{2})}^{2}]}} = - \frac{1}{2} \int \frac{(5 - 2 x) d x}{\sqrt{4 + 5 x - x^{2}}} + \frac{7}{2} \int \frac{d x}{\sqrt{4 - {(x - \frac{5}{2})}^{2} + \frac{25}{4}}} = - \frac{1}{2} \int (\frac{5 - 2 x}{\sqrt{4 + 5 x - x^{2}}}) d x + \frac{7}{2} \int \frac{d x}{\sqrt{\frac{41}{4} - {(x - \frac{5}{2})}^{2}}} = - \frac{1}{2} \int (\frac{5 - 2 x}{\sqrt{4 + 5 x - x^{2}}}) d x + \frac{7}{2} \int \frac{d x}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - {(x - \frac{5}{2})}^{2}}} let 4 + 5 x - x^{2} = t \Rightarrow (5 - 2 x) d x = d t Then, I = - \frac{1}{2} \int \frac{d t}{\sqrt{t}} + \frac{7}{2} \int \frac{d x}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - {(x - \frac{5}{2})}^{2}}} = - \frac{1}{2} \times 2 \sqrt{t} + \frac{7}{2} \times \sin^{- 1} (\frac{x - \frac{5}{2}}{\frac{\sqrt{41}}{2}}) + C = - \sqrt{t} + \frac{7}{2} \sin^{- 1} (\frac{2 x - 5}{\sqrt{41}}) + C = - \sqrt{4 + 5 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{2 x - 5}{\sqrt{41}}) + C$

Page No 18.110:

Question 4:

$\int \frac{6 x - 5}{\sqrt{3 x^{2} - 5 x + 1}} d x$

Answer:

$Let I = \int \frac{6 x - 5}{\sqrt{3 x^{2} - 5 x + 1}} d x Putting 3 x^{2} - 5 x + 1 = t \Rightarrow (6 x - 5) d x = d t Then, I = \int \frac{d t}{\sqrt{t}} = 2 \sqrt{t} + C = 2 \sqrt{3 x^{2} - 5 x + 1} + C$

Page No 18.110:

Question 5:

$\int \frac{3 x + 1}{\sqrt{5 - 2 x - x^{2}}} d x$

Answer:

$Let I = \int \frac{(3 x + 1) d x}{\sqrt{5 - 2 x - x^{2}}} Consider, 3 x + 1 = A \frac{d}{d x} (5 - 2 x - x^{2}) + B \Rightarrow 3 x + 1 = A (- 2 - 2 x) + B \Rightarrow 3 x + 1 = (- 2 A) x - 2 A + B Equating Coefficients of like terms - 2 A = 3 \Rightarrow A = - \frac{3}{2} And - 2 A + B = 1 \Rightarrow - 2 \times - \frac{3}{2} + B = 1 \Rightarrow B = 1 - 3 \Rightarrow B = - 2 ∴ I = \int [\frac{- \frac{3}{2} (- 2 - 2 x) - 2}{\sqrt{5 - 2 x - x^{2}}}] d x = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - 2 x - x^{2}}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - (x^{2} + 2 x)}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - (x^{2} + 2 x + 1 - 1)}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{6 - {(x + 1)}^{2}}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{{(\sqrt{6})}^{2} - {(x + 1)}^{2}}} let 5 - 2 x - x^{2} = t \Rightarrow (- 2 - 2 x) d x = d t ∴ I = - \frac{3}{2} \int \frac{d t}{\sqrt{t}} - 2 \int \frac{d x}{\sqrt{{(\sqrt{6})}^{2} - {(x + 1)}^{2}}} = - \frac{3}{2} \times 2 \sqrt{t} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C = - 3 \sqrt{5 - 2 x - x^{2}} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C$

Page No 18.110:

Question 6:

$\int \frac{x}{\sqrt{8 + x - x^{2}}} d x$

Answer:

$Let I = \int \frac{x d x}{\sqrt{8 + x - x^{2}}} Consider, x = A \frac{d}{d x} (8 + x - x^{2}) + B x = A (1 - 2 x) + B x = (- 2 A) x + A + B Equating Coefficients of like terms - 2 A = 1 \Rightarrow A = - \frac{1}{2} And A + B = 0 \Rightarrow - \frac{1}{2} + B = 0 \Rightarrow B = \frac{1}{2} ∴ x = - \frac{1}{2} (1 - 2 x) + \frac{1}{2} Then, I = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{8 + x - x^{2}}} = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{8 - (x^{2} - x)}} = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{8 - (x^{2} - x + \frac{1}{4} - \frac{1}{4})}} = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{8 + \frac{1}{4} - {(x - \frac{1}{2})}^{2}}} = - \frac{1}{2} \int \frac{(1 - 2 x) d x}{\sqrt{8 + x - x^{2}}} + \frac{1}{2} \int \frac{d x}{\sqrt{{(\frac{\sqrt{33}}{2})}^{2} - {(x - \frac{1}{2})}^{2}}} let 8 + x - x^{2} = t \Rightarrow (1 - 2 x) d x = d t ∴ I = - \frac{1}{2} \int \frac{d t}{\sqrt{t}} + \frac{1}{2} \int \frac{d x}{\sqrt{{(\frac{\sqrt{33}}{2})}^{2} - {(x - \frac{1}{2})}^{2}}} = - \frac{1}{2} \times 2 \sqrt{t} + \frac{1}{2} \sin^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{33}}{2}}) + C = - \sqrt{t} + \frac{1}{2} \sin^{- 1} (\frac{2 x - 1}{\sqrt{33}}) + C = - \sqrt{8 + x - x^{2}} + \frac{1}{2} \sin^{- 1} (\frac{2 x - 1}{\sqrt{33}}) + C$

Page No 18.110:

Question 7:

$\int \frac{x + 2}{\sqrt{x^{2} + 2 x - 1}} d x$

Answer:

$Let I = \int \frac{(x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} Consider, x + 2 = A \frac{d}{d x} (x^{2} + 2 x - 1) + B \Rightarrow x + 2 = A (2 x + 2) + B \Rightarrow x + 2 = (2 A) x + 2 A + B Equating Coefficients of like terms 2 A = 1 \Rightarrow A = \frac{1}{2} And 2 A + B = 2 \Rightarrow 2 \times \frac{1}{2} + B = 2 \Rightarrow B = 1 T h e n, I = \int \frac{[\frac{1}{2} (2 x + 2) + 1]}{\sqrt{x^{2} + 2 x - 1}} d x = \frac{1}{2} \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x - 1}} + \int \frac{d x}{\sqrt{x^{2} + 2 x - 1}} let x^{2} + 2 x - 1 = t \Rightarrow (2 x + 2) d x = d t ∴ I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} + \int \frac{d x}{\sqrt{x^{2} + 2 x - 1}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t + \int \frac{d x}{\sqrt{x^{2} + 2 x + 1 - 2}} = \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + \int \frac{d x}{\sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}} = \sqrt{t} + \log |x + 1 + \sqrt{{(x + 1)}^{2} - {(\sqrt{2})}^{2}}| + C = \sqrt{x^{2} + 2 x - 1} + \log |x + 1 + \sqrt{x^{2} + 2 x - 1}| + C$

Page No 18.110:

Question 8:

$\int \frac{x + 2}{\sqrt{x^{2} - 1}} d x$

Answer:

$Let I = \int \frac{x + 2}{\sqrt{x^{2} - 1}} d x = \int \frac{x}{\sqrt{x^{2} - 1}} d x + 2 \int \frac{d x}{\sqrt{x^{2} - 1}} let x^{2} - 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Then, I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} + 2 \int \frac{d x}{\sqrt{x^{2} - 1^{2}}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t + 2 \int \frac{d x}{\sqrt{x^{2} - 1^{2}}} = \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + 2 \log |x + \sqrt{x^{2} - 1}| + C = \sqrt{t} + 2 \log |x + \sqrt{x^{2} - 1}| + C = \sqrt{x^{2} - 1} + 2 \log |x + \sqrt{x^{2} - 1}| + C$

Page No 18.110:

Question 9:

$\int \frac{x - 1}{\sqrt{x^{2} + 1}} d x$

Answer:

$Let I = \int (\frac{x - 1}{\sqrt{x^{2} + 1}}) d x = \int \frac{x d x}{\sqrt{x^{2} + 1}} - \int \frac{d x}{\sqrt{x^{2} + 1}} Putting x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Then, I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} - \int \frac{d x}{\sqrt{x^{2} + 1^{2}}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t - \int \frac{d x}{\sqrt{x^{2} + 1^{2}}} = \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] - \int \frac{d x}{\sqrt{x^{2} + 1^{2}}} = \sqrt{t} - \log |x + \sqrt{x^{2} + 1}| + C = \sqrt{x^{2} + 1} - \log |x + \sqrt{x^{2} + 1}| + C$

Page No 18.110:

Question 10:

$\int \frac{x}{\sqrt{x^{2} + x + 1}} d x$

Answer:

$Let I = \int \frac{x d x}{\sqrt{x^{2} + x + 1}} Consider, x = A \frac{d}{d x} (x^{2} + x + 1) + B \Rightarrow x = A (2 x + 1) + B \Rightarrow x = (2 A) x + A + B Equating Coefficient of like terms 2 A = 1 \Rightarrow A = \frac{1}{2} And A + B = 0 \Rightarrow \frac{1}{2} + B = 0 \Rightarrow B = - \frac{1}{2} ∴ I = \int \frac{(\frac{1}{2} (2 x + 1) - \frac{1}{2})}{\sqrt{x^{2} + x + 1}} d x = \frac{1}{2} \int (\frac{2 x + 1}{\sqrt{x^{2} + x + 1}}) d x - \frac{1}{2} \int \frac{d x}{\sqrt{x^{2} + x + \frac{1}{4} - \frac{1}{4} + 1}} Putting x^{2} + x + 1 = t \Rightarrow (2 x + 1) d x = d t Then, I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} - \frac{1}{2} \int \frac{d x}{\sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t - \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}| + C = \frac{1}{2} |\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}| - \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C = \sqrt{t} - \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C = \sqrt{x^{2} + x + 1} - \frac{1}{2} \log |x + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C$

Page No 18.110:

Question 11:

$\int \frac{x + 1}{\sqrt{x^{2} + 1}} d x$

Answer:

$Let I = \int (\frac{x + 1}{\sqrt{x^{2} + 1}}) d x = \int \frac{x d x}{\sqrt{x^{2} + 1}} + \int \frac{d x}{\sqrt{x^{2} + 1}} Putting, x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Then, I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} + \int \frac{d x}{\sqrt{x^{2} + 1}} = \frac{1}{2} \int t^{- \frac{1}{2}} d t + \int \frac{d x}{\sqrt{x^{2} + 1}} = \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + \log |x + \sqrt{x^{2} + 1}| + C = \sqrt{t} + \log |x + \sqrt{x^{2} + 1}| + C = \sqrt{x^{2} + 1} + \log |x + \sqrt{x^{2} + 1}| + C$

Page No 18.110:

Question 12:

$\int \frac{2 x + 5}{\sqrt{x^{2} + 2 x + 5}} d x$

Answer:

$Let I = \int \frac{(2 x + 5) d x}{\sqrt{x^{2} + 2 x + 5}} Consider, 2 x + 5 = A \frac{d}{d x} (x^{2} + 2 x + 5) + B \Rightarrow 2 x + 5 = A (2 x + 2) + B \Rightarrow 2 x + 5 = (2 A) x + 2 A + B Equating Coefficients of like terms 2 A = 2 \Rightarrow A = 1 And 2 A + B = 5 \Rightarrow B = 3 ∴ I = \int (\frac{2 x + 2 + 3}{\sqrt{x^{2} + 2 x + 5}}) d x = \int \frac{(2 x + 2) d x}{\sqrt{x^{2} + 2 x + 5}} + 3 \int \frac{d x}{\sqrt{x^{2} + 2 x + 5}} let x^{2} + 2 x + 5 = t \Rightarrow (2 x + 2) d x = d t Then, I = \int \frac{d t}{\sqrt{t}} + 3 \int \frac{d x}{\sqrt{x^{2} + 2 x + 1 + 4}} = \int t^{- \frac{1}{2}} d t + 3 \int \frac{d x}{\sqrt{{(x + 1)}^{2} + 2^{2}}} = \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + 3 \log |x + 1 + \sqrt{{(x + 1)}^{2} + 4}| + C = 2 \sqrt{t} + 3 \log |x + 1 + \sqrt{x^{2} + 2 x + 5}| + C = 2 \sqrt{x^{2} + 2 x + 5} + 3 \log |x + 1 + \sqrt{x^{2} + 2 x + 5}| + C$

Page No 18.110:

Question 13:

$\int \frac{3 x + 1}{\sqrt{5 - 2 x - x^{2}}} d x$

Answer:

$Let I = \int \frac{(3 x + 1) d x}{\sqrt{5 - 2 x - x^{2}}} Consider, 3 x + 1 = A \frac{d}{d x} (5 - 2 x - x^{2}) + B \Rightarrow 3 x + 1 = A (- 2 - 2 x) + B \Rightarrow 3 x + 1 = (- 2 A) x + (- 2 A + B) Equating Coefficients of like terms - 2 A = 3 \Rightarrow A = - \frac{3}{2} And - 2 A + B = 1 \Rightarrow - 2 \times - \frac{3}{2} + B = 1 \Rightarrow B = - 2 ∴ I = \int [\frac{- \frac{3}{2} (- 2 - 2 x) - 2}{\sqrt{5 - 2 x - x^{2}}}] d x = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - 2 x - x^{2}}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - (x^{2} + 2 x)}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{5 - (x^{2} + 2 x + 1 - 1)}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{6 - {(x + 1)}^{2}}} = - \frac{3}{2} \int \frac{(- 2 - 2 x) d x}{\sqrt{5 - 2 x - x^{2}}} - 2 \int \frac{d x}{\sqrt{{(\sqrt{6})}^{2} - {(x + 1)}^{2}}} Putting, 5 - 2 x - x^{2} = t \Rightarrow (- 2 - 2 x) d x = d t Then, I = - \frac{3}{2} \int \frac{d t}{\sqrt{t}} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C_{1} = - \frac{3}{2} \times 2 \sqrt{t} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C = - 3 \sqrt{5 - 2 x - x^{2}} - 2 \sin^{- 1} (\frac{x + 1}{\sqrt{6}}) + C$

Page No 18.110:

Question 14:

$\int \sqrt{\frac{1 - x}{1 + x}} d x$

Answer:

$Let I = \int \sqrt{\frac{1 - x}{1 + x}} d x = \int \sqrt{\frac{(1 - x) (1 - x)}{(1 + x) (1 - x)}} d x = \int (\frac{1 - x}{\sqrt{1 - x^{2}}}) d x = \int \frac{d x}{\sqrt{1 - x^{2}}} - \int \frac{x d x}{\sqrt{1 - x^{2}}} Putting 1 - x^{2} = t \Rightarrow - 2 x d x = d t \Rightarrow x d x = - \frac{d t}{2} Then, I = \int \frac{d x}{\sqrt{1 - x^{2}}} + \frac{1}{2} \int \frac{d t}{\sqrt{t}} = \sin^{- 1} (x) + \frac{1}{2} \times 2 \sqrt{t} + C = \sin^{- 1} (x) + \sqrt{1 - x^{2}} + C$

Page No 18.111:

Question 15:

$\int \frac{2 x + 1}{\sqrt{x^{2} + 4 x + 3}} d x$

Answer:

$Let I = \int \frac{(2 x + 1) d x}{\sqrt{x^{2} + 4 x + 3}} Consider, 2 x + 1 = A \frac{d}{d x} (x^{2} + 4 x + 3) + B \Rightarrow 2 x + 1 = A (2 x + 4) + B \Rightarrow 2 x + 1 = (2 A) x + 4 A + B Equating Coefficients of like terms 2 A = 2 \Rightarrow A = 1 And 4 A + B = 1 \Rightarrow 4 + B = 1 \Rightarrow B = - 3 ∴ I = \int (\frac{2 x + 4 - 3}{\sqrt{x^{2} + 4 x + 3}}) d x = \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 3}} - 3 \int \frac{d x}{\sqrt{x^{2} + 4 x + 4 - 4 + 3}} = \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 3}} - 3 \int \frac{d x}{\sqrt{{(x + 2)}^{2} - 1^{2}}} Let x^{2} + 4 x + 3 = t \Rightarrow (2 x + 4) d x = d t Then, I = \int \frac{d t}{\sqrt{t}} - 3 \int \frac{d x}{\sqrt{{(x + 2)}^{2} - 1^{2}}} = \int t^{- \frac{1}{2}} d t - 3 \int \frac{d x}{\sqrt{{(x + 2)}^{2} - 1^{2}}} = [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] - 3 \log |x + 2 + \sqrt{{(x + 2)}^{2} - 1}| + C = 2 \sqrt{t} - 3 \log |x + 2 + \sqrt{x^{2} + 4 x + 3}| + C = 2 \sqrt{x^{2} + 4 x + 3} - 3 \log |x + 2 + \sqrt{x^{2} + 4 x + 3}| + C$

Page No 18.111:

Question 16:

$\int \frac{2 x + 3}{\sqrt{x^{2} + 4 x + 5}} d x$

Answer:

$Let I = \int \frac{(2 x + 3) d x}{\sqrt{x^{2} + 4 x + 5}} = \int \frac{(2 x + 4 - 1)}{\sqrt{x^{2} + 4 x + 5}} d x = \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 5}} - \int \frac{d x}{\sqrt{x^{2} + 4 x + 5}} = \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 5}} - \int \frac{d x}{\sqrt{{(x + 2)}^{2} + 1}} Consider, x^{2} + 4 x + 5 = t \Rightarrow (2 x + 4) d x = d t ∴ I = \int \frac{d t}{\sqrt{t}} - \int \frac{d x}{\sqrt{{(x + 2)}^{2} + 1^{2}}} = \int t^{- \frac{1}{2}} d t - \int \frac{d x}{\sqrt{{(x + 2)}^{2} + 1^{2}}} = \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} - \log |x + 2 + \sqrt{{(x + 2)}^{2} + 1}| + C = 2 \sqrt{x^{2} + 4 x + 5} - \log |x + 2 + \sqrt{x^{2} + 4 x + 5}| + C$

Page No 18.111:

Question 17:

$\int \frac{5 x + 3}{\sqrt{x^{2} + 4 x + 10}} d x$

Answer:

$Let I = \int \frac{(5 x + 3) d x}{\sqrt{x^{2} + 4 x + 10}} Consider, 5 x + 3 = A \frac{d}{d x} (x^{2} + 4 x + 10) + B \Rightarrow 5 x + 3 = A (2 x + 4) + B \Rightarrow 5 x + 3 = (2 A) x + 4 A + B Equating Coefficients of like terms 2 A = 5 \Rightarrow A = \frac{5}{2} And 4 A + B = 3 \Rightarrow 4 \times \frac{5}{2} + B = 3 \Rightarrow B = - 7 ∴ I = \frac{5}{2} \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 10}} - 7 \int \frac{d x}{\sqrt{x^{2} + 4 x + 10}} = \frac{5}{2} \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 10}} - 7 \int \frac{d x}{\sqrt{x^{2} + 4 x + 4 - 4 + 10}} = \frac{5}{2} \int \frac{(2 x + 4) d x}{\sqrt{x^{2} + 4 x + 10}} - 7 \int \frac{d x}{\sqrt{{(x + 2)}^{2} + {(\sqrt{6})}^{2}}} Putting, x^{2} + 4 x + 10 = t \Rightarrow (2 x + 4) d x = d t Then, I = \frac{5}{2} \int \frac{d t}{\sqrt{t}} - 7 \log |x + 2 + \sqrt{{(x + 2)}^{2} + 6}| + C = \frac{5}{2} \int t^{- \frac{1}{2}} d t - 7 \log |x + 2 + \sqrt{x^{2} + 4 x + 10}| + C = \frac{5}{2} \times 2 \sqrt{t} - 7 \log |x + 2 + \sqrt{x^{2} + 4 x + 10}| + C = 5 \sqrt{x^{2} + 4 x + 10} - 7 \log |x + 2 + \sqrt{x^{2} + 4 x + 10}| + C$

Page No 18.111:

Question 18:

Evaluate the following integrals:

$\int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x$

Answer:

$Let I = \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x We express x + 2 = A (\frac{d}{d x} (x^{2} + 2 x + 3)) + B x + 2 = A (2 x + 2) + B Equating the coefficients of x and constants, we get 1 = 2 A and 2 = 2 A + B or A = \frac{1}{2} and B = 1 ∴ I = \int \frac{\frac{1}{2} (2 x + 2) + 1}{\sqrt{x^{2} + 2 x + 3}} d x = \frac{1}{2} \int \frac{(2 x + 2)}{\sqrt{x^{2} + 2 x + 3}} d x + \int \frac{1}{\sqrt{x^{2} + 2 x + 3}} d x = \frac{1}{2} I_{1} + I_{2} . . . (1) Now, I_{1} = \int \frac{(2 x + 2)}{\sqrt{x^{2} + 2 x + 3}} d x Let x^{2} + 2 x + 3 = u On differentiating both sides, we get (2 x + 2) d x = d u ∴ I_{1} = \int \frac{1}{\sqrt{u}} d u = 2 \sqrt{u} + c_{1} = 2 \sqrt{x^{2} + 2 x + 3} + c_{1} . . . (2) And, I_{2} = \int \frac{1}{\sqrt{x^{2} + 2 x + 3}} d x = \int \frac{1}{\sqrt{x^{2} + 2 x + 1 - 1 + 3}} d x = \int \frac{1}{\sqrt{{(x + 1)}^{2} + {(\sqrt{2})}^{2}}} d x Let (x + 1) = u On differentiating both sides, we get d x = d u ∴ I_{2} = \int \frac{1}{\sqrt{{(u)}^{2} + {(\sqrt{2})}^{2}}} d u = \log |u + \sqrt{{(u)}^{2} + {(\sqrt{2})}^{2}}| + c_{2} = \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c_{2} . . . (3) From (1), (2) and (3), we get ∴ I = \frac{1}{2} (2 \sqrt{x^{2} + 2 x + 3} + c_{1}) + \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c_{2} = \sqrt{x^{2} + 2 x + 3} + \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c Hence, \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x = \sqrt{x^{2} + 2 x + 3} + \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c$

Page No 18.114:

Question 1:

$\int \frac{1}{4 \cos^{2} x + 9 \sin^{2} x} d x$

Answer:

$Let I = \int \frac{1}{4 \cos^{2} x + 9 \sin^{2} x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\frac{1}{\cos^{2} x}}{4 + 9 \tan^{2} x} d x = \int \frac{\sec^{2} x}{4 + 9 \tan^{2} x} d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{4 + 9 t^{2}} = \frac{1}{9} \int \frac{d t}{\frac{4}{9} + t^{2}} = \frac{1}{9} \int \frac{d t}{{(\frac{2}{3})}^{2} + t^{2}} = \frac{1}{9} \times \frac{3}{2} \tan^{- 1} (\frac{t}{\frac{2}{3}}) + C = \frac{1}{6} \tan^{- 1} (\frac{3 t}{2}) + C = \frac{1}{6} \tan^{- 1} (\frac{3 \tan x}{2}) + C$

Page No 18.114:

Question 2:

$\int \frac{1}{4 \sin^{2} x + 5 \cos^{2} x} d x$

Answer:

$Let I = \int \frac{1}{4 \sin^{2} x + 5 \cos^{2} x} d x Dividing numerator & denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{4 \tan^{2} x + 5} d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{4 t^{2} + 5} = \frac{1}{4} \int \frac{d t}{t^{2} + \frac{5}{4}} = \frac{1}{4} \int \frac{d t}{t^{2} + {(\frac{\sqrt{5}}{2})}^{2}} = \frac{1}{4} \times \frac{2}{\sqrt{5}} \tan^{- 1} (\frac{t}{\sqrt{5}} \times 2) + C = \frac{1}{2 \sqrt{5}} \tan^{- 1} (\frac{2 \tan x}{\sqrt{5}}) + C$

Page No 18.114:

Question 3:

$\int \frac{2}{2 + \sin 2 x} d x$

Answer:

$Let I = \int \frac{2}{2 + \sin (2 x)} d x = \int \frac{2}{2 + 2 \sin x \cos x} d x = \int \frac{1}{1 + \sin x \cos x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x d x}{\sec^{2} x + \tan x} = \int \frac{\sec^{2} x d x}{1 + \tan^{2} x + \tan x} Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{t^{2} + t + 1} = \int \frac{d t}{t^{2} + t + \frac{1}{4} - \frac{1}{4} + 1} = \int \frac{d t}{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 t + 1}{\sqrt{3}}) + C = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan x + 1}{\sqrt{3}}) + C$

Page No 18.114:

Question 4:

$\int \frac{\cos x}{\cos 3 x} d x$

Answer:

$Let I = \int \frac{\cos x}{\cos 3 x} d x = \int \frac{\cos x}{(4 \cos^{3} x - 3 \cos x)} d x [\cos 3 A = 4 \cos^{3} A - 3 \cos A] = \int \frac{1}{4 \cos^{2} x - 3} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{4 - 3 \sec^{2} x} d x = \int \frac{\sec^{2} x}{4 - 3 (1 + \tan^{2} x)} d x = \int \frac{\sec^{2} x}{1 - 3 \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 - {(\sqrt{3} \tan x)}^{2}} d x Let \sqrt{3} \tan x = t \Rightarrow \sqrt{3} \sec^{2} x d x = d t \Rightarrow \sec^{2} x d x = \frac{d t}{\sqrt{3}} ∴ I = \frac{1}{\sqrt{3}} \int \frac{d t}{1^{2} - t^{2}} = \frac{1}{\sqrt{3}} \times \frac{1}{2} \ln |\frac{1 + t}{1 - t}| + C = \frac{1}{2 \sqrt{3}} \ln |\frac{1 + \sqrt{3} \tan x}{1 - \sqrt{3} \tan x}| + C$

Page No 18.114:

Question 5:

$\int \frac{1}{1 + 3 \sin^{2} x} d x$

Answer:

$Let I = \int \frac{1}{1 + 3 \sin^{2} x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{\sec^{2} x + 3 \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + \tan^{2} x + 3 \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + 4 \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + {(2 \tan x)}^{2}} d x Let 2 \tan x = t \Rightarrow 2 \sec^{2} x d x = d t \Rightarrow \sec^{2} x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{1 + t^{2}} = \frac{1}{2} \tan^{- 1} (t) + C = \frac{1}{2} \tan^{- 1} (2 \tan x) + C$

Page No 18.114:

Question 6:

$\int \frac{1}{3 + 2 \cos^{2} x} d x$

Answer:

$Let I = \int \frac{1}{3 + 2 \cos^{2} x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{3 \sec^{2} x + 2} d x = \int \frac{\sec^{2} x}{3 (1 + \tan^{2} x) + 2} d x = \int \frac{\sec^{2} x}{3 \tan^{2} x + 5} d x = \int \frac{\sec^{2} x}{{(\sqrt{5})}^{2} + {(\sqrt{3} \tan x)}^{2}} d x Let \sqrt{3} \tan x = t \Rightarrow \sqrt{3} \sec^{2} x d x = d t \Rightarrow \sec^{2} x d x = \frac{d t}{\sqrt{3}} ∴ I = \frac{1}{\sqrt{3}} \int \frac{d t}{{(\sqrt{5})}^{2} + t^{2}} = \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{5}} \tan^{- 1} (\frac{t}{\sqrt{5}}) + C = \frac{1}{\sqrt{15}} \tan^{- 1} (\frac{\sqrt{3} \tan x}{\sqrt{5}}) + C$

Page No 18.114:

Question 7:

$\int \frac{1}{(\sin x - 2 \cos x) (2 \sin x + \cos x)} d x$

Answer:

$Let I = \int \frac{1}{(\sin x - 2 \cos x) (2 \sin x + \cos x)} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{(\frac{\sin x - 2 \cos x}{\cos x}) \times (\frac{2 \sin x + \cos x}{\cos x})} d x = \int \frac{\sec^{2} x}{(\tan x - 2) (2 \tan x + 1)} d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{(t - 2) (2 t + 1)} = \int \frac{d t}{2 t^{2} + t - 4 t - 2} = \int \frac{d t}{2 t^{2} - 3 t - 2} = \frac{1}{2} \int \frac{d t}{t^{2} - \frac{3}{2} t - 1} = \frac{1}{2} \int \frac{d t}{t^{2} - \frac{3}{2} t + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{2} - 1} = \frac{1}{2} \int \frac{d t}{{(t - \frac{3}{4})}^{2} - \frac{9}{16} - 1} = \frac{1}{2} \int \frac{d t}{{(t - \frac{3}{4})}^{2} - {(\frac{5}{4})}^{2}} = \frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \log |\frac{t - \frac{3}{4} - \frac{5}{4}}{t - \frac{3}{4} + \frac{5}{4}}| + C = \frac{1}{5} \ln |\frac{t - 2}{t + \frac{1}{2}}| + C = \frac{1}{5} \ln |\frac{{(t - 2)}^{2}}{2 t + 1}| + C = \frac{1}{5} \ln |\frac{t - 2}{2 t + 1}| + \frac{1}{5} \ln (2) + C = \frac{1}{5} \ln |\frac{t - 2}{2 t + 1}| + C' where C' = C + \frac{1}{5} \ln (2) = \frac{1}{5} \ln |\frac{\tan x - 2}{2 \tan x + 1}| + C$

Page No 18.114:

Question 8:

$\int \frac{\sin 2 x}{\sin^{4} x + \cos^{4} x} d x$

Answer:

$Let I = \int \frac{\sin (2 x)}{\sin^{4} x + \cos^{4} x} d x = \int \frac{2 \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x Dividing numerator & denominator by \cos^{4} x \Rightarrow I = \int \frac{(\frac{2 \sin x \cos x}{\cos^{4} x})}{\tan^{4} x + 1} d x = \int \frac{2 \tan x . \sec^{2} x}{{(\tan^{2} x)}^{2} + 1} d x Let \tan^{2} x = t \Rightarrow 2 \tan x \sec^{2} x d x = d t = \int \frac{d t}{t^{2} + 1} ∴ I = \tan^{- 1} (t) + C = \tan^{- 1} (\tan^{2} x) + C$

Page No 18.114:

Question 9:

$\int \frac{1}{\cos x (\sin x + 2 \cos x)} d x$

Answer:

$Let I = \int \frac{1}{\cos x (\sin x + 2 \cos x)} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{\frac{\cos x}{\cos x} \times (\frac{\sin x + 2 \cos x}{\cos x})} d x = \int \frac{\sec^{2} x}{(\tan x + 2)} d x Let \tan x + 2 = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{t} = \ln |t| + C = \ln |\tan x + 2| + C$

Page No 18.114:

Question 10:

$\int \frac{1}{\sin^{2} x + \sin 2 x} d x$

Answer:

$Let I = \int \frac{1}{\sin^{2} x + \sin (2 x)} d x = \int \frac{1}{\sin^{2} x + 2 \sin x \cos x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{\tan^{2} x + 2 \tan x} d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{t^{2} + 2 t} = \int \frac{d t}{t^{2} + 2 t + 1 - 1} = \int \frac{d t}{{(t + 1)}^{2} - {(- 1)}^{2}} = \frac{1}{2} \ln |\frac{t + 1 - 1}{t + 1 + 1}| + C = \frac{1}{2} \ln |\frac{t}{t + 2}| + C = \frac{1}{2} \ln |\frac{\tan x}{\tan x + 2}| + C$

Page No 18.114:

Question 11:

$\int \frac{1}{\cos 2 x + 3 \sin^{2} x} d x$

Answer:

$Let I = \int \frac{1}{\cos (2 x) + 3 \sin^{2} x} d x = \int \frac{1}{(1 - 2 \sin^{2} x) + 3 \sin^{2} x} d x = \int \frac{1}{1 + \sin^{2} x} d x Dividing numerator and denominator by \cos^{2} x \Rightarrow I = \int \frac{\sec^{2} x}{\sec^{2} x + \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + \tan^{2} x + \tan^{2} x} d x = \int \frac{\sec^{2} x}{1 + 2 \tan^{2}} d x = \int \frac{\sec^{2} x}{1 + {(\sqrt{2} \tan x)}^{2}} d x Let \sqrt{2} \tan x = t \Rightarrow \sqrt{2} \sec^{2} x d x = d t \Rightarrow \sec^{2} x d x = \frac{d t}{\sqrt{2}} ∴ I = \frac{1}{\sqrt{2}} \int \frac{d t}{1 + t^{2}} = \frac{1}{\sqrt{2}} \tan^{- 1} (t) + C = \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{2} \tan x) + C$

Page No 18.117:

Question 1:

$\int \frac{1}{5 + 4 \cos x} d x$

Answer:

$Let I = \int \frac{1}{5 + 4 \cos x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} I = \int \frac{1}{5 + 4 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) + 4 (1 - \tan^{2} \frac{x}{2})} d x = \int \frac{\sec^{2} \frac{x}{2} d x}{5 + 5 \tan^{2} \frac{x}{2} + 4 - 4 \tan^{2} \frac{x}{2}} = \int \frac{\sec^{2} (\frac{x}{2}) d x}{\tan^{2} (\frac{x}{2}) + 9} Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{t^{2} + 3^{2}} = \frac{2}{3} \tan^{- 1} (\frac{t}{3}) + C = \frac{2}{3} \tan^{- 1} (\frac{\tan (\frac{x}{2})}{3}) + C$

Page No 18.117:

Question 2:

$\int \frac{1}{5 - 4 \sin x} d x$

Answer:

$Let I = \int \frac{1}{5 - 4 \sin x} d x Putting \sin x = \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} \Rightarrow I = \int \frac{1}{5 - 4 \times \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) - 8 \tan \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{5 \tan^{2} (\frac{x}{2}) - 8 \tan (\frac{x}{2}) + 5} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{5 t^{2} - 8 t + 5} = \frac{2}{5} \int \frac{d t}{t^{2} - \frac{8}{5} t + 1} = \frac{2}{5} \int \frac{d t}{t^{2} - \frac{8}{5} t + {(\frac{4}{5})}^{2} - {(\frac{4}{5})}^{2} + 1} = \frac{2}{5} \int \frac{d t}{{(t - \frac{4}{5})}^{2} - \frac{16}{25} + 1} = \frac{2}{5} \int \frac{d t}{{(t - \frac{4}{5})}^{2} + {(\frac{3}{5})}^{2}} = \frac{2}{5} \times \frac{5}{3} \tan^{- 1} (\frac{t - \frac{4}{5}}{\frac{3}{5}}) + C = \frac{2}{3} \tan^{- 1} (\frac{5 t - 4}{3}) + C = \frac{2}{3} \tan^{- 1} (\frac{5 \tan (\frac{x}{2}) - 4}{3}) + C$

Page No 18.117:

Question 3:

$\int \frac{1}{1 - 2 \sin x} d x$

Answer:

$Let I = \int \frac{1}{1 - 2 \sin x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{1 - 2 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{(1 + \tan^{2} \frac{x}{2}) - 4 \tan (\frac{x}{2})} d x = \int \frac{\sec^{2} (\frac{x}{2})}{\tan^{2} (\frac{x}{2}) - 4 \tan (\frac{x}{2}) + 1} d x Let \tan (\frac{x}{2}) = t \Rightarrow \sec^{2} (\frac{x}{2}) \times \frac{1}{2} d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{t^{2} - 4 t + 1} = 2 \int \frac{d t}{t^{2} - 4 t + 4 - 4 + 1} = 2 \int \frac{d t}{{(t - 2)}^{2} - 3} = 2 \int \frac{d t}{{(t - 2)}^{2} - {(\sqrt{3})}^{2}} = 2 \times \frac{1}{2 \sqrt{3}} \ln |\frac{t - 2 - \sqrt{3}}{t - 2 + \sqrt{3}}| + C = \frac{1}{\sqrt{3}} \ln |\frac{\tan (\frac{x}{2}) - 2 - \sqrt{3}}{\tan (\frac{x}{2}) - 2 + \sqrt{3}}| + C$

Page No 18.117:

Question 4:

$\int \frac{1}{4 \cos x - 1} d x$

Answer:

$Let I = \int \frac{1}{4 \cos x - 1} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{4 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) - 1} d x = \int \frac{1}{\frac{4 (1 - \tan^{2} \frac{x}{2}) - (1 + \tan^{2} \frac{x}{2})}{(1 + \tan^{2} \frac{x}{2})}} = \int \frac{(1 + \tan^{2} \frac{x}{2}) d x}{4 - 4 \tan^{2} (\frac{x}{2}) - 1 - \tan^{2} (\frac{x}{2})} = \int \frac{\sec^{2} (\frac{x}{2}) d x}{3 - 5 \tan^{2} (\frac{x}{2})} Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{3 - 5 t^{2}} = \frac{2}{5} \int \frac{d t}{\frac{3}{5} - t^{2}} = \frac{2}{5} \int \frac{d t}{{(\frac{\sqrt{3}}{\sqrt{5}})}^{2} - t^{2}} = \frac{2}{5} \times \frac{\sqrt{5}}{2 \sqrt{3}} \ln |\frac{\frac{\sqrt{3}}{\sqrt{5}} + t}{\frac{\sqrt{3}}{\sqrt{5}} - t}| + C = \frac{1}{\sqrt{15}} \ln |\frac{\sqrt{3} + \sqrt{5} t}{\sqrt{3} - \sqrt{5} t}| + C = \frac{1}{\sqrt{15}} \ln |\frac{\sqrt{3} + \sqrt{5} \tan (\frac{x}{2})}{\sqrt{3} - \sqrt{5} \tan (\frac{x}{2})}| + C$

Page No 18.117:

Question 5:

$\int \frac{1}{1 - \sin x + \cos x} d x$

Answer:

$Let I = \int \frac{1}{1 - \sin x + \cos x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} = \int \frac{1}{1 - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{(1 + \tan^{2} \frac{x}{2}) - 2 \tan x (2 + 1 - \tan^{2} \frac{x}{2})} d x = \int \frac{\sec^{2} \frac{x}{2}}{2 - 2 \tan (\frac{x}{2})} d x = \frac{1}{2} \int \frac{\sec^{2} (\frac{x}{2})}{1 - \tan (\frac{x}{2})} d x Let [1 - \tan (\frac{x}{2})] = t \Rightarrow - \sec^{2} (\frac{x}{2}) \times \frac{1}{2} d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = - 2 d t ∴ I = \frac{1}{2} \int \frac{- 2 d t}{t} = - \int \frac{d t}{t} = - \ln |t| + C = - \ln |1 - \tan \frac{x}{2}| + C$

Page No 18.117:

Question 6:

$\int \frac{1}{3 + 2 \sin x + \cos x} d x$

Answer:

$Let I = \int \frac{1}{3 + 2 \sin x + \cos x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{3 + 2 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{3 (1 + \tan^{2} \frac{x}{2}) + 4 \tan (\frac{x}{2}) + 1 - \tan^{2} (\frac{x}{2})} d x = \int \frac{\sec^{2} (\frac{x}{2})}{3 + 3 \tan^{2} (\frac{x}{2}) + 4 \tan (\frac{x}{2}) + 1 - \tan^{2} (\frac{x}{2})} d x = \int \frac{\sec^{2} (\frac{x}{2})}{2 \tan^{2} (\frac{x}{2}) + 4 \tan (\frac{x}{2}) + 4} d x = \frac{1}{2} \int \frac{\sec^{2} (\frac{x}{2})}{\tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2}) + 2} d x Let \tan (\frac{x}{2}) = t \Rightarrow \sec^{2} (\frac{x}{2}) \times \frac{1}{2} d x = d t \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \frac{1}{2} \int \frac{2 d t}{t^{2} + 2 t + 2} = \int \frac{d t}{t^{2} + 2 t + 1 + 1} = \int \frac{d t}{{(t + 1)}^{2} + 1^{2}} = \tan^{- 1} (\frac{t + 1}{1}) + C = \tan^{- 1} (1 + \tan \frac{x}{2}) + C$

Page No 18.117:

Question 7:

$\int \frac{1}{13 + 3 \cos x + 4 \sin x} d x$

Answer:

$L e t I = \int \frac{1}{13 + 3 \cos x + 4 \sin x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \sin x = \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} ∴ I = \int \frac{1}{13 + 3 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) + 4 \times 2 \frac{\tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{13 (1 + \tan^{2} \frac{x}{2}) + 3 - 3 \tan^{2} \frac{x}{2} + 8 \tan (\frac{x}{2})} d x = \int \frac{\sec^{2} \frac{x}{2}}{13 \tan^{2} \frac{x}{2} - 3 \tan^{2} \frac{x}{2} + 16 + 8 \tan (\frac{x}{2})} d x = \int \frac{\sec^{2} \frac{x}{2}}{10 \tan^{2} (\frac{x}{2}) + 8 \tan (\frac{x}{2}) + 16} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \int \frac{2 d t}{10 t^{2} + 8 t + 16} = \int \frac{d t}{5 t^{2} + 4 t + 8} = \frac{1}{5} \int \frac{d t}{t^{2} + \frac{4}{5} t + \frac{8}{5}} = \frac{1}{5} \int \frac{d t}{t^{2} + \frac{4}{5} t + {(\frac{2}{5})}^{2} - {(\frac{2}{5})}^{2} + \frac{8}{5}} = \frac{1}{5} \int \frac{d t}{{(t + \frac{2}{5})}^{2} - \frac{4}{25} + \frac{8}{5}} = \frac{1}{5} \int \frac{d t}{{(t + \frac{2}{5})}^{2} + \frac{- 4 + 40}{25}} = \frac{1}{5} \int \frac{d t}{{(t + \frac{2}{5})}^{2} + {(\frac{6}{5})}^{2}} = \frac{1}{5} \times \frac{5}{6} \tan^{- 1} (\frac{t + \frac{2}{5}}{\frac{6}{5}}) + C = \frac{1}{6} \tan^{- 1} (\frac{5 t + 2}{6}) + C = \frac{1}{6} \tan^{- 1} (\frac{5 \tan \frac{x}{2} + 2}{6}) + C$

Page No 18.117:

Question 8:

$\int \frac{1}{\cos x - \sin x} d x$

Answer:

$Let I = \int \frac{d x}{\cos x - \sin x} Putting \cos x = \frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} and \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{d x}{\frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} = \int \frac{\sec^{2} (\frac{x}{2}) d x}{1 - \tan^{2} (\frac{x}{2}) - 2 \tan (\frac{x}{2})} Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \int \frac{2 d t}{1 - t^{2} - 2 t} = \int \frac{- 2 d t}{t^{2} + 2 t - 1} = \int \frac{- 2 d t}{t^{2} + 2 t + 1 - 2} = - \int \frac{2 d t}{{(t + 1)}^{2} - {(\sqrt{2})}^{2}} = \int \frac{2 d t}{{(\sqrt{2})}^{2} - {(t - 1)}^{2}} = \frac{2}{2 \sqrt{2}} \ln |\frac{\sqrt{2} + t + 1}{\sqrt{2} - t - 1}| + C = \frac{1}{\sqrt{2}} \ln |\frac{\sqrt{2} + \tan \frac{x}{2} + 1}{\sqrt{2} - \tan \frac{x}{2} - 1}| + C$

Page No 18.117:

Question 9:

$\int \frac{1}{\sin x + \cos x} d x$

Answer:

$Let I = \int \frac{1}{\sin x + \cos x} d x Putting \sin x = \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} and \cos x = \frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} = \int \frac{1}{\frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} + \frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}} d x = \int \frac{\sec^{2} (\frac{x}{2})}{1 - \tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2})} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{1 - t^{2} + 2 t} = - 2 \int \frac{d t}{t^{2} - 2 t - 1} = - 2 \int \frac{d t}{t^{2} - 2 t + 1 - 2} = 2 \int \frac{d t}{{(\sqrt{2})}^{2} - {(t - 1)}^{2}} = 2 \times \frac{1}{2 \sqrt{2}} \ln |\frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1}| + C = \frac{1}{\sqrt{2}} \ln |\frac{\sqrt{2} + \tan \frac{x}{2} - 1}{\sqrt{2} - \tan \frac{x}{2} + 1}| + C$

Page No 18.117:

Question 10:

$\int \frac{1}{5 - 4 \cos x} d x$

Answer:

$Let I = \int \frac{1}{5 - 4 \cos x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{5 - 4 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) - 4 + 4 \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} (\frac{x}{2})}{9 \tan^{2} \frac{x}{2} + 1} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{d t}{9 t^{2} + 1} = \frac{2}{9} \int \frac{d t}{t^{2} + \frac{1}{9}} = \frac{2}{9} \int \frac{d t}{t^{2} + {(\frac{1}{3})}^{2}} = \frac{2}{9} \times 3 \tan^{- 1} (\frac{t}{\frac{1}{3}}) + C = \frac{2}{3} \tan^{- 1} (3 t) + C = \frac{2}{3} \tan^{- 1} (3 \tan \frac{x}{2}) + C$

Page No 18.117:

Question 11:

$\int \frac{1}{2 + \sin x + \cos x} d x$

Answer:

$Let I = \int \frac{1}{2 + \sin x + \cos x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{2 + \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{1 + \tan^{2} \frac{x}{2}}{2 (1 + \tan^{2} \frac{x}{2}) + 2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{2 + 2 \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{\tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2} + 3} d x Let \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t ∴ I = 2 \int \frac{d t}{t^{2} + 2 t + 3} = 2 \int \frac{d t}{t^{2} + 2 t + 1 + 2} = 2 \int \frac{d t}{{(t + 1)}^{2} + {(\sqrt{2})}^{2}} = 2 \times \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t + 1}{\sqrt{2}}) + C = \sqrt{2} \tan^{- 1} (\frac{\tan \frac{x}{2} + 1}{\sqrt{2}}) + C$

Page No 18.117:

Question 12:

$\int \frac{1}{\sin x + \sqrt{3} \cos x} d x$

Answer:

$Let I = \int \frac{1}{\sin x + \sqrt{3} \cos x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \sqrt{3} \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{1 + \tan^{2} \frac{x}{2}}{2 \tan \frac{x}{2} + \sqrt{3} - \sqrt{3} \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{- \sqrt{3} \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2} + \sqrt{3}} d x$

$Let \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t ∴ I = 2 \int \frac{d t}{- \sqrt{3} t^{2} + 2 t + \sqrt{3}} = - \frac{2}{\sqrt{3}} \int \frac{d t}{t^{2} - \frac{2}{\sqrt{3}} t - 1} = - \frac{2}{\sqrt{3}} \int \frac{d t}{t^{2} - \frac{2}{\sqrt{3}} t + {(\frac{1}{\sqrt{3}})}^{2} - {(\frac{1}{\sqrt{3}})}^{2} - 1} = - \frac{2}{\sqrt{3}} \int \frac{d t}{{(t - \frac{1}{\sqrt{3}})}^{2} - {(\frac{2}{\sqrt{3}})}^{2}} = - \frac{2}{\sqrt{3}} \times \frac{1}{2 \frac{2}{\sqrt{3}}} \log |\frac{t - \frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}}{t - \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}}| + C$

$= - \frac{1}{2} \log |\frac{t - \frac{3}{\sqrt{3}}}{t + \frac{1}{\sqrt{3}}}| + C = - \frac{1}{2} \log |\frac{\sqrt{3} t - 3}{\sqrt{3} t + 1}| + C = \frac{1}{2} \log |\frac{\sqrt{3} t + 1}{\sqrt{3} t - 3}| + C = \frac{1}{2} \log |\frac{\sqrt{3} \tan \frac{x}{2} + 1}{\sqrt{3} \tan \frac{x}{2} - 3}| + C or, \frac{1}{2} \log |\frac{1 + \sqrt{3} \tan \frac{x}{2}}{3 - \sqrt{3} \tan \frac{x}{2}}| + C$

Page No 18.117:

Question 13:

$\int \frac{1}{\sqrt{3} \sin x + \cos x} d x$

Answer:

$Let I = \int \frac{d x}{\sqrt{3} \sin x + \cos x} Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{\sqrt{3} \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{1 + \tan^{2} \frac{x}{2}}{2 \sqrt{3} \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{- \tan^{2} \frac{x}{2} + 2 \sqrt{3} \tan \frac{x}{2} + 1} d x$

$Let \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t ∴ I = 2 \int \frac{d t}{- t^{2} + 2 \sqrt{3} t + 1} = - 2 \int \frac{d t}{t^{2} - 2 \sqrt{3} t - 1} = - 2 \int \frac{d t}{t^{2} - 2 \sqrt{3} t + {(\sqrt{3})}^{2} - {(\sqrt{3})}^{2} - 1} = - 2 \int \frac{d t}{{(t - \sqrt{3})}^{2} - {(2)}^{2}} = - \frac{2}{2 \times 2} \log |\frac{t - \sqrt{3} - 2}{t - \sqrt{3} + 2}| + C$

$= - \frac{1}{2} \log |\frac{\tan \frac{x}{2} - 2 - \sqrt{3}}{\tan \frac{x}{2} + 2 - \sqrt{3}}| + C = \frac{1}{2} \log |\frac{\tan \frac{x}{2} + 2 - \sqrt{3}}{\tan \frac{x}{2} + 2 - \sqrt{3}}| + C$

Page No 18.117:

Question 14:

$\int \frac{1}{\sin x - \sqrt{3} \cos x} d x$

Answer:

$Let I = \int \frac{d x}{\sin x - \sqrt{3} \cos x} Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} - \sqrt{3} \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{1 + \tan^{2} \frac{x}{2}}{2 \tan \frac{x}{2} - \sqrt{3} + \sqrt{3} \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{\sqrt{3} \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2} - \sqrt{3}} d x$

$Let \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t ∴ I = 2 \int \frac{d t}{\sqrt{3} t^{2} + 2 t - \sqrt{3}} = \frac{2}{\sqrt{3}} \int \frac{d t}{t^{2} + \frac{2}{\sqrt{3}} t - 1} = 2 \int \frac{d t}{t^{2} + \frac{2}{\sqrt{3}} t + {(\frac{1}{\sqrt{3}})}^{2} - 1 - {(\frac{1}{\sqrt{3}})}^{2}} = 2 \int \frac{d t}{{(t + \frac{1}{\sqrt{3}})}^{2} - {(\frac{2}{\sqrt{3}})}^{2}} = \frac{2}{2 \times \frac{2}{\sqrt{3}}} \log |\frac{t + \frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}}{t + \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}}| + C$
$= \frac{\sqrt{3}}{2} \log |\frac{\tan \frac{x}{2} - \frac{1}{\sqrt{3}}}{\tan \frac{x}{2} + \frac{3}{\sqrt{3}}}| + C = \frac{\sqrt{3}}{2} \log |\frac{\sqrt{3} \tan \frac{x}{2} - 1}{\sqrt{3} \tan \frac{x}{2} + 3}| + C$

Page No 18.117:

Question 15:

$\int \frac{1}{5 + 7 \cos x + \sin x} d x$

Answer:

$Let I = \int \frac{1}{5 + 7 \cos x + \sin x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow I = \int \frac{1}{5 + 7 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) + \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{\sec^{2} (\frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) + 7 - 7 \tan^{2} \frac{x}{2} + 2 \tan (\frac{x}{2})} d x = \int \frac{\sec^{2} (\frac{x}{2})}{- 2 \tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2}) + 12} d x Let \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \int \frac{2 d t}{- 2 t^{2} + 2 t + 12} = \int \frac{d t}{- t^{2} + t + 6} = \int \frac{- d t}{t^{2} - t - 6} = \int \frac{- d t}{t^{2} - t + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 6} = \int \frac{- d t}{{(t - \frac{1}{2})}^{2} - \frac{1}{4} - 6} = \int \frac{- d t}{{(t - \frac{1}{2})}^{2} - {(\frac{5}{2})}^{2}} = \int \frac{d t}{{(\frac{5}{2})}^{2} - {(t - \frac{1}{2})}^{2}} = \frac{1}{2 \times \frac{5}{2}} \log |\frac{\frac{5}{2} + t - \frac{1}{2}}{\frac{5}{2} - t + \frac{1}{2}}| + C = \frac{1}{5} \log |\frac{2 + t}{3 - t}| + C = \frac{1}{5} \log |\frac{2 + \tan \frac{x}{2}}{3 - \tan \frac{x}{2}}| + C$

Page No 18.122:

Question 1:

$\int \frac{1}{1 - \cot x} d x$

Answer:

$Let I = \int \frac{1}{1 - \cot x} d x = \int \frac{1}{1 - \frac{\cos x}{\sin x}} d x = \int \frac{\sin x}{\sin x - \cos x} d x = \frac{1}{2} \int \frac{2 \sin x}{\sin x - \cos x} d x = \frac{1}{2} \int [\frac{\sin x + \cos x + \sin x - \cos x}{\sin x - \cos x}] d x = \frac{1}{2} \int (\frac{\sin x + \cos x}{\sin x - \cos x}) d x + \frac{1}{2} \int d x Putting \sin x - \cos x = t \Rightarrow (\cos x + \sin x) d x = d t ∴ I = \frac{1}{2} \int \frac{1}{t} d t + \frac{1}{2} \int d x = \frac{1}{2} \ln |t| + \frac{x}{2} + C = \frac{x}{2} + \frac{1}{2} \ln |\sin x - \cos x| + C$

Page No 18.122:

Question 2:

$\int \frac{1}{1 - \tan x} d x$

Answer:

$Let I = \int \frac{1}{1 - \tan x} d x = \int \frac{1}{1 - \frac{\sin x}{\cos x}} d x = \int \frac{\cos x}{\cos x - \sin x} d x = \frac{1}{2} \int \frac{2 \cos x}{\cos x - \sin x} d x = \frac{1}{2} \int (\frac{\cos x + \sin x + \cos x - \sin x}{\cos x - \sin x}) d x = \frac{1}{2} \int (\frac{\cos x + \sin x}{\cos x - \sin x}) d x + \frac{1}{2} \int d x Putting \cos x - \sin x = t \Rightarrow (- \sin x - \cos x) d x = d t \Rightarrow (\sin x + \cos x) d x = - d t ∴ I = - \frac{1}{2} \int \frac{d t}{t} + \frac{x}{2} + C = - \frac{1}{2} \ln |\cos x - \sin x| + \frac{x}{2} + C = \frac{x}{2} - \frac{1}{2} \ln |\cos x - \sin x| + C$

Page No 18.122:

Question 3:

$\int \frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3} d x$

Answer:

$Let I = \int (\frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3}) d x Let 3 + 2 \cos x + 4 \sin x = A (2 \sin x + \cos x + 3) + B (2 \cos x - \sin x) + C \Rightarrow 3 + 2 \cos x + 4 \sin x = (2 A - B) \sin x + (A + 2 B) \cos x + 3 A + C$

Comparing the coefficients of like terms
$2 A - B = 4 . . . (1) A + 2 B = 2 . . . (2) 3 A + C = 3 . . . (3)$

Multiplying eq (1) by 2 and adding it to eq (2) we get ,

$\Rightarrow 4 A - 2 B + A + 2 B = 8 + 2 \Rightarrow 5 A = 10 \Rightarrow A = 2$

Putting value of A = 2 in eq (1)

$\Rightarrow 2 \times 2 - B = 4 \Rightarrow B = 0 Putting value of A in eq (3) \Rightarrow 3 \times 2 + C = 3 \Rightarrow C = - 3$

$∴ I = \int [\frac{2 (2 \sin x + \cos x + 3) - 3}{2 \sin x + \cos x + 3}] d x = 2 \int d x - 3 \int \frac{1}{2 \sin x + \cos x + 3} d x Substituting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} ∴ I = 2 \int d x - 3 \int \frac{1}{2 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + 3} d x = 2 \int d x - 3 \int \frac{(1 + \tan^{2} \frac{x}{2})}{4 \tan (\frac{x}{2}) + 1 - \tan^{2} (\frac{x}{2}) + 3 (1 + \tan^{2} \frac{x}{2})} d x = 2 \int d x - 3 \int \frac{\sec^{2} (\frac{x}{2})}{2 \tan^{2} (\frac{x}{2}) + 4 \tan (\frac{x}{2}) + 4} d x = 2 \int d x - \frac{3}{2} \int \frac{\sec^{2} (\frac{x}{2})}{\tan^{2} (\frac{x}{2}) + 2 \tan (\frac{x}{2}) + 2} d x Putting \tan (\frac{x}{2}) = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int d x - \frac{3}{2} \int \frac{2}{t^{2} + 2 t + 2} d t = 2 \int d x - 3 \int \frac{1}{t^{2} + 2 t + 1 + 1} d t = 2 \int d x - 3 \int \frac{1}{{(t + 1)}^{2} + {(1)}^{2}} d t = 2 x - \frac{3}{1} \tan^{- 1} (\frac{t + 1}{1}) + C = 2 x - 3 \tan^{- 1} (\tan \frac{x}{2} + 1) + C [∵ t = \tan \frac{x}{2}]$

Page No 18.122:

Question 4:

$\int \frac{1}{p + q \tan x} d x$

Answer:

$Let I = \int \frac{d x}{p + q \tan x} = \int \frac{1}{p + \frac{q \sin x}{\cos x}} d x = \int \frac{\cos x}{q \sin x + p \cos x} d x Let \cos x = A (q \sin x + p \cos x) + B (q \cos x - p \sin x) \Rightarrow \cos x = (A p + B q) \cos x + (A q - B p) \sin x$

Comparing coefficients of like terms

$A p + B q = 1 . . . (1) A q - B p = 0 . . . (2)$

Multiplying eq (1) by p and eq (2) by q and then adding

$\Rightarrow A p^{2} + B p q = p \Rightarrow A q^{2} - B p q = 0 \Rightarrow A = \frac{p}{p^{2} q^{2}}$

Putting value of A in eq (1)

\frac{p^{2}}{p^{2} + q^{2}} + B q = 1 \Rightarrow B q = 1 - \frac{p^{2}}{p^{2} + q^{2}} \Rightarrow B q = \frac{p^{2} + q^{2} - p^{2}}{p^{2} + q^{2}} \Rightarrow B = \frac{q}{p^{2} + q^{2}} ∴ I = \int [\frac{p}{p^{2} + q^{2}} \times \frac{(q \sin x + p \cos x)}{(q \sin x + p \cos x)} + \frac{q}{p^{2} + q^{2}} \times \frac{(q \cos x - p \sin x)}{(q \sin x + p \cos x)}] d x = \frac{p}{p^{2} + q^{2}} \int d x + \frac{q}{p^{2} + q^{2}} \int (\frac{q \cos x - p \sin x}{q \sin x + p \cos x}) d x Putting q \sin x + p \cos x = t \Rightarrow (q \cos x - p \sin x) d x = d t ∴ I = \frac{p}{p^{2} + q^{2}} \int d x + \frac{q}{p^{2} + q^{2}} \int \frac{1}{t} d t = \frac{p}{p^{2} + q^{2}} x + \frac{q}{p^{2} + q^{2}} \ln |q \sin x + p \cos x| + C

Page No 18.122:

Question 5:

$\int \frac{5 \cos x + 6}{2 \cos x + \sin x + 3} d x$

Answer:

$Let I = \int (\frac{5 \cos x + 6}{2 \cos x + \sin x + 3}) d x & let 5 \cos x + 6 = A (2 \cos x + \sin x + 3) + B (- 2 \sin x + \cos x) + C . . . . (1) \Rightarrow 5 \cos x + 6 = (A - 2 B) \sin x + (2 A + B) \cos x + 3 A + C$

Comparing coefficients of like terms

$A - 2 B = 0 . . . (2) 2 A + B = 5 . . . (3) 3 A + C = 6 . . . (4)$

Multiplying eq (3) by 2 and then adding to eq (2)

4A + 2B + A – 2B = 10
$\Rightarrow$ A = 2

Putting value of A in eq (2) and eq (4) we get,
B = 1& C = 0

$By putting the values of A, B and C in eq (1) we get, ∴ I = \int [\frac{2 (2 \cos x + \sin x + 3) + (- 2 \sin x + \cos x)}{(2 \cos x + \sin x + 3)}] d x = 2 \int d x + \int (\frac{- 2 \sin x + \cos x}{2 \cos x + \sin x + 3}) d x Putting 2 \cos x + \sin x + 3 = t \Rightarrow (- 2 \sin x + \cos x) d x = d t ∴ I = 2 \int d x + \int \frac{1}{t} d t = 2 x + \ln |2 \cos x + \sin x + 3| + C$

Page No 18.122:

Question 6:

$\int \frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} d x$

Answer:

$Let I = \int (\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x}) d x & let 2 \sin x + 3 \cos x = A (3 \sin x + 4 \cos x) + B (3 \cos x - 4 \sin x) . . . (1) \Rightarrow 2 \sin x + 3 \cos x = (3 A - 4 B) \sin x + (4 A + 3 B) \cos x$

By comparing the coefficients of like terms we get,

$3 A - 4 B = 2 . . . (2) 4 A - 3 B = 3 . . . (3)$

Multiplying eq (2) by 3 and eq (3) by 4 and then adding,

$9 A - 12 B + 16 A + 12 B = 6 + 12 \Rightarrow 25 A = 18 \Rightarrow A = \frac{18}{25} Putting value of A = \frac{18}{25} in eq (2) we get, 3 \times \frac{18}{25} - 4 B = 2 \Rightarrow \frac{54}{25} - 2 = 4 B \Rightarrow \frac{4}{25 \times 4} = B \Rightarrow B = \frac{1}{25}$
Thus, substituting the values of A,B and C in eq (1) we get ,

$I = \int [\frac{\frac{18}{25} (3 \sin x + 4 \cos x) + \frac{1}{25} (3 \cos x - 4 \sin x)}{(3 \sin x + 4 \cos x)}] d x = \frac{18}{25} \int d x + \frac{1}{25} \int (\frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x}) d x Putting 3 \sin x + 4 \cos x = t \Rightarrow (3 \cos x - 4 \sin x) d x = d t ∴ I = \frac{18}{25} \int d x + \frac{1}{25} \int \frac{1}{t} d t = \frac{18 x}{25} + \frac{1}{25} \ln |t| + C = \frac{18 x}{25} + \frac{1}{25} \ln |3 \sin x + 4 \cos x| + C$

Page No 18.122:

Question 7:

$\int \frac{1}{3 + 4 \cot x} d x$

Answer:

$Let I = \int \frac{1}{3 + 4 \cot x} d x = \int \frac{1}{3 + \frac{4 \cos x}{\sin x}} d x = \int \frac{\sin x}{3 \sin x + 4 \cos x} d x Let \sin x = A (3 \sin x + 4 \cos x) + B (3 \cos x - 4 \sin x) . . . (1) \Rightarrow \sin x = (3 A - 4 B) \sin x + (4 A + 3 B) \cos x By comparing the coefficients of both sides we get, 3 A - 4 B = 1 . . . (2) 4 A + 3 B = 0 . . . (3)$

Multiplying eq (2) by 3 and equation (3) by 4 , then by adding them we get

$9 A - 12 B + 16 A + 12 B = 3 + 0 \Rightarrow 25 A = 3 \Rightarrow A = \frac{3}{25} Putting value of A in eq (3) we get, 4 \times \frac{3}{25} + 3 B = 0 \Rightarrow 3 B = - \frac{12}{25} \Rightarrow B = - \frac{4}{25}$
$Thus, by substituting the value of A and B in eq (1) we get I = \int [\frac{\frac{3}{25} (3 \sin x + 4 \cos x) - \frac{4}{25} (3 \cos x - 4 \sin x)}{3 \sin x + 4 \cos x}] d x = \frac{3}{25} \int d x - \frac{4}{25} \int (\frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x}) d x Putting 3 \sin x + 4 \cos x = t \Rightarrow (3 \cos x - 4 \sin x) d x = d t ∴ I = \frac{3}{25} \int d x - \frac{4}{25} \int \frac{d t}{t} = \frac{3}{25} x - \frac{4}{25} \ln |t| + C = \frac{3 x}{25} - \frac{4}{25} \ln |3 \sin x + 4 \cos x| + C$

Page No 18.122:

Question 8:

$\int \frac{2 \tan x + 3}{3 \tan x + 4} d x$

Answer:

$Let I = \int (\frac{2 \tan x + 3}{3 \tan x + 4}) d x = \int (\frac{\frac{2 \sin x}{\cos x} + 3}{\frac{3 \sin x}{\cos x} + 4}) d x = \int (\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x}) d x Let (2 \sin x + 3 \cos x) = A (3 \sin x + 4 \cos x) + B (3 \cos x - 4 \sin x) . . . . (1) \Rightarrow 2 \sin x + 3 \cos x = (3 A - 4 B) \sin x + (4 A + 3 B) \cos x Equating the coefficients of like terms 3 A - 4 B = 2 . . . (2) 4 A + 3 B = 3 . . . (3)$

Multiplying equation (2) by 3 and equation (3) by 4 ,then by adding them we get

$\begin{matrix} 9 A & - & 12 B & = & 6 \\ 16 A & + & 12 B & = & 12 \end{matrix} \begin{matrix} 25 A & = & 18 \end{matrix} \Rightarrow A = \frac{18}{25} Putting value of A in eq (2) we get, \Rightarrow B = \frac{1}{25}$
$Thus, by substituting the values of A and B in eq (1) we get, I = \int \{\frac{\frac{18}{25} (3 \sin x + 4 \cos x) + \frac{1}{25} (3 \cos x - 4 \sin x)}{3 \sin x + 4 \cos x}\} d x = \frac{18}{25} \int d x + \frac{1}{25} \int (\frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x}) d x Putting 3 \sin x + 4 \cos x = t \Rightarrow (3 \cos x - 4 \sin x) d x = d t ∴ I = \frac{18}{25} x + \frac{1}{25} \int \frac{1}{t} d t = \frac{18 x}{25} + \frac{1}{25} \ln |t| + C = \frac{18 x}{25} + \frac{1}{25} \ln |3 \sin x + 4 \cos x| + C$

Page No 18.122:

Question 9:

$\int \frac{1}{4 + 3 \tan x} d x$

Answer:

$Let I = \int \frac{d x}{4 + 3 \tan x} = \int \frac{d x}{4 + \frac{3 \sin x}{\cos x}} = \int \frac{\cos x d x}{4 \cos x + 3 \sin x} Consider, \cos x = A (4 \cos x + 3 \sin x) + B \frac{d}{dx} (4 \cos x + 3 \sin x) \Rightarrow \cos x = A (4 \cos x + 3 \sin x) + B (- 4 \sin x + 3 \cos x) \Rightarrow \cos x = (4 A + 3 B) \cos x + (3 A - 4 B) \sin x Equating the coefficients of like terms 4 A + 3 B = 1 . . . . . (1) 3 A - 4 B = 0 . . . . . (2)$
Solving (1) and (2), we get
$A = \frac{4}{25} and B = \frac{3}{25}$
$\int [\frac{\frac{4}{25} (4 \cos x + 3 \sin x) + (- 4 \sin x + 3 \cos x) \frac{3}{25}}{4 \cos x + 3 \sin x}] d x = \frac{4}{25} \int d x + \frac{3}{25} \int (\frac{- 4 \sin x + 3 \cos x}{4 \cos x + 3 \sin x}) d x let 4 \cos x + 3 \sin x = t \Rightarrow (- 4 \sin x + 3 \cos x) d x = d t Then, I = \frac{4}{25} \int d x + \frac{3}{25} \int \frac{d t}{t} = \frac{4 x}{25} + \frac{3}{25} \log |t| + C = \frac{4 x}{25} + \frac{3}{25} \log |4 \cos x + 3 \sin x| + C$

Page No 18.122:

Question 10:

$\int \frac{8 \cot x + 1}{3 \cot x + 2} d x$

Answer:

$L e t I = \int (\frac{8 \cot x + 1}{3 \cot x + 2}) d x = \int (\frac{8 \frac{\cos x}{\sin x} + 1}{\frac{3 \cos x}{\sin x} + 2}) d x = \int (\frac{8 \cos x + \sin x}{3 \cos x + 2 \sin x}) d x Now, let 8 \cos x + \sin x = A (3 \cos x + 2 \sin x) + B (- 3 \sin x + 2 \cos x) . . . (1) \Rightarrow 8 \cos x + \sin x = 3 A \cos x + 2 A \sin x - 3 B \sin x + 2 B \cos x \Rightarrow 8 \cos x + \sin x = (3 A + 2 B) \cos x + (2 A - 3 B) \sin x Equating the coefficients of like terms we get, 2 A - 3 B = 1 . . . (2) 3 A + 2 B = 8 . . . (3)$

Solving eq (2) and eq (3) we get,
A = 2, B = 1
Thus, by substituting the values of A and B in eq (1) we get ,

$I = \int [\frac{2 (3 \cos x + 2 \sin x) + 1 (- 3 \sin x + 2 \cos x)}{(3 \cos x + 2 \sin x)}] d x = 2 \int (\frac{3 \cos x + 2 \sin x}{3 \cos x + 2 \sin x}) d x + \int (\frac{- 3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x}) d x = 2 \int d x + \int (\frac{- 3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x}) d x Putting 3 \cos x + 2 \sin x = t \Rightarrow (- 3 \sin x + 2 \cos x) d x = d t ∴ I = 2 \int d x + \int \frac{1}{t} d t = 2 x + \ln |t| + C = 2 x + \ln |3 \cos x + 2 \sin x| + C$

Page No 18.122:

Question 11:

$\int \frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x} d x$

Answer:

$Let I = \int (\frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x}) d x & let 4 \sin x + 5 \cos x = A (5 \sin x + 4 \cos x) + B (5 \cos x - 4 \sin x) . . . (1) \Rightarrow 4 \sin x + 5 \cos x = (5 A - 4 B) \sin x + (4 A + 5 B) \cos x By equating the coefficients of like terms we get, 5 A - 4 B = 4 . . . (2) 4 A + 5 B = 5 . . . (3)$

By solving eq (2) and eq (3) we get,
$A = \frac{40}{41}, B = \frac{9}{41} Thus, by substituting the values of A and B in eq (1), we get I = \int [\frac{\frac{40}{41} (5 \sin x + 4 \cos x) + \frac{9}{41} (5 \cos x - 4 \sin x)}{(5 \sin x + 4 \cos x)}] d x = \frac{40}{41} \int d x + \frac{9}{41} \int (\frac{5 \cos x - 4 \sin x}{5 \sin x + 4 \cos x}) d x Putting 5 \sin x + 4 \cos x = t \Rightarrow (5 \cos x - 4 \sin x) d x = d t ∴ I = \frac{40}{41} x + \frac{9}{41} \int \frac{1}{t} d t = \frac{40}{41} x + \frac{9}{41} \ln |t| + C = \frac{40}{41} x + \frac{9}{41} \ln |5 \sin x + 4 \cos x| + C$

Page No 18.133:

Question 1:

$\int x \cos x d x$

Answer:

$\int x \cos x d x Taking x as the first function and \cos x as the second function . = x \int \cos x d x - \int \{\frac{d}{d x} (x) \int \cos x d x\} d x = x \sin x - \int \sin x d x = x \sin x + \cos x + C$

Page No 18.133:

Question 2:

$\int \log (x + 1) d x$

Answer:

$\int \log (x + 1) d x = \int 1 . \log (x + 1) d x Taking \log (x + 1) as the first function and 1 as the second function . = \log (x + 1) \int 1 d x - \int [\frac{d}{d x} \{\log (x + 1)\} \int 1 d x] d x = x \log (x + 1) - \int \frac{x}{x + 1} d x = x \log (x + 1) - \int \frac{x + 1}{x + 1} - \frac{1}{x + 1} d x = x \log (x + 1) - x + \log |x + 1| + C$

Page No 18.133:

Question 3:

$\int x^{3} \log x d x$

Answer:

$\int x^{3} \log x d x Taking \log x as the first function and x^{3} as the second function . = \log x \int x^{3} d x - \int (\frac{d}{d x} \log x \int x^{3} d x) d x = (\log x) \frac{x^{4}}{4} - \int \frac{1}{x} (\frac{x^{4}}{4}) d x = \frac{x^{4}}{x} \log x - \frac{x^{4}}{16} + C$

Page No 18.133:

Question 4:

$\int x e^{x} d x$

Answer:

$\int x e^{x} d x Taking x as the first function and e^{x} as the second function . = x \int e^{x} d x - \int \{\frac{d}{d x} (x) \int e^{x} d x\} d x = x e^{x} - \int 1 (e^{x}) d x = x e^{x} - e^{x} + C = (x - 1) e^{x} + C$

Page No 18.133:

Question 5:

$\int x e^{2 x} d x$

Answer:

$\int x e^{2 x} d x Taking x as the first function and e^{2 x} as the second function . = x \int e^{2 x} d x - \int \{\frac{d}{d x} (x) \int e^{2 x} d x\} d x = \frac{x e^{2 x}}{2} - \int (\frac{e^{2 x}}{2}) d x = \frac{x}{2} e^{2 x} - \frac{e^{2 x}}{4} + C = e^{2 x} (\frac{x}{2} - \frac{1}{4}) + C$

Page No 18.133:

Question 6:

$\int x^{2} e^{- x} d x$

Answer:

$\int x^{2} e^{- x} d x Taking x^{2} as the first function and e^{- x} as the second function . = x^{2} \int e^{- x} d x - \int (\frac{d}{d x} x^{2} \int e^{- x} d x) d x = - x^{2} e^{- x} - \int 2 x (e^{- x}) (- 1) d x = - x^{2} e^{- x} + 2 \int x e^{- x} d x = - x^{2} e^{- x} + 2 [- x e^{- x} + \int e^{- x} d x] = - x^{2} e^{- x} + 2 [- x e^{- x} - e^{- x}] + C = - e^{- x} [x^{2} + 2 x + 2] + C$

Page No 18.133:

Question 7:

$\int x^{2} \cos x d x$

Answer:

$\int x^{2} \cos x d x Taking x^{2} as the first function and \cos x as the second function . = x^{2} \int \cos x d x - \int (\frac{d}{d x} x^{2} \int \cos x d x) d x = x^{2} \sin x - \int 2 x \sin x d x = x^{2} \sin x - 2 [x \int \sin x - \int \{\frac{d}{d x} (x) \int \sin x d x\} d x] = x^{2} \sin x - 2 [- x \cos x + \int \cos x d x] = x^{2} \sin x + 2 x \cos x - 2 \sin x + C$

Page No 18.133:

Question 8:

$\int x^{2} \cos 2 x d x$

Answer:

$\int x^{2} \cos 2 x d x Taking x^{2} as the first function and \cos 2 x as the second function . = x^{2} \int \cos 2 x d x - \int (2 x \int \cos 2 x d x) d x = \frac{x^{2} \sin 2 x}{2} - \int \frac{2 x \sin 2 x}{2} d x = \frac{x^{2}}{2} \sin 2 x - \int x \sin 2 x d x = \frac{x^{2}}{2} \sin 2 x - [x \int \sin 2 x - \int (\int \sin 2 x d x) d x] = \frac{x^{2}}{2} \sin 2 x - [\frac{- x \cos 2 x}{2} + \int \frac{\cos 2 x}{2} d x] = \frac{x^{2}}{2} \sin 2 x + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4} + C$

Page No 18.133:

Question 9:

$\int x \sin 2 x d x$

Answer:

$\int x \sin 2 x d x Taking x as the first function and \sin 2 x as the second function . = x \int \sin 2 x d x - \int \{\frac{d}{d x} (x) \int \sin 2 x d x\} d x = \frac{- x \cos 2 x}{2} + \int \frac{\cos 2 x}{2} d x = \frac{- x \cos 2 x}{2} + \frac{\sin 2 x}{4} + C$

Page No 18.133:

Question 10:

$\int \frac{\log (\log x)}{x} d x$

Answer:

$\int \frac{\log (\log x)}{x} d x Taking \log \log x as the first function and \frac{1}{x} as the second function . = \log \log x \int \frac{1}{x} d x - \int \{\frac{d}{d x} \log (\log x) \int \frac{1}{x} d x\} d x = \log x . \log (\log x) - \int \frac{1}{x \log x} (\log x) d x = \log x . \log (\log x) - \int \frac{1}{x} d x = \log x . \log (\log x) - \log x + C$

$= \log x [\log (\log x) - 1] + C$

Page No 18.133:

Question 11:

$\int x^{2} \cos x d x$

Answer:

$\int x^{2} \cos x d x Taking x^{2} as the first function and \cos x as the second function . = x^{2} \int \cos x d x - \int \{\frac{d}{d x} (x^{2}) \int \cos x d x\} d x = x^{2} \sin x - \int 2 x \sin x d x = x^{2} \sin x - 2 [x \int \sin x - \int \{\frac{d}{d x} (x) \int \sin x d x\} d x] = x^{2} \sin x + 2 x \cos x - 2 \int \cos x d x = x^{2} \sin x + 2 x \cos x - 2 \sin x + C$

Page No 18.133:

Question 12:

$\int x {cosec}^{2} x d x$

Answer:

$\int x {cosec}^{2} x d x Taking x as the first function and {cosec}^{2} x as the second function . = x \int {cosec}^{2} x d x - \int \{\frac{d}{d x} (x) \int {cosec}^{2} x d x\} d x = - x \cot x + \int \cot x d x = - x \cot x + \log |\sin x| + c$

Page No 18.133:

Question 13:

$\int x \cos^{2} x d x$

Answer:

$\int x \cos^{2} x d x Taking x as the first function and \cos^{2} x as the second function . = x \int \frac{1 + \cos 2 x}{2} d x - \int \{\frac{d}{d x} (x) \int \frac{1 + \cos 2 x}{2} d x\} d x = \frac{x}{2} [x + \frac{\sin 2 x}{2}] - \int \frac{1}{2} (x + \frac{\sin 2 x}{2}) d x = \frac{x}{2} [x + \frac{\sin 2 x}{2}] - [\frac{x^{2}}{4} - \frac{\cos 2 x}{8}] + C = \frac{x^{2}}{2} + \frac{x \sin 2 x}{2} - \frac{x^{2}}{4} + \frac{\cos 2 x}{8} + C = \frac{x^{2}}{4} + \frac{x \sin 2 x}{2} + \frac{\cos 2 x}{8} + C$

Page No 18.133:

Question 14:

$\int x^{n} \cdot \log x d x$

Answer:

$\int x^{n} \log x d x Taking \log x as the first function and x^{n} as the second function . = \log x \int x^{n} d x - \int (\frac{d}{d x} \log x \int x^{n} d x) d x = \log x (\frac{x^{n + 1}}{n + 1}) - \int \frac{1}{x} (\frac{x^{n + 1}}{n + 1}) d x = \log x (\frac{x^{n + 1}}{n + 1}) - \int \frac{x^{n}}{n + 1} d x = \log x (\frac{x^{n + 1}}{n + 1}) - \frac{x^{n + 1}}{{(n + 1)}^{2}} + C$

Page No 18.133:

Question 15:

$\int \frac{\log x}{x^{n}} d x$

Answer:

$\int \frac{1}{x^{n}} \log x d x Taking \log x as the first function and \frac{1}{x^{n}} as the second function . = \log x \int \frac{1}{x^{n}} d x - \int (\frac{d}{d x} \log x \int \frac{1}{x^{n}} d x) d x = \log x (\frac{x^{- n + 1}}{- n + 1}) - \int \frac{1}{x} (\frac{x^{- n + 1}}{- n + 1}) d x = \log x (\frac{x^{- n + 1}}{- n + 1}) - \int \frac{x^{- n}}{- n + 1} d x = \log x (\frac{x^{- n + 1}}{- n + 1}) - \frac{x^{- n + 1}}{{(- n + 1)}^{2}} + C = \log x (\frac{x^{1 - n}}{1 - n}) - \frac{x^{1 - n}}{{(1 - n)}^{2}} + C$

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Question 16:

$\int x^{2} \sin^{2} x d x$

Answer:

$\int x^{2} \sin^{2} x d x Taking x^{2} as the first function and \sin^{2} x as the second function . = x^{2} \int \frac{1 - \cos 2 x}{2} - \int \{\frac{d}{d x} (x^{2}) \int \frac{1 - \cos 2 x}{2} d x\} d x = \frac{x^{2}}{2} [x - \frac{\sin 2 x}{2}] - \int 2 x (\frac{x - \frac{\sin 2 x}{2}}{2}) d x = \frac{x^{2}}{2} (x - \frac{\sin 2 x}{2}) - \int x^{2} d x + \int \frac{x \sin 2 x}{2} d x [Here, taking x as the first function and \sin 2 x as the second function .] = \frac{x^{3}}{2} - \frac{x^{2} \sin 2 x}{4} - \frac{x^{3}}{3} + \frac{1}{2} [x \int \sin 2 x - \int \{\frac{d}{d x} (x) \int \sin 2 x d x\} d x] = \frac{x^{3}}{2} - \frac{x^{2} \sin 2 x}{4} - \frac{x^{3}}{3} + \frac{1}{2} [\frac{- x \cos 2 x}{2} + \int \frac{\cos 2 x d x}{4}] = \frac{x^{3}}{6} - \frac{x^{2} \sin 2 x}{4} - \frac{x \cos 2 x}{4} + \frac{\sin 2 x}{8} + C$

Page No 18.133:

Question 17:

$\int 2 x^{3} e^{x^{2}} d x$

Answer:

$\int 2 x^{3} \cdot e^{x^{2}} d x = \int x^{2} \cdot (e^{x^{2}}) \cdot 2 x d x L et x^{2} = t \Rightarrow 2 x d x = d t = \int \underset{I}{t} \cdot {\underset{II}{e}}^{t} d t = t \cdot e^{t} - \int 1 \cdot e^{t} d t = t e^{t} - e^{t} + C = x^{2} e^{x^{2}} - e^{x^{2}} + C = e^{x^{2}} (x^{2} - 1) + C$

Page No 18.133:

Question 18:

$\int x^{3} \cos x^{2} d x$

Answer:

$\int x^{3} \cos x^{2} d x Let x^{2} = t \Rightarrow 2 x = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{2 x} = \frac{1}{2} [\int t \cos t d t] Taking t as the first function and \cos t as the second function . = \frac{1}{2} [t \sin t - \int \sin t d t] = \frac{1}{2} [t \sin t + \cos t] . . . (1) Substituting the value of t in eq (1) = \frac{x^{2} \sin x^{2}}{2} + \frac{\cos x^{2}}{2} + c$

Page No 18.133:

Question 19:

$\int x \sin x \cos x d x$

Answer:

$\int x \sin x \cdot \cos x d x = \frac{1}{2} \int x (2 \sin x \cos x) d x = \frac{1}{2} \int \underset{}{x} \cdot \underset{}{\sin (2 x)} d x Taking x as the first function and \sin 2 x as the second function . = \frac{1}{2} [x \int \sin 2 x dx - \int \{\frac{d}{d x} (x) \int \sin 2 x d x\} d x] = \frac{1}{2} [x \times \frac{- \cos (2 x)}{2} - \int 1 \cdot (\frac{- \cos 2 x}{2}) d x] = \frac{1}{2} [\frac{- x \cos (2 x)}{2} + \frac{\sin (2 x)}{4}] + C = \frac{- x \cos (2 x)}{4} + \frac{\sin (2 x)}{8} + C$

Page No 18.133:

Question 20:

$\int \sin x \log (\cos x) d x$

Answer:

$Let I = \int \sin x \cdot \log (\cos x) d x Let \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = - \int \log t d t = - \int \underset{}{1} \cdot \underset{}{\log t} d t Taking \log t as the first function and 1 as the second function . = \log t \int 1 d t - \int \{\frac{d}{d t} (\log t) \int 1 d t\} d t = - [\log t \cdot t - \int \frac{1}{t} \times t d t] = - [\log t \cdot t - t] + C = - t (\log t - 1) + C . . . . (1) Substituting the value of t in eq (1) = - \cos x \{\log (\cos x) - 1\} + C = \cos x \{1 - \log (\cos x)\} + C$

Page No 18.133:

Question 21:

$\int {(\log x)}^{2} \cdot x d x$

Answer:

$\int {\underset{}{(\log x)}}^{2} \underset{}{x \cdot} d x Taking {(\log x)}^{2} as the first function and x as the second function . = {(\log x)}^{2} \int x d x - \int \{\frac{d}{d x} {(\log x)}^{2} \int x d x\} d x = {(\log x)}^{2} \cdot \frac{x^{2}}{2} - \int \frac{(2 \log x)}{x} \times \frac{x^{2}}{2} d x = {(\log x)}^{2} \times \frac{x^{2}}{2} - \int \underset{II}{x} \underset{I}{\log x} d x = {(\log x)}^{2} \times \frac{x^{2}}{2} - [\log x \int x d x - \int \{\frac{d}{d x} (\log x) \int x d x\} d x] = {(\log x)}^{2} \times \frac{x^{2}}{2} - [\log x \cdot \frac{x^{2}}{2} - \int \frac{1}{x} \times \frac{x^{2}}{2} d x] = {(\log x)}^{2} \times \frac{x^{2}}{2} - \log x \cdot \frac{x^{2}}{2} + \frac{x^{2}}{4} + C = \frac{x^{2}}{2} [{(\log x)}^{2} - \log x + \frac{1}{2}] + C$

Page No 18.133:

Question 22:

$\int e^{\sqrt{x}} d x$

Answer:

$Let I = \int e^{\sqrt{x}} d x = \int \sqrt{x} \cdot \frac{e^{\sqrt{x}}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t ∴ I = 2 \int \underset{}{t} \cdot \underset{}{e^{t}} d t Taking t as the first function and e^{t} as the second function . = 2 [t \int e^{t} d t - \int \{\frac{d}{d t} (t) \int e^{t} d t\} d t] = 2 [t \cdot e^{t} - \int 1 \cdot e^{t} d t] + C . . . (1) Substituting the value of t in eq (1) = 2 [\sqrt{x} e^{\sqrt{x}} - e^{\sqrt{x}}] + C = 2 e^{\sqrt{x}} (\sqrt{x} - 1) + C$

Page No 18.133:

Question 23:

$\int \frac{\log (x + 2)}{{(x + 2)}^{2}} d x$

Answer:

$Let I = \int \frac{\log (x + 2) d x}{{(x + 2)}^{2}} Let \log (x + 2) = t \Rightarrow x + 2 = e^{t} \Rightarrow \frac{1}{(x + 2)} d x = d t ∴ I = \int \frac{t}{e^{t}} d t = \int t e^{- t} d t Taking t as the first function and e^{- t} as the second function . = t \int e^{- t} - \int \{\frac{d}{d t} (t) \int e^{- 2 t} d t\} d t = t \times \frac{e^{- t}}{- 1} - \int 1 \cdot e^{- t} d t = - t e^{- t} + \frac{e^{- t}}{- 1} + C = - e^{- t} (t + 1) + C = - \frac{(t + 1)}{e^{t}} + C . . . (1) Substituting the value of t in eq (1) = \frac{- [\log (x + 2) + 1]}{x + 2} + C = - \frac{\log (x + 2)}{x + 2} - \frac{1}{(x + 2)} + C$

Page No 18.133:

Question 24:

$\int \frac{x + \sin x}{1 + \cos x} d x$

Answer:

$\int (\frac{x + \sin x}{1 + \cos x}) d x = \int [\frac{x}{1 + \cos x} + \frac{\sin x}{1 + \cos x}] d x = \int [\frac{x}{2 \cos^{2} \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}] d x = \frac{1}{2} \int \underset{I}{x} \cdot \underset{II}{\sec^{2}} \frac{x}{2} d x + \int \tan \frac{x}{2} d x = \frac{1}{2} [x \cdot \frac{\tan (\frac{x}{2})}{\frac{1}{2}} - \int 1 \times 2 \tan (\frac{x}{2}) d x] + \frac{\log |s e c \frac{x}{2}|}{\frac{1}{2}} + C = x \tan (\frac{x}{2}) - \frac{\log |\sec \frac{x}{2}|}{\frac{1}{2}} + \log \frac{|\sec \frac{x}{2}|}{\frac{1}{2}} + C = x \tan (\frac{x}{2}) + C$

Page No 18.133:

Question 25:

$\int \log_{10} x d x$

Answer:

$\int \log_{10} x d x = \int \frac{\log x}{\log 10} d x = \frac{1}{\log 10} \int \underset{}{1} \cdot \underset{}{\log x} d x Taking \log x as the first function and 1 as the second function = \frac{1}{\log 10} [\log x \int 1 d x - \int \{\frac{d}{d x} (\log x) \int 1 d x\} d x] = \frac{1}{\log 10} [\log x \cdot x - \int \frac{1}{x} \cdot x d x] = \frac{1}{\log 10} [x \log x - x] + C = \frac{1}{\log 10} [x (\log x - 1)] + C$

Page No 18.133:

Question 26:

$\int \cos \sqrt{x} d x$

Answer:

$Let I = \int \cos \sqrt{x} d x = \int \frac{\sqrt{x} \cdot \cos \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t ∴ I = 2 \int \underset{}{t} \cdot \underset{}{\cos (t)} \cdot d t Taking t as the first function and \cos t as the second function . = 2 [t \cdot \sin t - \int 1 \cdot \sin t d t] = 2 [t \cdot \sin t + \cos t] + C . . . . (1) Substituting the value of t in eq (1) = 2 [\sqrt{x} \cdot \sin \sqrt{x} + \cos \sqrt{x}] + C$

Page No 18.133:

Question 27:

Evaluate the following integrals:

$\int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x$

Answer:

$Let I = \int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x Let the first function be \cos^{- 1} x and second function be \frac{x}{\sqrt{1 - x^{2}}} . First we find the integral of the second function, i . e ., \int \frac{x}{\sqrt{1 - x^{2}}} d x . Put t = 1 - x^{2} . Then d t = - 2 x d x Therefore, \int \frac{x}{\sqrt{1 - x^{2}}} d x = - \frac{1}{2} \int \frac{1}{\sqrt{t}} d t = - \sqrt{t} = - \sqrt{1 - x^{2}} Hence, using integration by parts, we get \int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x = (\cos^{- 1} x) \int \frac{x}{\sqrt{1 - x^{2}}} d x - \int [(\frac{d (\cos^{- 1} x)}{d x}) \int (\frac{x}{\sqrt{1 - x^{2}}} d x)] d x = (\cos^{- 1} x) (- \sqrt{1 - x^{2}}) - \int (\frac{- 1}{\sqrt{1 - x^{2}}}) (- \sqrt{1 - x^{2}}) d x = - \sqrt{1 - x^{2}} \cos^{- 1} x - x + c Hence, \int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x = - \sqrt{1 - x^{2}} \cos^{- 1} x - x + c$

Page No 18.133:

Question 28:

Evaluate the following integrals:

$\int \frac{\log x}{{(x + 1)}^{2}} d x$

Answer:

$Let I = \int \frac{\log x}{{(x + 1)}^{2}} d x Let the first function be (\log x) and second function be \frac{1}{{(x + 1)}^{2}} . First we find the integral of the second function, i . e ., \int \frac{1}{{(x + 1)}^{2}} d x . Put t = (x + 1) . Then d t = d x Therefore, \int \frac{1}{{(x + 1)}^{2}} d x = \int t^{- 2} d t = - \frac{1}{t} = - \frac{1}{1 + x} Hence, using integration by parts, we get \int \frac{\log x}{{(x + 1)}^{2}} d x = (\log x) \int \frac{1}{{(x + 1)}^{2}} d x - \int [(\frac{d (\log x)}{d x}) \int \frac{1}{{(x + 1)}^{2}} d x] d x = (\log x) (- \frac{1}{1 + x}) - \int (\frac{1}{x}) (- \frac{1}{1 + x}) d x = - \frac{\log x}{1 + x} + \int (\frac{1}{x^{2} + x}) d x = - \frac{\log x}{1 + x} + \int \frac{1}{x^{2} + x + \frac{1}{4} - \frac{1}{4}} d x = - \frac{\log x}{1 + x} + \int \frac{1}{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d x = - \frac{\log x}{1 + x} + \frac{1}{2 \times \frac{1}{2}} \log |\frac{x + \frac{1}{2} - \frac{1}{2}}{x + \frac{1}{2} + \frac{1}{2}}| + c = - \frac{\log x}{1 + x} + \log |\frac{x}{x + 1}| + c Hence, \int \frac{\log x}{{(x + 1)}^{2}} d x = - \frac{\log x}{1 + x} + \log |\frac{x}{x + 1}| + c$

Page No 18.133:

Question 29:

$\int {cosec}^{3} x d x$

Answer:

$Let I = \int {cosec}^{3} x d x = \int {cosec}^{2} x \cdot cosec x d x = \int {cosec}^{2} x \cdot \sqrt{1 + \cot^{2} x} d x Let \cot x = t \Rightarrow - {cosec}^{2} x d x = d t ∴ I = - \int \sqrt{1 + t^{2}} d t = - \frac{t}{2} \sqrt{1 + t^{2}} - \frac{1^{2}}{2} \log |t + \sqrt{1 + t^{2}}| + C . . . (1) Substituting the value of t in eq (1) = - \frac{\cot x}{2} \cdot cosec x - \frac{1}{2} \log |\cot x + cosec x| + C = - \frac{1}{2} cosec x \cot x - \frac{1}{2} \log |\frac{\cos x}{\sin x} + \frac{1}{\sin x}| + C = - \frac{1}{2} cosec x \cot x - \frac{1}{2} \log |\frac{2 \cos^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}| + C = - \frac{1}{2} cosec x \cot x - \frac{1}{2} \log |\cot \frac{x}{2}| + C = - \frac{1}{2} cosec x \cot x + \frac{1}{2} \log |\tan \frac{x}{2}| + C (∵ \log |\cot \frac{x}{2}| = \log |\frac{1}{\tan \frac{x}{2}}| \Rightarrow - \log |\tan \frac{x}{2}|)$

Page No 18.133:

Question 30:

$\int \sec^{- 1} \sqrt{x} d x$

Answer:

$\int \underset{II}{1} . \underset{I}{\sec^{- 1} \sqrt{x}} d x = \underset{}{\sec^{- 1} \sqrt{x}} \int 1 d x - \int \{\frac{d}{d x} (\sec^{- 1} \sqrt{x}) \int 1 d x\} d x = \sec^{- 1} \sqrt{x} . x - \int \frac{1}{\sqrt{x} \sqrt{1 - x}} \times \frac{1}{2 \sqrt{x}} \times x d x = x \sec^{- 1} \sqrt{x} - \frac{1}{2} \int {(1 - x)}^{- \frac{1}{2}} d x = x \sec^{- 1} x - \frac{1}{2} [\frac{{(1 - x)}^{- \frac{1}{2} + 1}}{(- \frac{1}{2} + 1) (- 1)}] + C = x \sec^{- 1} x + {(1 - x)}^{\frac{1}{2}} + C$

Page No 18.134:

Question 31:

$\int \sin^{- 1} \sqrt{x} d x$

Answer:

$Let I = \int \sin^{- 1} \sqrt{x} d x = \int \frac{\sqrt{x} . \sin^{- 1} \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t ∴ I = \int \underset{II}{t} . \underset{I}{\sin^{- 1} t} d t = \sin^{- 1} t \int t d t - \int \{\frac{d}{d t} (\sin^{- 1} t) \int t d t\} d t = 2 [\sin^{- 1} t . \frac{t^{2}}{2} - \int \frac{1}{\sqrt{1 - t^{2}}} \times \frac{t^{2}}{2} d t] = \sin^{- 1} t . t^{2} - \int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t = \sin^{- 1} t . t^{2} + \int (\frac{1 - t^{2} - 1}{\sqrt{1 - t^{2}}}) d t = \sin^{- 1} t . t^{2} + \int \sqrt{1 - t^{2}} d t - \int \frac{d t}{\sqrt{1 - t^{2}}} = \sin^{- 1} t . t^{2} + \frac{t}{2} \sqrt{1 - t^{2}} + \frac{1}{2} \sin^{- 1} t - \sin^{- 1} t + C = \sin^{- 1} t . t^{2} + \frac{t}{2} \sqrt{1 - t^{2}} - \frac{1}{2} \sin^{- 1} t + C = x . \sin^{- 1} \sqrt{x} + \frac{\sqrt{x}}{2} \sqrt{1 - x} - \frac{1}{2} \sin^{- 1} (\sqrt{x}) + C (∵ \sqrt{x} = t) = \frac{(2 x - 1) \sin^{- 1} \sqrt{x}}{2} + \frac{\sqrt{x - x^{2}}}{2} + C$

Page No 18.134:

Question 32:

$\int x \tan^{2} x d x$

Answer:

$\int$ x tan² x dx
= ∫ x (sec² x – 1) dx
$= \int \underset{I}{x} . \underset{II}{\sec^{2} x} d x - \int x d x = x \int \sec^{2} x - \int \{\frac{d}{d x} (x) \int \sec^{2} x d x\} d x - \frac{x^{2}}{2} + C_{1} = x . \tan x - \int 1 . \tan x d x - \frac{x^{2}}{2} + C_{1} = x \tan x - \log |\sec x| - \frac{x^{2}}{2} + C_{1} + C_{2} = x \tan x - \log |\sec x| - \frac{x^{2}}{2} + C (where C = C_{1} + C_{2})$

Page No 18.134:

Question 33:

$\int x (\frac{\sec 2 x - 1}{\sec 2 x + 1}) d x$

Answer:

$\int x (\frac{\sec 2 x - 1}{\sec 2 x + 1}) d x = \int x (\frac{\frac{1}{\cos 2 x} - 1}{\frac{1}{\cos 2 x} + 1}) d x = \int x (\frac{1 - \cos 2 x}{1 + \cos 2 x}) d x = \int x (\frac{2 \sin^{2} x}{2 \cos^{2} x}) d x [∵ (1 - \cos 2 x) = 2 \sin^{2} x and (1 + \cos 2 x) = 2 \cos^{2} x] = \int x . \tan^{2} x d x = \int x . (\sec^{2} x - 1) d x = \int \underset{I}{x} . \underset{II}{\sec^{2} x} d x - \int x d x = x \int \sec^{2} x d x - \int \{\frac{d}{d x} (x) \int \sec^{2} x d x\} d x - \frac{x^{2}}{2} + C_{1} = x \tan x - \int 1 . \tan x d x - \frac{x^{2}}{2} + C_{1} = x \tan x - \log |\sec x| - \frac{x^{2}}{2} + C_{2} + C_{1} = x \tan x - \log |\sec x| - \frac{x^{2}}{2} + C (where C = C_{1} + C_{2})$

Page No 18.134:

Question 34:

$\int (x + 1) e^{x} \log (x e^{x}) d x$

Answer:

$\int (x + 1) e^{x} . \log (x e^{x}) d x Let x e^{x} = t \Rightarrow (x . e^{x} + 1 . e^{x}) d x = d t ∴ \int (x + 1) e^{x} . \log (x e^{x}) d x = \int \underset{II}{1} . \underset{I}{\log (t)} d t = \log t \int 1 d t - \int \{\frac{d}{d t} (\log t) - \int 1 d t\} d t = \log (t) \times t - \int \frac{1}{t} \times t d t = t \log (t) - t + C . . . (1) Substituting the value of t in eq (1) \Rightarrow \int (x + 1) e^{x} . \log (x e^{x}) d x = (x e^{x}) . \log (x e^{x}) - x e^{x} + C = x e^{x} \{\log (x e^{x}) - 1\} + C$

Page No 18.134:

Question 35:

$\int \sin^{- 1} (3 x - 4 x^{3}) d x$

Answer:

$\int$ sin^–1 (3x – 4x³)dx
Let x = sin θ
⇒ dx = cos θ.dθ
& θ = sin^–1 x
$\int$ sin^–1 (3x – 4x³)dx = $\int$ sin^–1 (3 sin θ – 4 sin³ θ) . cos θ dθ
= ∫ sin^–1 (sin 3θ) . cos θ dθ
$= 3 \int \underset{I}{θ} . \underset{II}{\cos θ} d θ = 3 [θ \int \cos θ d θ - \int \{\frac{d}{d θ} (θ) - \int \cos θ d θ\} d θ] = 3 [θ . \sin θ - \int 1 . \sin θ d θ] = 3 [θ . \sin θ + \cos θ] + C = 3 [θ . \sin θ + \sqrt{1 - \sin^{2} θ}] + C = 3 [(\sin^{- 1} x) . x + \sqrt{1 - x^{2}}] + C (∵ θ = \sin^{- 1} x)$

Page No 18.134:

Question 36:

$\int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) d x$

Answer:

$\int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) d x Let x = \tan θ d x = \sec^{2} θ d θ ∴ \int \sin^{- 1} (\frac{2 x}{1 + x^{2}}) d x = \int \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}) . \sec^{2} θ d θ = \int \sin^{- 1} (\sin 2 θ) . \sec^{2} θ d θ = \int (2 θ) \sec^{2} θ d θ = 2 \int \underset{I}{θ} \underset{II}{\sec^{2} θ} d θ = 2 [θ \int \sec^{2} θ d θ - \int \{\frac{d}{d θ} (θ) \int s {ec}^{2} θ d θ\} d θ] = 2 [θ . \tan θ - \int 1 . \tan θ d θ] = 2 [θ \tan θ - \log |\sec θ|] + C = 2 [θ \tan θ - \log {|1 + \tan^{2} θ|}^{\frac{1}{2}}] + C = 2 [(\tan^{- 1} x) \times x - \log {(1 + x^{2})}^{\frac{1}{2}}] + C = 2 x \tan^{- 1} x - 2 \times \frac{1}{2} \log |1 + x^{2}| + C = 2 x \tan^{- 1} x - \log |1 + x^{2}| + C$

Page No 18.134:

Question 37:

$\int \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) d x$

Answer:

$Let I = \int \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) d x = \int 3 \tan^{- 1} (x) d x = 3 \int [\tan^{- 1} (x) \times 1] d x = 3 [\tan^{- 1} x \times x - \int \frac{1}{1 + x^{2}} \times x d x] = 3 x \tan^{- 1} x - 3 \int \frac{x}{1 + x^{2}} d x let 1 + x^{2} = t \Rightarrow 2 x d x = d t Then, I = 3 x \tan^{- 1} x - \frac{3}{2} \int \frac{d t}{t} = 3 x \tan^{- 1} x - \frac{3}{2} \log |t| + C = 3 x \tan^{- 1} x - \frac{3}{2} \log |1 + x^{2}| + C$

Page No 18.134:

Question 38:

$\int x^{2} \sin^{- 1} x d x$

Answer:

$\int \underset{II}{x^{2}} . \underset{I}{\sin^{- 1}} x d x = \underset{}{\sin^{- 1}} x \int x^{2} d x - \int \{\frac{d}{d x} (\underset{}{\sin^{- 1}} x) \int x^{2} d x\} d x = \sin^{- 1} x . \frac{x^{3}}{3} - \int \frac{1}{\sqrt{1 - x^{2}}} \frac{x^{3}}{3} d x Let 1 - x^{2} = t \Rightarrow x^{2} = 1 - t \Rightarrow - 2 x d x = d t \Rightarrow x d x = - \frac{d t}{2} ∴ \int \underset{}{x^{2}} . \underset{}{\sin^{- 1}} x d x = \sin^{- 1} x . \frac{x^{3}}{3} - \frac{1}{3} \int \frac{x^{2} . x}{\sqrt{1 - x^{2}}} d x = \sin^{- 1} x . \frac{x^{3}}{3} - \frac{1}{6} \int \frac{(1 - t)}{\sqrt{t}} d t = \sin^{- 1} x . \frac{x^{3}}{3} + \frac{1}{6} \int t^{- \frac{1}{2}} d t - \frac{1}{6} \int t^{\frac{1}{2}} d t = \sin^{- 1} x . \frac{x^{3}}{3} + \frac{1}{6} \times 2 \sqrt{t} - \frac{1}{6} \times \frac{2}{3} t^{\frac{3}{2}} + C = \sin^{- 1} x . \frac{x^{3}}{3} + \frac{\sqrt{1 - x^{2}}}{3} - \frac{1}{9} {(1 - x^{2})}^{\frac{3}{2}} + C (∵ 1 - x^{2} = t)$

Page No 18.134:

Question 39:

$\int \frac{\sin^{- 1} x}{x^{2}} d x$

Answer:

$Let I = \int \frac{\sin^{- 1} x}{x^{2}} d x Putting x = \sin θ \Rightarrow θ = \sin^{- 1} x & d x = \cos θ d θ ∴ I = \int \frac{θ . \cos θ}{\sin^{2} θ} d θ = \int θ . (\frac{\cos θ}{\sin θ}) \times \frac{1}{\sin θ} d θ = \int \underset{I}{θ} . \underset{II}{cosec θ} \cot θ d θ = θ \int cosec θ \cot θ d θ - \int \{\frac{d}{d θ} (θ) \int cosec θ \cot θ d θ\} d θ = θ (- cosec θ) - \int 1 . (- cosec θ) d θ = - θ cosec θ + \int cosec θ d θ = - θ cosec θ + \ln |cosec θ - \cot θ| + C = \frac{- θ}{\sin θ} + \ln |\frac{1 - \cos θ}{\sin θ}| + C = \frac{- θ}{\sin θ} + \ln |\frac{1 - \sqrt{1 - \sin^{2} θ}}{\sin θ}| + C = \frac{- \sin^{- 1} x}{x} + \ln |\frac{1 - \sqrt{1 - x^{2}}}{x}| + C [∵ θ = \sin^{- 1} x]$

Page No 18.134:

Question 40:

$\int \frac{x^{2} \tan^{- 1} x}{1 + x^{2}} d x$

Answer:

$Let I = \int (\frac{x^{2} \tan^{- 1} x}{1 + x^{2}}) d x = \int (\frac{x^{2} + 1 - 1}{x^{2} + 1}) \tan^{- 1} x d x = \int (1 - \frac{1}{x^{2} + 1}) \tan^{- 1} x d x = \int \underset{II}{1} . \underset{I}{\tan^{- 1}} x d x - \int \frac{\tan^{- 1} x}{x^{2} + 1} d x = [\tan^{- 1} x \int 1 d x - \int \{\frac{d}{d x} (\tan^{- 1} x) \int 1 d x\} d x] - \int \frac{\tan^{- 1} x}{x^{2} + 1} d x = t [{an}^{- 1} x \times x - \int \frac{x}{1 + x^{2}} d x] - \int \frac{\tan^{- 1} x}{x^{2} + 1} d x Putting x^{2} + 1 = t in the first integral and \tan^{- 1} x = p in the second integral \Rightarrow 2 x d x = d t and \frac{1}{1 + x^{2}} d x = d p \Rightarrow x d x = \frac{d t}{2} and \frac{1}{1 + x^{2}} d x = d p ∴ I = \tan^{- 1} x . x - \frac{1}{2} \int \frac{d t}{t} - \int p . d p = x \tan^{- 1} x - \frac{1}{2} \ln |t| - \frac{p^{2}}{2} + C = x \tan^{- 1} x - \frac{1}{2} \ln |1 + x^{2}| - \frac{{(\tan^{- 1} x)}^{2}}{2} + C [∵ t = x^{2} + 1 and p = \tan^{- 1} x]$

Page No 18.134:

Question 41:

$\int \cos^{- 1} (4 x^{3} - 3 x) d x$

Answer:

$\int \cos^{- 1} (4 x^{3} - 3 x) d x Let x = \cos θ \Rightarrow θ = \cos^{- 1} x & d x = - \sin θ d θ ∴ \int \cos^{- 1} (4 x^{3} - 3 x) d x = \int \cos^{- 1} (4 \cos^{3} θ - 3 \cos θ) . (- \sin θ) d θ = \int \cos^{- 1} (\cos 3 θ) . (- \sin θ) d θ (∵ \cos 3 θ = 4 \cos^{3} θ - 3 \cos θ) = - 3 \int \underset{I}{θ} \underset{II}{\sin θ} d θ = θ \int \sin θ d θ - \int \{\frac{d}{d θ} (θ) \int \sin θ d θ\} d θ = 3 [θ (- \cos θ) - \int 1 . (- \cos θ) d θ] = 3 θ \cos θ - 3 \sin θ + C = 3 \cos^{- 1} x . x - 3 \sqrt{1 - x^{2}} + C (∵ x = \cos θ)$

Page No 18.134:

Question 42:

$\int \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x$

Answer:

$Let I = \int \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x = 2 \int \underset{II}{1} . \underset{I}{\tan^{- 1}} x d x [∵ \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 \tan^{- 1} x] = 2 [\tan^{- 1} x \int 1 d x - \int \{\frac{d}{d x} (\tan^{- 1} x) \int 1 d x\} d x] = 2 [\tan^{- 1} x . x - \int \frac{1}{1 + x^{2}} \times x d x] = 2 x \tan^{- 1} x - \int \frac{2 x}{1 + x^{2}} d x Putting 1 + x^{2} = t \Rightarrow 2 x d x = d t ∴ I = 2 x \tan^{- 1} x - \int \frac{d t}{t} = 2 x \tan^{- 1} x - \ln |t| + C = 2 x \tan^{- 1} x - \ln |1 + x^{2}| + C [∵ t = 1 + x^{2}]$

Page No 18.134:

Question 43:

$\int \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x$

Answer:

$Let I = \int \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x = 2 \int \underset{II}{1} . \underset{I}{\tan^{- 1} x} d x = 2 [\tan^{- 1} x \int 1 d x - \int \{\frac{d}{d x} \{\tan^{- 1} x\} \int 1 d x\} d x] = 2 [\tan^{- 1} x . x - \int \frac{1}{1 + x^{2}} \times x d x] = 2 \tan^{- 1} x . x - \int \frac{2 x}{1 + x^{2}} d x Putting 1 + x^{2} = t \Rightarrow 2 x d x = d t ∴ I = 2 x \tan^{- 1} x - \int \frac{d t}{t} = 2 x \tan^{- 1} x - \ln |t| + C = 2 x \tan^{- 1} x - \ln |1 + x^{2}| + C [∵ t = 1 + x^{2}]$

Page No 18.134:

Question 44:

$\int (x + 1) \log x d x$

Answer:

$\int \underset{II}{(x + 1)} . \underset{I}{\log x} d x = \log x \int (x + 1) d x - \int \{\frac{d}{d x} (\log x) \int (x + 1) d x\} d x = \log x [\frac{x^{2}}{2} + x] - \int \frac{1}{x} (\frac{x^{2}}{2} + x) d x = \log x (\frac{x^{2}}{2} + x) - \int (\frac{x}{2} + 1) d x = \log x (\frac{x^{2}}{2} + x) - (\frac{x^{2}}{4} + x) + C$

Page No 18.134:

Question 45:

$\int x^{2} \tan^{- 1} x d x$

Answer:

$Let I = \int \underset{II}{x^{2}} . \underset{I}{\tan^{- 1} x} d x = \tan^{- 1} x \int x^{2} d x - \int \{\frac{d}{d x} (\tan^{- 1} x) \int x^{2} d x\} d x = \tan^{- 1} x \times \frac{x^{3}}{3} - \int (\frac{1}{1 + x^{2}}) \times \frac{x^{3}}{3} d x = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{1}{3} \int \frac{x^{2} . x}{1 + x^{2}} d x Let 1 + x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{1}{6} \int \frac{(t - 1)}{t} . d t = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{1}{6} \int d t + \frac{1}{6} \int \frac{d t}{t} = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{t}{6} + \frac{1}{6} \log |t| + C = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{(1 + x^{2})}{6} + \frac{1}{6} \log (1 + x^{2}) + C = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{x^{2}}{6} + \frac{1}{6} \log (1 + x^{2}) - \frac{1}{6} + C = \tan^{- 1} x . \frac{x^{3}}{3} - \frac{x^{2}}{6} + \frac{1}{6} \log |1 + x^{2}| + C' where C' = C - \frac{1}{6}$

Page No 18.134:

Question 46:

$\int (e^{\log x} + \sin x) \cos x d x$

Answer:

$\int (e^{\log x} + \sin x) \cos x d x = \int (x + \sin x) \cos x d x (∵ e^{\log x} = x) = \int (x \cos x + \sin x \cos x) d x = \int x \cos x d x + \frac{1}{2} \int 2 \sin x \cos x d x = \int \underset{I}{x} \underset{II}{\cos x} d x + \frac{1}{2} \int \sin 2 x d x = [x \int \cos x d x - \int \{\frac{d}{d x} (x) \int \cos x d x\} d x] + \frac{1}{2} \int \sin 2 x d x = x \sin x - \int 1 . \sin x d x + \frac{1}{2} [\frac{- \cos 2 x}{2}] + C = x \sin x - (- \cos x) - \frac{1}{4} \cos 2 x + C = x \sin x + \cos x - \frac{1}{4} (1 - 2 \sin^{2} x) + C = x \sin x + \cos x + \frac{\sin^{2} x}{2} - \frac{1}{4} + C = x \sin x + \cos x + \frac{\sin^{2} x}{2} + C' where C' = C - \frac{1}{4}$

Page No 18.134:

Question 47:

$\int \frac{(x \tan^{- 1} x)}{{(1 + x^{2})}^{3 / 2}} d x$

Answer:

$Let I = \int \frac{x \tan^{- 1} x}{{(1 + x^{2})}^{\frac{3}{2}}} d x Putting x = \tan θ \Rightarrow d x = \sec^{2} θ d θ & θ = \tan^{- 1} x ∴ I = \int \frac{(\tan θ) . θ . \sec^{2} θ d θ}{{(1 + \tan^{2} θ)}^{\frac{3}{2}}} = \int \frac{θ . \tan θ \sec^{2} θ d θ}{{(\sec^{2} θ)}^{\frac{3}{2}}} = \int \frac{θ \tan θ . \sec^{2} θ d θ}{\sec^{3} θ} = \int \frac{θ . \tan θ}{\sec θ} d θ = \int \underset{I}{θ} . \underset{II}{\sin θ} d θ = θ \int \sin θ d θ - \int \{\frac{d}{d θ} (θ) \int \sin d θ\} d θ = θ (- \cos θ) - \int 1 . (- \cos θ) d θ = - θ \cos θ + \sin θ + C = \frac{- θ}{\sec θ} + \frac{1}{cose c θ} + C = \frac{- θ}{\sqrt{1 + \tan^{2} θ}} + \frac{1}{\sqrt{1 + \cot^{2} θ}} + C = \frac{- θ}{\sqrt{1 + \tan^{2} θ}} + \frac{\tan θ}{\sqrt{\tan^{2} θ + 1}} + C = \frac{- \tan^{- 1} x}{\sqrt{1 + x^{2}}} + \frac{x}{\sqrt{x^{2} + 1}} + C$

Page No 18.134:

Question 48:

$\int \tan^{- 1} (\sqrt{x}) d x$

Answer:

$Let I = \int \tan^{- 1} \sqrt{x} d x = \int \frac{\sqrt{x} . \tan^{- 1} \sqrt{x} d x}{\sqrt{x}} Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t = \frac{d x}{\sqrt{x}} = 2 d t ∴ I = 2 \int \underset{II}{t} . \underset{I}{\tan^{- 1} (t)} d t = 2 [\tan^{- 1} t \int t d t - \int \{\frac{d}{d t} (\tan^{- 1} t) \int t d t\} d t] = 2 [\tan^{- 1} (t) . \frac{t^{2}}{2} - \int \frac{1}{1 + t^{2}} . \frac{t^{2}}{2} d t] = \tan^{- 1} (t) . t^{2} - \int \frac{t^{2}}{1 + t^{2}} d t = \tan^{- 1} (t) . t^{2} - \int (\frac{1 + t^{2} - 1}{1 + t^{2}}) d t = \tan^{- 1} (t) . t^{2} - \int d t + \int \frac{d t}{1 + t^{2}} = \tan^{- 1} (t) . t^{2} - t + \tan^{- 1} (t) + C (∵ \sqrt{x} = t) = \tan^{- 1} (\sqrt{x}) . x - \sqrt{x} + \tan^{- 1} \sqrt{x} + C = (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C$

Page No 18.134:

Question 49:

$\int x^{3} \tan^{- 1} x d x$

Answer:

$\int \underset{II}{x^{3}} . \underset{I}{\tan^{- 1} x} d x = \tan^{- 1} x \int x^{3} d x - \int \{\frac{d}{d x} (\tan^{- 1} x) \int x^{3} d x\} d x = \tan^{- 1} x . \frac{x^{4}}{4} - \int \frac{1}{1 + x^{2}} \times \frac{x^{4}}{4} d x = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int \frac{x^{4} d x}{x^{2} + 1} = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int (\frac{x^{4} - 1 + 1}{x^{2} + 1}) d x = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int (\frac{x^{4} - 1}{x^{2} + 1}) d x - \frac{1}{4} \int \frac{1}{x^{2} + 1} d x = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int \frac{(x^{2} - 1) (x^{2} + 1)}{(x^{2} + 1)} d x - \frac{1}{4} \int \frac{1}{x^{2} + 1} d x = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} \int (x^{2} - 1) d x - \frac{1}{4} \tan^{- 1} x + C = \tan^{- 1} x . \frac{x^{4}}{4} - \frac{1}{4} (\frac{x^{3}}{3} - x) - \frac{1}{4} \tan^{- 1} x + C = (\frac{x^{4} - 1}{4}) \tan^{- 1} x - \frac{1}{12} (x^{3} - 3 x) + C$

Page No 18.134:

Question 50:

$\int x \sin x \cos 2 x d x$

Answer:

$\int x . \cos 2 x \sin x d x = \frac{1}{2} \int x (2 \cos 2 x \sin x) d x [∵ 2 \cos A \sin B = \sin (A + B) - \sin (A - B)] = \frac{1}{2} \int x (\sin 3 x - \sin x) d x = \frac{1}{2} \int x . \sin 3 x d x - \frac{1}{2} \int x \sin x d x = \frac{1}{2} [x \int \sin 3 x d x - \int \{\frac{d}{d x} (x) \int \sin 3 x d x\} d x] - \frac{1}{2} [x \int \sin x d x - \int \{\frac{d}{d x} (x) \int \sin x d x\} d x] = \frac{1}{2} [x . (\frac{- \cos 3 x}{3}) - \int 1 . (\frac{- \cos 3 x}{3}) d x] - \frac{1}{2} [x . (- \cos x) - \int 1 . (- \cos x) d x] = \frac{1}{2} [x . (\frac{- \cos 3 x}{3}) + \frac{1}{9} \sin 3 x] - \frac{1}{2} [x . (- \cos x) + \sin x] = - \frac{x \cos 3 x}{6} + \frac{\sin 3 x}{18} + \frac{x \cos x}{2} - \frac{\sin x}{2} + C$

Page No 18.134:

Question 51:

$\int (\tan^{- 1} x^{2}) x d x$

Answer:

Let I = $\int$ (tan^–1 x²) x dx
Putting x² = t
⇒ 2x dx = dt
$\Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \underset{II}{1} . \underset{I}{\tan^{- 1} t} . d t = \frac{1}{2} \tan^{- 1} t \int 1 d t - \int \{\frac{d}{d t} (\tan^{- 1} t) \int 1 d t\} d t = \frac{1}{2} [\tan^{- 1} t . t - \int \frac{t}{1 + t^{2}} d t] Now putting 1 + t^{2} = p \Rightarrow 2 t d t = d p \Rightarrow t d t = \frac{d p}{2} ∴ I = \frac{1}{2} t . \tan^{- 1} t - \frac{1}{2} \int \frac{t d t}{1 + t^{2}} = \frac{t . \tan^{- 1} t}{2} - \frac{1}{2 x^{2}} \int \frac{d p}{p} = \frac{t . \tan^{- 1} t}{2} - \frac{1}{4} \ln p + C = \frac{x^{2} . \tan^{- 1} x^{2}}{2} - \frac{1}{4} \ln |1 + x^{4}| + C [∵ p = 1 + t^{2}]$

Page No 18.134:

Question 52:

$\int \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x$

Answer:

$\int \frac{x . \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x Let \sin^{- 1} x = θ x = \sin θ d x = \cos θ d θ ∴ \int \frac{x . \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x = \int \frac{(\sin θ) . θ}{\sqrt{1 - \sin^{2} θ}} . \cos θ d θ = \int \frac{(\sin θ) . θ}{\cos θ} . \cos θ d θ = \int \underset{I}{θ} . \underset{II}{\sin θ} d θ = θ \int \sin θ d θ - \int \{\frac{d}{d θ} (θ) \int \sin θ d θ\} d θ = θ (- \cos θ) - \int 1 . (- \cos θ) d θ = - θ \cos θ + \sin θ + C = - θ \sqrt{1 - \sin^{2} θ} + \sin θ + C = - \sin^{- 1} x \sqrt{1 - x^{2}} + x + C (∵ \sin^{- 1} x = θ)$

Page No 18.134:

Question 53:

$\int \sin^{- 1} 2 x d x$

Answer:

$\int \sin^{- 1} 2 x d x$

$= \int \sin^{- 1} 2 x \times 1 d x$

$= \sin^{- 1} 2 x \int d x - \int (\frac{d}{d x} \sin^{- 1} 2 x \int d x) d x$

$= x \sin^{- 1} 2 x - \int \frac{2}{\sqrt{1 - 4 x^{2}}} \times x d x$

$= x \sin^{- 1} 2 x - \int \frac{2 x}{\sqrt{1 - 4 x^{2}}} d x$

$= x \sin^{- 1} 2 x + \frac{1}{4} \int {(1 - 4 x^{2})}^{- \frac{1}{2}} (- 8 x) d x$

$= x \sin^{- 1} 2 x + \frac{1}{4} \times \frac{{(1 - 4 x^{2})}^{\frac{1}{2}}}{\frac{1}{2}} + C [\begin{matrix} \int {[f (x)]}^{n} f' (x) d x = \frac{{[f (x)]}^{n + 1}}{n + 1} + C \\ f (x) = 1 - 4 x^{2} \Rightarrow f' (x) = - 8 x \end{matrix}]$

$= x \sin^{- 1} 2 x + \frac{1}{2} \sqrt{1 - 4 x^{2}} + C$

Page No 18.134:

Question 54:

$\int \sin^{3} \sqrt{x} d x$

Answer:

$Let, I = \int \sin^{3} \sqrt{x} d x . . . . . (1) Consider, \sqrt{x} = t . . . . . (2) Differentiating both sides we get, \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow d x = 2 \sqrt{x} d t \Rightarrow d x = 2 t d t Therefore, (1) becomes, I = \int \sin^{3} t 2 t d t = 2 \int t \sin^{3} t d t = 2 \int t (\frac{3 \sin t - \sin 3 t}{4}) d t (Since, \sin 3 A = 3 \sin A - 4 \sin^{3} A) = \frac{3}{2} \int t si n t d t - \frac{1}{2} \int t \sin 3 t d t = \frac{3}{2} [t \int \sin t d t - \int (\frac{d t}{d t} \int \sin t d t) d t] - \frac{1}{2} [t \int \sin 3 t d t - \int (\frac{d t}{d t} \int \sin 3 t d t) d t] = \frac{3}{2} [- t co s t + \int \cos t d t] - \frac{1}{2} [- \frac{t \cos 3 t}{3} + \frac{1}{3} \int \cos 3 t d t] = \frac{3}{2} [- t \cos t + \sin t] - \frac{1}{2} [- \frac{t \cos 3 t}{3} + \frac{1}{9} \sin 3 t] + C = - \frac{3}{2} t \cos t + \frac{3}{2} \sin t + \frac{1}{6} t \cos 3 t - \frac{1}{18} \sin 3 t + C = - \frac{3}{2} \sqrt{x} \cos \sqrt{x} + \frac{3}{2} \sin \sqrt{x} + \frac{1}{6} \sqrt{x} \cos (3 \sqrt{x}) - \frac{1}{18} \sin (3 \sqrt{x}) + C$

Note: The answer in indefinite integration may vary depending on the integral constant.

Page No 18.134:

Question 55:

$\int x \sin^{3} x d x$

Answer:

Let I = $\int x \sin^{3} x d x$
sin (3A) = 3 sin A – 4 sin³ A
$\sin^{3} A = \frac{1}{4} [3 \sin A - \sin 3 A] ∴ I = \frac{1}{4} \int x . (3 \sin x - \sin 3 x) d x = \frac{3}{4} \int \underset{I}{x} . \underset{II}{\sin x} d x - \frac{1}{4} \int \underset{I}{x} \underset{II}{. \sin (3 x)} d x = \frac{3}{4} [x (- \cos x) - \int 1 . (- \cos x) d x] - \frac{1}{4} [x (- \frac{\cos 3 x}{3}) - \int 1 . (- \frac{\cos 3 x}{3}) d x] = - \frac{3 x \cos x}{4} + \frac{3}{4} \sin x + \frac{x \cos 3 x}{12} - \frac{1}{36} \sin 3 x + C$

Page No 18.134:

Question 56:

$\int \cos^{3} \sqrt{x} d x$

Answer:

$Let, I = \int \cos^{3} \sqrt{x} d x . . . . . (1) Consider, \sqrt{x} = t . . . . . (2) Differentiating both sides we get, \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow d x = 2 \sqrt{x} d t \Rightarrow d x = 2 t d t Therefore, (1) becomes, I = \int \cos^{3} t 2 t d t = 2 \int t \cos^{3} t d t = 2 \int t (\frac{3 \cos t + \cos 3 t}{4}) d t (Since, \cos 3 A = 4 \cos^{3} A - 3 \cos A) = \frac{3}{2} \int t \cos t d t + \frac{1}{2} \int t \cos 3 t d t = \frac{3}{2} [t \int \cos t d t - \int (\frac{d t}{d t} \int \cos t d t) d t] + \frac{1}{2} [t \int \cos 3 t d t - \int (\frac{d t}{d t} \int \cos 3 t d t) d t] = \frac{3}{2} [t \sin t - \int \sin t d t] + \frac{1}{2} [\frac{t \sin 3 t}{3} - \frac{1}{3} \int \sin 3 t d t] = \frac{3}{2} [t \sin t + \cos t] + \frac{1}{2} [\frac{t \sin 3 t}{3} + \frac{1}{9} \cos 3 t] + C = \frac{3}{2} t \sin t + \frac{3}{2} \cos t + \frac{1}{6} t \sin 3 t + \frac{1}{18} \cos 3 t + C = \frac{3}{2} \sqrt{x} \sin \sqrt{x} + \frac{3}{2} \cos \sqrt{x} + \frac{1}{6} \sqrt{x} \sin (3 \sqrt{x}) + \frac{1}{18} \cos (3 \sqrt{x}) + C$

Note: The final answer in indefinite integration may vary based on the integration constant.

Page No 18.134:

Question 57:

$\int x \cos^{3} x d x$

Answer:

Let I= $\int x \cos^{3} x d x$

$As we know, \cos 3 x = 4 \cos^{3} x - 3 \cos x \Rightarrow \cos^{3} x = \frac{1}{4} (\cos 3 x + 3 \cos x)$

$∴ I = \frac{1}{4} \int x . (\cos 3 x + 3 \cos x) d x = \frac{1}{4} \int \underset{I}{x} . \underset{II}{\cos} (3 x) d x + \frac{3}{4} \int \underset{I}{x} . \underset{II}{\cos x} d x = \frac{1}{4} [x . \int \cos 3 x d x - \int \{\frac{d}{d x} (x) . \int \cos 3 x d x\} d x] + \frac{3}{4} [x \int \cos x - \int \{\frac{d}{d x} (x) . \int \cos x d x\} d x] = \frac{1}{4} [x . \frac{\sin 3 x}{3} - \int 1 . \frac{\sin 3 x}{3} d x] + \frac{3}{4} [x (\sin x) - \int 1 . \sin x d x] = \frac{x \sin 3 x}{12} + \frac{\cos 3 x}{36} + \frac{3}{4} x \sin x + \frac{3}{4} \cos x + C$

Page No 18.134:

Question 58:

$\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} d x$

Answer:

$Let I = \int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} d x Putting x = \cos θ \Rightarrow d x = - \sin θ d θ & θ = \cos^{- 1} x ∴ I = \int \tan^{- 1} \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} (- \sin θ) d θ = \int \tan^{- 1} \sqrt{\frac{2 \sin^{2} \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}}} (- \sin θ) d θ = \int \tan^{- 1} (\tan \frac{θ}{2}) (- \sin θ) d θ = - \frac{1}{2} \int \underset{I}{θ} \underset{II}{\sin θ} d θ = - \frac{1}{2} [θ \int \sin θ d θ - \int \{(\frac{d}{d θ} θ) \int \sin θ d θ\} d θ] = - \frac{1}{2} [θ (- \cos θ) - \int 1 . (- \cos θ) d θ] = - \frac{1}{2} [- θ \cos θ + \sin θ] + C = - \frac{1}{2} [- θ . \cos θ + \sqrt{1 - \cos^{2} θ}] + C = - \frac{1}{2} [- \cos^{- 1} x . x + \sqrt{1 - x^{2}}] + C [∵ θ = \cos^{- 1} x] = \frac{x \cos^{- 1} x}{2} - \frac{\sqrt{1 - x^{2}}}{2} + C$

Page No 18.134:

Question 59:

$\int \sin^{- 1} \sqrt{\frac{x}{a + x}} d x$

Answer:

$Let I = \int \sin^{- 1} \sqrt{\frac{x}{a + x}} d x Putting x = a \tan^{2} θ \Rightarrow \sqrt{\frac{x}{a}} = \tan θ \Rightarrow d x = a (2 \tan θ) \sec^{2} θ d θ ∴ I = \int \sin^{- 1} \sqrt{\frac{a \tan^{2} θ}{a + a \tan^{2} θ}} (2 a \tan θ) \sec^{2} θ d θ = \int \sin^{- 1} \sqrt{\frac{\tan^{2} θ}{\sec^{2} θ}} (2 a \tan θ \sec^{2} θ) d θ = 2 a \int [\sin^{- 1} (\sin θ) \tan θ \sec^{2} θ] d θ$

$= 2 a \int \underset{I}{θ} \underset{II}{\tan θ \sec^{2} θ} d θ = 2 a [θ \frac{\tan^{2} θ}{2} - \int 1 \frac{\tan^{2} θ}{2} d θ] = 2 a [\frac{θ . \tan^{2} θ}{2} - \frac{1}{2} \int (s e c^{2} θ - 1) d θ] = a θ \tan^{2} θ - a \tan θ + a θ + C = a (\frac{x}{a}) \tan^{- 1} (\frac{\sqrt{x}}{\sqrt{a}}) - a \sqrt{\frac{x}{a}} + a \tan^{- 1} \sqrt{\frac{x}{a}} + C = x \tan^{- 1} \sqrt{\frac{x}{a}} - \sqrt{a x} + a \tan^{- 1} \sqrt{\frac{x}{a}} + C$

Page No 18.134:

Question 60:

$\int \frac{x^{3} \sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} d x$

Answer:

$Let I = \int \frac{x^{3} \times \sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} d x Putting \sin^{- 1} x^{2} = t \Rightarrow x^{2} = \sin t \Rightarrow \frac{1 \times 2 x d x}{\sqrt{1 - {(x^{2})}^{2}}} = d t \Rightarrow \frac{x d x}{\sqrt{1 - x^{4}}} = \frac{d t}{2} ∴ I = \int x^{2} . \frac{\sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} . x d x = \int (\sin t) . t . \frac{d t}{2} = \frac{1}{2} \int \underset{I}{t} . \underset{II}{\sin t} d t = \frac{1}{2} [t \int \sin t d t - \int \{\frac{d}{d t} (t) \int \sin t d t\} d t] = \frac{1}{2} [t . (- \cos t) - \int 1 . (- \cos t) d t] = \frac{1}{2} [- t \cos t + \sin t] + C = \frac{1}{2} [- t \sqrt{1 - \sin^{2} t} + \sin t] + C = \frac{1}{2} [- \sin^{- 1} (x^{2}) \sqrt{1 - x^{4}} + x^{2}] + C$

Page No 18.134:

Question 61:

$\int \frac{x^{2} \sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} d x$

Answer:

$Let I = \int \frac{x^{2} . \sin^{- 1} x d x}{{(1 - x^{2})}^{\frac{3}{2}}} Putting x = \sin θ \Rightarrow d x = \cos θ d θ & θ = \sin^{- 1} x ∴ I = \int \frac{\sin^{2} θ . θ . \cos θ d θ}{{(1 - \sin^{2} θ)}^{\frac{3}{2}}} = \int \frac{\sin^{2} θ . θ . \cos θ d θ}{{(\cos^{2} θ)}^{\frac{3}{2}}} = \int \frac{\sin^{2} θ . θ . \cos θ d θ}{\cos^{3} θ} = \int \tan^{2} θ . θ . d θ = \int (\sec^{2} θ - 1) θ . d θ = \int \underset{I}{θ} . \underset{II}{\sec^{2} θ} d θ - \int θ . d θ = θ \int \sec^{2} θ d θ - \int \{\frac{d}{d θ} (θ) \int \sec^{2} θ d θ\} d θ - \int θ . d θ = θ \tan θ - \int 1 . \tan θ d θ - \frac{θ^{2}}{2} = θ . \tan θ - \ln |\sec θ| - \frac{θ^{2}}{2} + C = θ . \frac{\sin θ}{\cos θ} + \ln |\cos θ| - \frac{θ^{2}}{2} + C = θ . \frac{\sin θ}{\cos θ} + \ln |\sqrt{1 - \sin^{2} θ}| - \frac{θ^{2}}{2} + C = \frac{θ . \sin θ}{\sqrt{1 - \sin^{2} θ}} + \frac{1}{2} \ln |1 - \sin^{2} θ| - \frac{θ^{2}}{2} + C = \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{1}{2} \ln (1 - x^{2}) - \frac{1}{2} {(\sin^{- 1} x)}^{2} + C [∵ θ = \sin^{- 1} x]$

Page No 18.14:

Question 1:

$\int (3 x \sqrt{x} + 4 \sqrt{x} + 5) d x$

Answer:

$\int (3 x \sqrt{x} + 4 \sqrt{x} + 5) d x = \int (3 x^{1} \cdot x^{\frac{1}{2}} + 4 x^{\frac{1}{2}} + 5) d x = 3 \int x^{\frac{3}{2}} d x + 4 \int x^{\frac{1}{2}} d x + 5 \int d x = 3 [\frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + 4 [\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + 5 x + C = 3 \times \frac{2}{5} x^{\frac{5}{2}} + 4 \times \frac{2}{3} x^{\frac{3}{2}} + 5 x + C = \frac{6}{5} x^{\frac{5}{2}} + \frac{8}{3} x^{\frac{3}{2}} + 5 x + C$

Page No 18.14:

Question 2:

$\int (2^{x} + \frac{5}{x} - \frac{1}{x^{1 / 3}}) d x$

Answer:

$\int (2^{x} + \frac{5}{x} - \frac{1}{x^{\frac{1}{3}}}) d x = \int 2^{x} d x + 5 \int \frac{d x}{x} - \int \frac{d x}{x^{\frac{1}{3}}} = \int 2^{x} d x + 5 \int \frac{d x}{x} - \int x^{- \frac{1}{3}} d x = \frac{2^{x}}{\ln 2} + 5 \ln x - [\frac{x^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1}] + C = \frac{2^{x}}{\ln 2} + 5 \ln x - \frac{3}{2} x^{\frac{2}{3}} + C$

Page No 18.14:

Question 3:

$\int \{\sqrt{x} (a x^{2} + b x + c)\} d x$

Answer:

$\int \sqrt{x} (a x^{2} + b x + c) d x = \int x^{\frac{1}{2}} (a x^{2} + b x + c) d x = \int (a x^{2 + \frac{1}{2}} + b x^{\frac{1}{2} + 1} + c x^{\frac{1}{2}}) d x = a \int x^{\frac{5}{2}} d x + b \int x^{\frac{3}{2}} d x + c \int x^{\frac{1}{2}} d x = a [\frac{x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1}] + b [\frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + c [\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{2 a}{7} x^{\frac{7}{2}} + \frac{2 b}{5} x^{\frac{3}{2}} + \frac{2 c}{3} x^{\frac{3}{2}} + C$

Page No 18.14:

Question 4:

$\int (2 - 3 x) (3 + 2 x) (1 - 2 x) d x$

Answer:

$\int (2 - 3 x) (3 + 2 x) (1 - 2 x) d x = \int (2 - 3 x) (3 - 6 x + 2 x - 4 x^{2}) d x = \int (2 - 3 x) (- 4 x^{2} - 4 x + 3) d x = \int (- 8 x^{2} - 8 x + 6 + 12 x^{3} + 12 x^{2} - 9 x) d x = \int (12 x^{3} + 4 x^{2} - 17 x + 6) d x = \frac{12 x^{4}}{4} + \frac{4 x^{3}}{3} - \frac{17 x^{2}}{2} + 6 x + C = 3 x^{4} + \frac{4}{3} x^{3} - \frac{17}{2} x^{2} + 6 x + C$

Page No 18.14:

Question 5:

$\int (\frac{m}{x} + \frac{x}{m} + m^{x} + x^{m} + m x) d x$

Answer:

$\int (\frac{m}{x} + \frac{x}{m} + m^{x} + x^{m} + m x) d x = m \int \frac{1}{x} d x + \frac{1}{m} \int x d x + \int m^{x} d x + \int x^{m} d x + m \int x d x = m \ln |x| + \frac{1}{m} [\frac{x^{1 + 1}}{1 + 1}] + [\frac{m^{x}}{\ln m}] + [\frac{x^{m + 1}}{m + 1}] + m [\frac{x^{1 + 1}}{1 + 1}] = m \ln |x| + \frac{x^{2}}{2 m} + \frac{m^{x}}{\ln m} + \frac{x^{m + 1}}{m + 1} + \frac{m x^{2}}{2} + C$

Page No 18.14:

Question 6:

$\int {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} d x$

Answer:

$\int {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} d x = \int (x + \frac{1}{x} - 2) d x = \int x d x + \int \frac{d x}{x} - 2 \int d x = \frac{x^{2}}{2} + \ln |x| - 2 x + C$

Page No 18.14:

Question 7:

$\int \frac{{(1 + x)}^{3}}{\sqrt{x}} d x$

Answer:

$\int \frac{{(1 + x)}^{3}}{\sqrt{x}} d x = \int (\frac{1 + x^{3} + 3 {(1)}^{2} x + 3 (1) x^{2}}{\sqrt{x}}) d x = \int (\frac{1 + x^{3} + 3 x + 3 x^{2}}{\sqrt{x}}) d x = \int (\frac{1}{\sqrt{x}} + \frac{x^{3}}{\sqrt{x}} + \frac{3 x}{\sqrt{x}} + \frac{3 x^{2}}{\sqrt{x}}) d x = \int (x^{- \frac{1}{2}} + x^{\frac{5}{2}} + 3 x^{\frac{1}{2}} + 3 x^{\frac{3}{2}}) d x = [\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + \frac{x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} + 3 \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + 3 \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + C = 2 \sqrt{x} + \frac{2}{7} x^{\frac{7}{2}} + 2 x^{\frac{3}{2}} + \frac{6}{5} x^{\frac{5}{2}} + C$

Page No 18.14:

Question 8:

$\int \{x^{2} + e^{\log x} + {(\frac{e}{2})}^{x}\} d x$

Answer:

$\int (x^{2} + e^{\log x} + {(\frac{e}{2})}^{x}) d x = \int x^{2} d x + \int x d x + \int {(\frac{e}{2})}^{x} d x = \frac{x^{3}}{3} + \frac{x^{2}}{2} + \frac{{(\frac{e}{2})}^{x}}{\ln (\frac{e}{2})} + C$

Page No 18.14:

Question 9:

$\int (x^{e} + e^{x} + e^{e}) d x$

Answer:

$\int (x^{e} + e^{x} + e^{e}) d x = \int x^{e} d x + \int e^{x} d x + e^{e} \int 1 d x = \frac{x^{e + 1}}{e + 1} + e^{x} + x \cdot e^{e} + C$

Page No 18.14:

Question 10:

$\int \sqrt{x} (x^{3} - \frac{2}{x}) d x$

Answer:

$\int \sqrt{x} (x^{3} - \frac{2}{x}) d x = \int (x^{\frac{7}{2}} - \frac{2}{\sqrt{x}}) d x = \int (x^{\frac{7}{2}} - 2 x^{- \frac{1}{2}}) d x = \frac{x^{\frac{7}{2} + 1}}{\frac{7}{2} + 1} - 2 \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C = \frac{2}{9} x^{\frac{9}{2}} - 4 x^{\frac{1}{2}} + C = \frac{2}{9} x^{\frac{9}{2}} - 4 \sqrt{x} + C$

Page No 18.14:

Question 11:

$\int \frac{1}{\sqrt{x}} (1 + \frac{1}{x}) d x$

Answer:

$\int \frac{1}{\sqrt{x}} (1 + \frac{1}{x}) d x = \int x^{- \frac{1}{2}} (1 + \frac{1}{x}) d x = \int (x^{- \frac{1}{2}} + \frac{1}{x^{\frac{3}{2}}}) d x = \int x^{- \frac{1}{2}} d x + \int x^{- \frac{3}{2}} d x = [\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + [\frac{x^{- \frac{3}{2} + 1}}{- \frac{3}{2} + 1}] = 2 \sqrt{x} - \frac{2}{\sqrt{x}} + C$

Page No 18.14:

Question 12:

$\int \frac{x^{6} + 1}{x^{2} + 1} d x$

Answer:

$\int (\frac{x^{6} + 1}{x^{2} + 1}) d x = \int [\frac{{(x^{2})}^{3} + 1^{3}}{x^{2} + 1}] d x A^{3} + B^{3} = (A + B) (A^{2} - AB + B^{2}) = \int \frac{(x^{2} + 1) (x^{4} - x^{2} + 1)}{(x^{2} + 1)} d x = \int (x^{4} - x^{2} + 1) d x = \int x^{4} d x + \int x^{2} d x + \int 1 d x = \frac{x^{4 + 1}}{4 + 1} - \frac{x^{2 + 1}}{2 + 1} + x + C = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x + C$

Page No 18.14:

Question 13:

$\int \frac{x^{- 1 / 3} + \sqrt{x} + 2}{\sqrt[3]{x}} d x$

Answer:

$\int (\frac{x^{- \frac{1}{3}} + \sqrt{x} + 2}{x^{\frac{1}{3}}}) d x = \int (\frac{x^{- \frac{1}{3}}}{x^{\frac{1}{3}}} + \frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}} + \frac{2}{x^{\frac{1}{3}}}) d x = \int (x^{- \frac{2}{3}} + x^{\frac{1}{6}} + 2 x^{- \frac{1}{3}}) d x = [\frac{x^{- \frac{2}{3} + 1}}{- \frac{2}{3} + 1} + \frac{x^{\frac{1}{6} + 1}}{\frac{1}{6} + 1} + 2 \frac{x^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1}] = [\frac{x^{\frac{1}{3}}}{\frac{1}{3}} + \frac{x^{\frac{7}{6}}}{\frac{7}{6}} + 3 x^{\frac{2}{3}}] + C = 3 x^{\frac{1}{3}} + \frac{6}{7} x^{\frac{7}{6}} + 3 x^{\frac{2}{3}} + C$

Page No 18.14:

Question 14:

$\int \frac{{(1 + \sqrt{x})}^{2}}{\sqrt{x}} d x$

Answer:

$\int \frac{{(1 + \sqrt{x})}^{2}}{\sqrt{x}} d x = \int (\frac{1}{\sqrt{x}} + \frac{x}{\sqrt{x}} + 2 \frac{\sqrt{x}}{\sqrt{x}}) d x = \int (x^{- \frac{1}{2}} + x^{\frac{1}{2}} + 2) d x = [\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + [\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + 2 x + C = 2 \sqrt{x} + \frac{2}{3} x^{\frac{3}{2}} + 2 x + C$

Page No 18.143:

Question 1:

$\int e^{x} (\cos x - \sin x) d x$

Answer:

$Let I = \int e^{x} (\cos x - \sin x) d x let e^{x} \cos x = t Diff both sides w . r . t x e^{x} \cdot \cos x + e^{x} (- \sin x) = \frac{d t}{d x} Put e^{x} f (x) = t \Rightarrow e^{x} (\cos x - \sin x) d x = d t ∴ \int e^{x} (\cos x - \sin x) d x = \int d t \Rightarrow I = t + C = e^{x} \cos x + C$

Page No 18.143:

Question 2:

$\int e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) d x$

Answer:

$Let I = \int e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) d x Also let e^{x} \times \frac{1}{x^{2}} = t Diff both sides w . r . t x e^{x} \times \frac{1}{x^{2}} + e^{x} (\frac{- 2}{x^{3}}) = \frac{d t}{d x} \Rightarrow e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) d x = d t ∴ \int e^{x} (\frac{1}{x^{2}} - \frac{2}{x^{3}}) d x = \int d t = t + C = \frac{e^{x}}{x^{2}} + C$

Page No 18.143:

Question 3:

$\int e^{x} (\frac{1 + \sin x}{1 + \cos x}) d x$

Answer:

$Let I = \int e^{x} (\frac{1 + \sin x}{1 + \cos x}) d x = \int e^{x} (\frac{1}{1 + \cos x} + \frac{\sin x}{1 + \cos x}) d x = \int e^{x} (\frac{1}{2 \cos^{2} \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}) d x = \int e^{x} (\frac{1}{2} \sec^{2} \frac{x}{2} + \tan \frac{x}{2}) d x Putting e^{x} \tan \frac{x}{2} = t Diff both sides w . r . t . x e^{x} \cdot \tan (\frac{x}{2}) + e^{x} \times \frac{1}{2} \sec^{2} \frac{x}{2} = \frac{d t}{d x} \Rightarrow e^{x} [\tan \frac{x}{2} + \frac{1}{2} \sec^{2} (\frac{x}{2})] d x = d t ∴ \int e^{x} (\frac{1}{2} \sec^{2} \frac{x}{2} + \tan \frac{x}{2}) d x = \int d t = t + C = e^{x} \tan (\frac{x}{2}) + C$

Page No 18.143:

Question 4:

$\int e^{x} (\cot x - {cosec}^{2} x) d x$

Answer:

$Let I = \int e^{x} (\cot x - {cosec}^{2} x) d x here f (x) = \cot x put e^{x} f (x) = t f' (x) = - {cosec}^{2} x let e^{x} \cot x = t Diff both sides w . r . t x e^{x} \cot x + e^{x} (- {cosec}^{2} x) = \frac{d t}{d x} \Rightarrow e^{x} (\cot x - {cosec}^{2} x) d x = d t ∴ \int e^{x} (\cot x - {cosec}^{2} x) d x = \int d t = t + C = e^{x} \cot x + C$

Page No 18.143:

Question 5:

$\int e^{x} (\frac{x - 1}{2 x^{2}}) d x$

Answer:

$Let I = \int e^{x} (\frac{x - 1}{2 x^{2}}) d x = \frac{1}{2} \int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x here \frac{1}{x} = f (x) Put e^{x} f (x) = t \Rightarrow - \frac{1}{x^{2}} = f' (x) let e^{x} \frac{1}{x} = t Diff both sides w . r . t x (e^{x} \frac{1}{x} + e^{x} \frac{- 1}{x^{2}}) = \frac{d t}{d x} \Rightarrow e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x = d t ∴ I = \frac{1}{2} \int d t = \frac{t}{2} + C = \frac{e^{x}}{2 x} + C$

Page No 18.143:

Question 6:

$\int e^{x} \sec x (1 + \tan x) d x$

Answer:

$Let I = \int e^{x} \sec x (1 + \tan x) d x = \int e^{x} (\sec x + \sec x \tan x) d x Here, f (x) = \sec x Put e^{x} f (x) = t \Rightarrow f' (x) = \sec x \tan x let e^{x} \sec x = t Diff both sides w . r . t x e^{x} \sec x + e^{x} \sec x \tan x = \frac{d t}{d x} \Rightarrow e^{x} (\sec x + \tan x) d x = d t ∴ \int e^{x} (\sec x + \sec x \tan x) d x = \int d t = t + C = e^{x} \sec x + C$

Page No 18.143:

Question 7:

$\int e^{x} (\tan x - \log \cos x) d x$

Answer:

$Let I = \int e^{x} (\tan x - \log \cos x) d x here f (x) = - \log \cos x Put e^{x} f (x) = t \Rightarrow f' (x) = \tan x let - e^{x} \log \cos x = t Diff both sides w . r . t x - [e^{x} \log (\cos x) + e^{x} \frac{1}{\cos x} \times (- \sin x)] = \frac{d t}{d x} \Rightarrow [- e^{x} \log (\cos x) + e^{x} \tan x] d x = d t ∴ \int e^{x} (\tan x - \log \cos x) d x = \int d t = t + C = - e^{x} \log (\cos x) + C = e^{x} \log (\sec x) + C$

Page No 18.143:

Question 8:

$\int e^{x} [\sec x + \log (\sec x + \tan x)] d x$

Answer:

$Let I = \int e^{x} [\sec x + \log (\sec x + \tan x)] d x Here, f (x) = \log (\sec x + \tan x) Put e^{x} f (x) = t \Rightarrow f' (x) = \sec x let e^{x} \log (\sec x + \tan x) = t Diff both sides w . r . t x e^{x} \log (\sec x + \tan x) + e^{x} \frac{1}{\sec x + \tan x} (\sec x + \tan x + \sec^{2} x) = \frac{d t}{d x} \Rightarrow [e^{x} \log (\sec x + \tan x) + e^{x} (\sec x)] d x = d t \Rightarrow e^{x} [\sec x + \log (\sec x + \tan x)] d x = d t ∴ \int e^{x} [\sec x + \log (\sec x + \tan x)] d x = \int d t = t + C = e^{x} \log (\sec x + \tan x) + C$

Page No 18.143:

Question 9:

$\int e^{x} (\cot x + \log \sin x) d x$

Answer:

$Let I = \int e^{x} (\cot x + \log \sin x) d x Here, f (x) = \log \sin x Put e^{x} f (x) = t \Rightarrow f' (x) = \cot x let e^{x} \log \sin x = t Diff both sides w . r . t x e^{x} \log (\sin x) + e^{x} \times \frac{1}{\sin x} \times \cos x = \frac{d t}{d x} \Rightarrow [e^{x} \log (\sin x) + e^{x} \cot x] d x = d t \Rightarrow e^{x} (\cot x + \log \sin x) d x = d t ∴ \int e^{x} (\cot x + \log \sin x) d x = \int d t = t + C = e^{x} \log \sin x + C$

Page No 18.143:

Question 10:

$\int e^{x} \frac{x - 1}{{(x + 1)}^{3}} d x$

Answer:

$Let I = \int e^{x} [\frac{x - 1}{{(x - 1)}^{3}}] d x = \int e^{x} [\frac{x + 1 - 2}{{(x + 1)}^{3}}] d x = \int e^{x} [\frac{1}{{(x - 1)}^{2}} - \frac{2}{{(x + 1)}^{3}}] d x Here, f (x) = \frac{1}{{(x + 1)}^{2}} \Rightarrow f' (x) = \frac{- 2}{{(x + 1)}^{2}} Put e^{x} f (x) = t let e^{x} \frac{1}{{(x + 1)}^{2}} = t Diff both sides e^{x} \frac{1}{{(x + 1)}^{2}} + e^{x} \frac{(- 2)}{{(x + 1)}^{3}} = \frac{d t}{d x} \Rightarrow e^{x} [\frac{1}{{(x + 1)}^{2}} - \frac{2}{{(x + 1)}^{3}}] d x = d t ∴ \int e^{x} [\frac{1}{{(x + 1)}^{2}} - \frac{2}{{(x + 1)}^{3}}] d x = \int d t = t + C = \frac{e^{x}}{{(x + 1)}^{2}} + C$

Page No 18.143:

Question 11:

$\int e^{x} (\frac{\sin 4 x - 4}{1 - \cos 4 x}) d x$

Answer:

$Let I = \int e^{x} [\frac{\sin 4 x - 4}{1 - \cos 4 x}] d x = \int e^{x} [\frac{2 \sin 2 x \cos 2 x}{2 \sin^{2} (2 x)} - \frac{4}{2 \sin^{2} 2 x}] d x = \int e^{x} [\cot (2 x) - 2 {cosec}^{2} (2 x)] d x Here, f (x) = \cot (2 x) \Rightarrow f' (x) = - 2 {cosec}^{2} (2 x) Put e^{x} f (x) = t let e^{x} \cot (2 x) = t Diff both sides w . r . t x e^{x} \cot (2 x) + e^{x} \times [- 2 {cosec}^{2} (2 x)] = \frac{d t}{d x} \Rightarrow e^{x} [\cot (2 x) - 2 {cosec}^{2} (2 x)] d x = d t ∴ \int e^{x} [\cot 2 x - 2 {cosec}^{2} 2 x] d x = \int d t \Rightarrow I = t + C = e^{x} \cot (2 x) + C$

Page No 18.143:

Question 12:

$\int e^{x} \frac{{(1 - x)}^{2}}{{(1 + x^{2})}^{2}} d x$

Answer:

$Let I = \int e^{x} [\frac{{(1 - x)}^{2}}{{(1 + x^{2})}^{2}}] d x = \int e^{x} [\frac{1 + x^{2} - 2 x}{{(1 + x^{2})}^{2}}] d x = \int e^{x} [\frac{1}{1 + x^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x Here, f (x) = \frac{1}{1 + x^{2}} \Rightarrow f' (x) = \frac{- 2 x}{{(1 + x^{2})}^{2}} Put e^{x} f (x) = t \Rightarrow e^{x} \frac{1}{1 + x^{2}} = t Diff both sides w . r . t x e^{x} \frac{1}{1 + x^{2}} + e^{x} \frac{- 1}{{(1 + x^{2})}^{2}} 2 x = \frac{d t}{d x} \Rightarrow e^{x} [\frac{1}{1 + x^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x = d t ∴ \int e^{x} [\frac{1}{1 + x^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x = \int d t \Rightarrow I = t + C = \frac{e^{x}}{1 + x^{2}} + C$

Page No 18.143:

Question 13:

$\int e^{x} \frac{1 + x}{{(2 + x)}^{2}} d x$

Answer:

$Let I = \int e^{x} [\frac{1 + x}{{(2 + x)}^{2}}] d x = \int e^{x} (\frac{2 + x - 1}{{(2 + x)}^{2}}) d x = \int e^{x} [\frac{1}{(2 + x)} - \frac{1}{{(2 + x)}^{2}}] d x Here, f (x) = \frac{1}{2 + x} \Rightarrow f' (x) = \frac{- 1}{{(2 + x)}^{2}} Put e^{x} f (x) = t \Rightarrow e^{x} \frac{1}{x + 2} = t Diff both sides w . r . t x e^{x} \frac{1}{x + 2} + e^{x} \frac{- 1}{{(x + 2)}^{2}} = \frac{d t}{d x} \Rightarrow e^{x} [\frac{1}{x + 2} - \frac{1}{{(x + 2)}^{2}}] d x = d t ∴ \int e^{x} [\frac{1}{(2 + x)} - \frac{1}{{(2 + x)}^{2}}] d x = \int d t \Rightarrow I = t + C = \frac{e^{x}}{2 + x} + C$

Page No 18.143:

Question 14:

$\int \frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x / 2} d x$

Answer:

$Let I = \int (\frac{\sqrt{1 - \sin x}}{1 + \cos x}) e^{\frac{- x}{2}} d x = \int (\frac{\sqrt{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}}{2 \cos^{2} \frac{x}{2}}) e^{\frac{- x}{2}} d x = \int \frac{\sqrt{{(\cos \frac{x}{2} - \sin \frac{x}{2})}^{2}}}{2 \cos^{2} \frac{x}{2}} e^{\frac{- x}{2}} d x = \int (\frac{\sin \frac{x}{2} - \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}) e^{\frac{- x}{2}} d x = \int [\frac{1}{2} \sec \frac{x}{2} \tan \frac{x}{2} - \frac{1}{2} \sec (\frac{x}{2})] e^{\frac{- x}{2}} d x = \frac{1}{2} \int (\sec \frac{x}{2} \tan \frac{x}{2} - \sec \frac{x}{2}) e^{\frac{- x}{2}} d x let e^{\frac{- x}{2}} \sec (\frac{x}{2}) = t Diff both sides w . r . t x e^{\frac{- x}{2}} \frac{\sec (\frac{x}{2}) \tan (\frac{x}{2})}{2} + \sec (\frac{x}{2}) \times e^{\frac{- x}{2}} \times \frac{- 1}{2} = \frac{d t}{d x} \Rightarrow {\frac{e}{2}}^{\frac{- x}{2}} [\sec \frac{x}{2} \tan \frac{x}{2} - \sec \frac{x}{2}] d x = d t ∴ \frac{1}{2} \int (\sec \frac{x}{2} \tan \frac{x}{2} - \sec \frac{x}{2}) e^{\frac{- x}{2}} d x = \int d t \Rightarrow I = \int t + C = e^{\frac{- x}{2}} \sec (\frac{x}{2}) + C$

Page No 18.143:

Question 15:

$\int e^{x} (\log x + \frac{1}{x}) d x$

Answer:

$Let I = \int e^{x} (\log x + \frac{1}{x}) d x Here, f (x) = \log x \Rightarrow f' (x) = \frac{1}{x} put e^{x} f (x) = t \Rightarrow e^{x} \log x = t Diff both sides w . r . t x e^{x} \log x + e^{x} \frac{1}{x} = \frac{d t}{d x} \Rightarrow e^{x} (\log x + \frac{1}{x}) d x = d t ∴ \int e^{x} [\log x + \frac{1}{x}] d x = \int d t \Rightarrow I = t + C = e^{x} \log x + C$

Page No 18.143:

Question 16:

$\int e^{x} (\log x + \frac{1}{x^{2}}) d x$

Answer:

$Let I = \int (\log x + \frac{1}{x^{2}}) e^{x} d x = \int e^{x} (\log x + \frac{1}{x} - \frac{1}{x} + \frac{1}{x^{2}}) d x = \int e^{x} (\log x + \frac{1}{x}) d x + \int e^{x} (- \frac{1}{x} + \frac{1}{x^{2}}) d x \begin{matrix} let e^{x} \log x = t \\ Diff both sides \\ (e^{x} \log x + e^{x} \frac{1}{x}) d x = d t \end{matrix} \begin{matrix} let e^{x} (- \frac{1}{x}) = p \\ Diff both sides \\ (e^{x} \frac{- 1}{x} + e^{x} \frac{1}{x^{2}}) d x = d p \end{matrix} ∴ I = \int d t + \int d p = t + p + C = e^{x} \log x + e^{x} \frac{- 1}{x} + C = e^{x} (\log x - \frac{1}{x}) + C$

Page No 18.143:

Question 17:

$\int \frac{e^{x}}{x} \{x {(\log x)}^{2} + 2 \log x\} d x$

Answer:

$Let I = \int \frac{e^{x}}{x} [x {(\log x)}^{2} + 2 \log x] d x = \int e^{x} [{(\log x)}^{2} + \frac{2 \log x}{x}] d x Here, f (x) = {(\log x)}^{2} \Rightarrow f' (x) = \frac{2 \log x}{x} put e^{x} f (x) = t \Rightarrow e^{x} {(\log x)}^{2} = t Diff both sides w . r . t x [e^{x} {(\log x)}^{2} + e^{x} \frac{2 \log x}{x}] d x = d t ∴ I = \int d t = t + C = e^{x} {(\log x)}^{2} + C$

Page No 18.143:

Question 18:

$\int e^{x} \cdot \frac{\sqrt{1 - x^{2}} \sin^{- 1} x + 1}{\sqrt{1 - x^{2}}} d x$

Answer:

$Let I = \int e^{x} [\frac{\sqrt{1 - x^{2}} \sin^{- 1} x + 1}{\sqrt{1 - x^{2}}}] d x = \int e^{x} [\sin^{- 1} x + \frac{1}{\sqrt{1 - x^{2}}}] d x Here, f (x) = \sin^{- 1} x \Rightarrow f' (x) = \frac{1}{\sqrt{1 - x^{2}}} Put e^{x} f (x) = t \Rightarrow e^{x} \sin^{- 1} x = t D i f f b o t h s i d e s w . r . t x (e^{x} \sin^{- 1} x + e^{x} \times \frac{1}{\sqrt{1 - x^{2}}}) d x = d t ∴ I = \int d t = t + C = e^{x} \sin^{- 1} x + C$

Page No 18.143:

Question 19:

$\int e^{2 x} (- \sin x + 2 \cos x) d x$

Answer:

$Let I = \int e^{2 x} (- \sin x + 2 \cos x) d x Put e^{2 x} \cos x = t Diff both sides w . r . t x [2 e^{2 x} \cos x + e^{2 x} \times (- \sin x)] d x = d t ∴ I = \int d t = t + C = e^{2 x} \cos x + C$

Page No 18.143:

Question 20:

$\int (\frac{1}{\log x} - \frac{1}{{(\log x)}^{2}}) d x$

Answer:

$Let I = \int (\frac{1}{\log x} - \frac{1}{{(\log x)}^{2}}) d x Put \log x = t \Rightarrow x = e^{t} \Rightarrow d x = e^{t} d t ∴ I = \int e^{t} (\frac{1}{t} - \frac{1}{t^{2}}) d t Here, f (t) = \frac{1}{t} \Rightarrow f' (t) = \frac{- 1}{t^{2}} let e^{t} \times \frac{1}{t} = p Diff both sides w . r . t t (e^{t} \times \frac{1}{t} + e^{t} \times \frac{- 1}{t^{2}}) d t = d p ∴ I = \int d p = p + C = \frac{e^{t}}{t} + C = \frac{x}{\log x} + C$

Page No 18.143:

Question 21:

$\int e^{x} (\frac{\sin x \cos x - 1}{\sin^{2} x}) d x$

Answer:

$Let I = \int e^{x} (\frac{\sin x \cos x - 1}{\sin^{2} x}) d x = \int e^{x} [\cot x - {cosec}^{2} x] d x Here, f (x) = \cot x \Rightarrow f' (x) = - {cosec}^{2} x Put e^{x} f (x) = t \Rightarrow e^{x} \cot x = t Diff both sides w . r . t x e^{x} (\cot x - {cosec}^{2} x) d x = d t ∴ I = \int d t = t + C = e^{x} \cot x + C$

Page No 18.143:

Question 22:

$\int \{\tan (\log x) + \sec^{2} (\log x)\} d x$

Answer:

$Let I = \int [\tan (\log x) + \sec^{2} (\log x)] d x Put \log x = t \Rightarrow x = e^{t} \Rightarrow d x = e^{t} d t ∴ I = \int (\tan t + \sec^{2} t) e^{t} d t Here, f (t) = \tan t \Rightarrow f' (t) = \sec^{2} t let e^{t} \tan (t) = p Diff both sides w . r . t t e^{t} [\tan t + \sec^{2} t] = \frac{d p}{d t} \Rightarrow e^{t} [\tan t + \sec^{2} t] d t = d p ∴ I = \int d p = p + C = e^{t} \tan t + C = x \tan (\log x) + C$

Page No 18.143:

Question 23:

$\int \frac{e^{x} (x - 4)}{{(x - 2)}^{3}} d x$

Answer:

$Let I = \int e^{x} (\frac{x - 4}{{(x - 2)}^{3}}) d x = \int e^{x} [\frac{x - 2 - 2}{{(x - 2)}^{3}}] d x = \int e^{x} [\frac{1}{{(x - 2)}^{2}} - \frac{2}{{(x - 2)}^{3}}] d x Here, f (x) = \frac{1}{{(x - 2)}^{2}} \Rightarrow f' (x) = \frac{- 2}{{(x - 2)}^{3}} Put e^{x} f (x) = t \Rightarrow e^{x} \frac{1}{{(x - 2)}^{2}} = t Diff both sides w . r . t x [e^{x} \frac{1}{{(x - 2)}^{2}} + e^{x} \frac{- 2}{{(x - 2)}^{3}}] d x = d t ∴ I = \int d t = t + C = \frac{e^{x}}{{(x - 2)}^{2}} + C$

Page No 18.143:

Question 24:

Evaluate the following integrals:

$\int e^{2 x} (\frac{1 - \sin 2 x}{1 - \cos 2 x}) d x$

Answer:

$We have, I = \int e^{2 x} (\frac{1 - \sin 2 x}{1 - \cos 2 x}) d x = \int e^{2 x} (\frac{1 - 2 sinx \cos x}{2 \sin^{2} x}) d x Put t = 2 x . Then d t = 2 d x Therefore, I = \frac{1}{2} \int e^{t} (\frac{1 - 2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \sin^{2} \frac{t}{2}}) d t = \frac{1}{4} \int e^{t} (\frac{1 - 2 \sin \frac{t}{2} \cos \frac{t}{2}}{\sin^{2} \frac{t}{2}}) d t = \frac{1}{4} \int e^{t} (\frac{1}{\sin^{2} \frac{t}{2}} - \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{\sin^{2} \frac{t}{2}}) d t = \frac{1}{4} \int e^{t} ({cosec}^{2} \frac{t}{2} - 2 \cot \frac{t}{2}) d t = - \frac{1}{4} \int e^{t} (2 \cot \frac{t}{2} - {cosec}^{2} \frac{t}{2}) d t Consider, f (x) = 2 \cot \frac{t}{2}, then f^{'} (x) = - {cosec}^{2} \frac{t}{2} Thus, the given integrand is of the form e^{x} [f (x) + f^{'} (x)] . Therefore, I = - \frac{1}{4} (2 \cot \frac{t}{2}) e^{t} + c = - \frac{1}{4} (2 \cot \frac{2 x}{2}) e^{2 x} + c Hence, \int e^{2 x} (\frac{1 - \sin 2 x}{1 - \cos 2 x}) d x = - \frac{1}{2} (\cot x) e^{2 x} + c$

Page No 18.149:

Question 1:

$\int e^{a x} \cos b x d x$

Answer:

$Let I = \int e^{a x} \cos (b x) d x Considering \cos (b x) as first function and e^{a x} as second function I = \cos (b x) \frac{e^{a x}}{a} - \int - \sin (b x) \times b \times \frac{e^{a x}}{a} d x \Rightarrow I = \frac{e^{a x} \cos (b x)}{a} + \frac{b}{a} \int \sin (b x) e^{a x} d x \Rightarrow I = \frac{e^{a x}}{a} \cos (b x) + \frac{b}{a} \int e^{a x} \times \sin (b x) d x \Rightarrow I = \frac{e^{a x}}{a} \cos (b x) + \frac{b}{a} I_{1} . . . . . (1) where I_{1} = \int e^{a x} \sin (b x) d x Now, I_{1} = \int e^{a x} \sin (b x) d x Considering \sin (b x) as first function e^{a x} as second function I_{1} = \sin (b x) \frac{e^{a x}}{a} - \int \cos (b x) b \frac{e^{a x}}{a} d x \Rightarrow I_{1} = \frac{\sin (b x) e^{a x}}{a} - \frac{b}{a} \int e^{a x} \cos (b x) d x \Rightarrow I_{1} = \frac{e^{a x} \sin (b x)}{a} - \frac{b}{a} I . . . . . (2) From (1) and (2) I = \frac{e^{a x}}{a} \cos (b x) + \frac{b}{a} [\frac{e^{a x} \sin (b x)}{a} - \frac{b}{a} I] \Rightarrow I = \frac{e^{a x} \cos (b x)}{a} + \frac{b e^{a x} \sin (b x)}{a^{2}} - \frac{b^{2}}{a^{2}} I \Rightarrow I (1 + \frac{b^{2}}{a^{2}}) = e^{a x} [\frac{a \cos (b x) + b \sin (b x)}{a^{2}}] ∴ I = \frac{e^{a x} [a \cos b x + b \sin (b x)]}{(a^{2} + b^{2})} + C$

Page No 18.149:

Question 2:

$\int e^{a x} \sin (b x + C) d x$

Answer:

$Let I = \int e^{a x} \sin (b x + C) d x Considering \sin (b x + C) as first function and e^{a x} as second function I = \sin (b x + C) \frac{e^{a x}}{a} - \int \cos (b x + C) b \frac{e^{a x}}{a} d x \Rightarrow I = \frac{e^{a x} \sin (b x + C)}{a} - \frac{b}{a} \int e^{a x} \cos (b x + C) d x \Rightarrow I = \frac{e^{a x} \sin (b x + C)}{a} - \frac{b}{a} I_{1} . . . (1) where I_{1} = \int e^{a x} \cos (b x + C) d x Now, I_{1} = \int e^{a x} \cos (b x + C) d x Consider \cos (b x + C) as first function e^{a x} as second funciton I_{1} = \cos (b x + C) \frac{e^{a x}}{a} - \int - \sin (b x + C) b \frac{e^{a x}}{a} d x \Rightarrow I_{1} = \frac{e^{a x} \cos (b x + C)}{a} + \frac{b}{a} \int e^{a x} \sin (b x + C) d x \Rightarrow I_{1} = \frac{e^{a x} \cos (b x + C)}{a} + \frac{b}{a} I . . . . . (2) From (1) & (2) I = \frac{e^{a x} \sin (b x + C)}{a} - \frac{b}{a} [\frac{e^{a x} \cos (b x + C)}{a} + \frac{b}{a} I] \Rightarrow I = \frac{e^{a x} \sin (b x + C)}{a} - \frac{b}{a^{2}} e^{a x} co s (b x + C) - \frac{b^{2}}{a^{2}} I \Rightarrow I (1 + \frac{b^{2}}{a^{2}}) = \frac{e^{a x} a \sin (b x + C) - b e^{a x} \cos (b x + C)}{a^{2}} + C_{1} \Rightarrow I = e^{a x} \frac{[a \sin (b x + C) - b \cos (b x + C)]}{a^{2} + b^{2}} + C_{1} Where C_{1} is integration constant$

Page No 18.149:

Question 3:

$\int \cos (\log x) d x$

Answer:

$Let I = \int \cos (\log x) d x Let \log x = t \Rightarrow x = e^{t} \Rightarrow d x = e^{t} d t I = \int e^{t} \cos (t) d t Considering \cos (t) as first function and e^{t} as second function I = \cos t e^{t} - \int (- \sin t) e^{t} d t \Rightarrow I = \cos t e^{t} + \int \sin t e^{t} d t \Rightarrow I = \cos t e^{t} + I_{1} . . . . . (1) where I_{1} = \int e^{t} \sin t d t I_{1} = \int e^{t} \sin t d t Cosidering \sin t as first function and e^{t} as second function I_{1} = \sin t e^{t} - \int \cos t e^{t} d t \Rightarrow I_{1} = \sin t e^{t} - I . . . . . (2) From (1) & (2) I = \cos t e^{t} + \sin t e^{t} - I \Rightarrow 2 I = e^{t} (\sin t + \cos t) \Rightarrow I = \frac{e^{t} (\sin t + \cos t)}{2} + C \Rightarrow I = \frac{e^{\log x} [\sin (\log x) + \cos (\log x)]}{2} + C \Rightarrow I = \frac{x}{2} [\sin (\log x) + \cos (\log x)] + C$

Page No 18.149:

Question 4:

$\int e^{2 x} \cos (3 x + 4) d x$

Answer:

$Let I = \int e^{2 x} \cos (3 x + 4) d x Considering \cos (3 x + 4) as first function and e^{2 x} as second function I = \cos (3 x + 4) \frac{e^{2 x}}{2} - \int - \sin (3 x + 4) \times 3 \frac{e^{2 x}}{2} d x \Rightarrow I = \frac{e^{2 x} \cos (3 x + 4)}{2} + \frac{3}{2} \int e^{2 x} \sin (3 x + 4) d x \Rightarrow I = \frac{e^{2 x} \cos (3 x + 4)}{2} + \frac{3}{2} I_{1} . . . . . (1) where I_{1} = \int e^{2 x} \sin (3 x + 4) d x considering \sin (3 x + 4) as first function and e^{2 x} as second function I_{1} = \sin (3 x + 4) \frac{e^{2 x}}{2} - \int 3 \cos (3 x + 4) \frac{e^{2 x}}{2} d x \Rightarrow I_{1} = \frac{e^{2 x} \sin (3 x + 4)}{2} - \frac{3}{2} \int e^{2 x} \cos (3 x + 4) d x \Rightarrow I_{1} = \frac{e^{2 x} \sin (3 x + 4)}{2} - \frac{3}{2} I . . . . . (2) From (1) & (2) I = \frac{e^{2 x} \cos (3 x + 4)}{2} + \frac{3}{4} e^{2 x} \sin (3 x + 4) - \frac{9}{4} I \Rightarrow I + \frac{9}{4} I = \frac{2 e^{2 x} \cos (3 x + 4) + 3 e^{2 x} \sin (3 x + 4)}{4} \Rightarrow I = \frac{e^{2 x}}{13} [2 \cos (3 x + 4) + 3 \sin (3 x + 4)] + C$

Page No 18.149:

Question 5:

$\int e^{2 x} \sin x \cos x d x$

Answer:

$Let I = \int e^{2 x} \sin x \cos x d x I = \frac{1}{2} \int e^{2 x} (2 \sin x \cos x) d x \Rightarrow I = \frac{1}{2} \int e^{2 x} \sin 2 x d x Considering \sin 2 x as first function and e^{2 x} as second function . I = \frac{1}{2} [\sin 2 x \frac{e^{2 x}}{2} - \int 2 \cos 2 x \frac{e^{2 x}}{2} d x] \Rightarrow I = \frac{e^{2 x} \sin 2 x}{4} - \frac{1}{2} \int e^{2 x} \cos 2 x d x \Rightarrow I = \frac{e^{2 x} \sin 2 x}{4} - \frac{1}{2} I_{1} . . . . . (1) Where I_{1} = \int e^{2 x} \cos 2 x d x Considering \cos 2 x as first function and e^{2 x} as second function I_{1} = \cos 2 x \frac{e^{2 x}}{2} - \int - 2 \sin 2 x \frac{e^{2 x}}{2} d x \Rightarrow I_{1} = \frac{e^{2 x} \cos 2 x}{2} + \int e^{2 x} \sin 2 x d x \Rightarrow I_{1} = \frac{e^{2 x} \cos 2 x}{4} + 2 I . . . . . (2) \Rightarrow I = \frac{e^{2 x} \sin 2 x}{4} - \frac{1}{2} [\frac{e^{2 x} \cos 2 x}{2} + 2 I] \Rightarrow I = \frac{e^{2 x} \sin 2 x}{4} - \frac{e^{2 x} \cos 2 x}{4} - \frac{I}{2} \times 2 \Rightarrow 2 I = \frac{e^{2 x} (\sin 2 x - \cos 2 x)}{4} + C \Rightarrow I = \frac{e^{2 x}}{8} (\sin 2 x - \cos 2 x) + C$

Page No 18.149:

Question 6:

$\int e^{2 x} \sin x d x$

Answer:

$Let I = \int e^{2 x} \sin x d x Considering \sin x as first function and e^{2 x} as second function I = \sin x \frac{e^{2 x}}{2} - \int \cos x \frac{e^{2 x}}{2} d x \Rightarrow I = \sin x \frac{e^{2 x}}{2} - \frac{1}{2} \int \cos x e^{2 x} d x \Rightarrow I = \frac{\sin x e^{2 x}}{2} - \frac{1}{2} [\cos x \frac{e^{2 x}}{2} - \int (- \sin x) \frac{e^{2 x}}{2} d x] \Rightarrow I = \frac{\sin x e^{2 x}}{2} - \frac{\cos x e^{2 x}}{4} - \frac{1}{2} \int \frac{e^{2 x} \sin x}{2} d x I = \frac{e^{2 x} (2 \sin x - \cos x)}{4} - \frac{I}{4} \Rightarrow 5 I = e^{2 x} (2 \sin x - \cos x) \Rightarrow I = \frac{e^{2 x} (2 \sin x - \cos x)}{5} + C$

Page No 18.149:

Question 7:

Evaluate the following integrals:

$\int e^{2 x} \sin (3 x + 1) d x$

Answer:

$We have, I = \int e^{2 x} \sin (3 x + 1) d x Let the first function be \sin (3 x + 1) and the second function be e^{2 x} . First we find the integral of the second function, i . e ., \int e^{2 x} dx . \int e^{2 x} dx = \frac{1}{2} e^{2 x} Now, using integration by parts, we get I = \sin (3 x + 1) \int e^{2 x} d x - \int [(\frac{d (\sin (3 x + 1))}{d x}) \int e^{2 x} d x] d x = \frac{1}{2} \sin (3 x + 1) e^{2 x} - \frac{3}{2} \int [\cos (3 x + 1) e^{2 x}] d x = \frac{1}{2} \sin (3 x + 1) e^{2 x} - \frac{3}{2} \{\cos (3 x + 1) \int e^{2 x} d x - \int [(\frac{d (\cos (3 x + 1))}{d x}) \int e^{2 x} d x] d x\} = \frac{1}{2} \sin (3 x + 1) e^{2 x} - \frac{3}{2} \{\frac{1}{2} \cos (3 x + 1) e^{2 x} + \frac{3}{2} \int \sin (3 x + 1) e^{2 x} d x\} = \frac{1}{2} \sin (3 x + 1) e^{2 x} - \frac{3}{4} \cos (3 x + 1) e^{2 x} - \frac{9}{4} I + c I + \frac{9}{4} I = \frac{1}{2} \sin (3 x + 1) e^{2 x} - \frac{3}{4} \cos (3 x + 1) e^{2 x} + c \frac{13}{4} I = \frac{e^{2 x}}{2} [\sin (3 x + 1) - \frac{3}{2} \cos (3 x + 1)] + c I = \frac{2}{13} e^{2 x} [\sin (3 x + 1) - \frac{3}{2} \cos (3 x + 1)] + c = \frac{e^{2 x}}{13} [2 \sin (3 x + 1) - 3 \cos (3 x + 1)] + c Hence, \int e^{2 x} \sin (3 x + 1) d x = \frac{e^{2 x}}{13} [2 \sin (3 x + 1) - 3 \cos (3 x + 1)] + c$

Page No 18.149:

Question 8:

$\int e^{x} \sin^{2} x d x$

Answer:

$Let I = \int e^{x} \sin^{2} x d x = \int e^{x} (\frac{1 - \cos 2 x}{2}) d x = \frac{1}{2} \int e^{x} d x - \frac{1}{2} \int e^{x} \cos 2 x d x = \frac{e^{x}}{2} - \frac{1}{2} \int e^{x} \cos (2 x) d x . . . . . (1) Let I_{1} = \int e^{x} \cos (2 x) d x Considering \cos (2 x) as first function and e^{x} as second function I_{1} = \cos (2 x) e^{x} - \int - 2 \sin (2 x) e^{x} d x \Rightarrow I_{1} = \cos (2 x) e^{x} + 2 \int \sin (2 x) e^{x} d x \Rightarrow I_{1} = \cos (2 x) e^{x} + 2 [\sin (2 x) e^{x} - \int 2 \cos (2 x) e^{x} d x] \Rightarrow I_{1} = \cos (2 x) e^{x} + 2 \sin (2 x) e^{x} - 4 I_{1} \Rightarrow 5 I_{1} = e^{x} (\cos 2 x + 2 \sin 2 x) \Rightarrow I_{1} = \frac{e^{x}}{5} (\cos 2 x + 2 \sin 2 x) + C . . . . . (2) From (1) & (2) I = \frac{e^{x}}{2} - \frac{e^{x}}{10} (\cos 2 x + 2 \sin 2 x) + C$

Page No 18.149:

Question 9:

$\int \frac{1}{x^{3}} \sin (\log x) d x$

Answer:

$Let I = \int \frac{1}{x^{3}} \sin (\log x) d x Putting \log x = t \Rightarrow x = e^{t} \Rightarrow d x = e^{t} d t ∴ I = \int \frac{1}{e^{3 t}} \sin t e^{t} d t = \int e^{- 2 t} \sin t d t Considering \sin t as first function and e^{- 2 t} as second function I = \sin t [\frac{e^{- 2 t}}{- 2}] - \int \cos t \frac{e^{- 2 t}}{- 2} d t \Rightarrow I = \frac{\sin t e^{- 2 t}}{- 2} + \frac{1}{2} \int \cos t e^{- 2 t} d t \Rightarrow I = \frac{\sin t e^{- 2 t}}{- 2} + \frac{1}{2} [\cos t \frac{e^{- 2 t}}{- 2} - \int (- \sin t) \frac{e^{- 2 t}}{- 2} d t] \Rightarrow I = \frac{\sin t e^{- 2 t}}{- 2} - \frac{1}{4} \cos t e^{- 2 t} - \int \frac{e^{- 2 t} \sin t d t}{4} \Rightarrow I = e^{- 2 t} [\frac{- 2 \sin t - \cos t}{4}] - \frac{I}{4} \Rightarrow \frac{5 I}{4} = e^{- 2 t} [\frac{- 2 \sin t - \cos t}{4}] \Rightarrow I = \frac{e^{- 2 t}}{5} [- 2 \sin t - \cos t] + C \Rightarrow I = \frac{- x^{- 2}}{5} [2 \sin (\log x) + \cos (\log x)] + C \Rightarrow I = \frac{- 1}{5 x^{2}} [\cos (\log x) + 2 \sin (\log x)] + C$

Page No 18.149:

Question 10:

$\int e^{2 x} \cos^{2} x d x$

Answer:

$Let I = \int e^{2 x} \cos^{2} x d x = \int e^{2 x} (\frac{1 + \cos 2 x}{2}) d x = \frac{1}{2} \int e^{2 x} d x + \frac{1}{2} \int e^{2 x} \cos (2 x) d x = \frac{e^{2 x}}{4} + \frac{1}{2} I_{1} . . . . . (1) Where I_{1} = \int e^{2 x} \cos 2 x d x Considering \cos (2 x) as first function and e^{2 x} as second function I_{1} = \cos (2 x) \frac{e^{2 x}}{2} - \int (- 2 \sin 2 x \times \frac{e^{2 x}}{2}) d x \Rightarrow I_{1} = \frac{\cos (2 x) e^{2 x}}{2} + \int e^{2 x} \sin (2 x) d x Now considering \sin (2 x) as first function and e^{2 x} as second function I_{1} = \frac{\cos (2 x) e^{2 x}}{2} + \sin (2 x) \frac{e^{2 x}}{2} - \int 2 \cos 2 x \frac{e^{2 x}}{2} d x \Rightarrow I_{1} = \frac{e^{2 x} (\cos 2 x + \sin 2 x)}{2} - I_{1} \Rightarrow 2 I_{1} = \frac{e^{2 x} (\cos 2 x + \sin 2 x)}{2} \Rightarrow I_{1} = \frac{e^{2 x} (\cos 2 x + \sin 2 x)}{4} . . . . . (2) From (1) & (2) I = \frac{e^{2 x}}{4} + \frac{e^{2 x}}{8} (\cos 2 x + \sin 2 x) + C$

Page No 18.149:

Question 11:

$\int e^{- 2 x} \sin x d x$

Answer:

$Let I = \int e^{- 2 x} \sin x d x Considering \sin x as first function and e^{- 2 x} as second function I = \sin x \frac{e^{- 2 x}}{- 2} - \int \cos x (\frac{e^{- 2 x}}{- 2}) d x \Rightarrow I = - \frac{e^{- 2 x} \sin x}{2} + \frac{1}{2} \int e^{- 2 x} \cos x d x \Rightarrow I = - \frac{e^{- 2 x} \sin x}{2} + \frac{I_{1}}{2} where . . . . . (1) Where, I_{1} = \int e^{- 2 x} \cos x d x Considering \cos x as first and e^{- 2 x} as second function I_{1} = \frac{\cos x e^{- 2 x}}{- 2} - \int (- \sin x) \frac{e^{- 2 x}}{- 2} d x \Rightarrow I_{1} = \frac{e^{- 2 x} \cos x}{- 2} - \int \frac{\sin x e^{- 2 x} d x}{2} \Rightarrow I_{1} = \frac{- e^{- 2 x} \cos x}{2} - \frac{I}{2} . . . . . (2) From (1) & (2) I = \frac{- e^{- 2 x} \sin x}{2} + \frac{1}{2} [\frac{- e^{- 2 x} \cos x}{2} - \frac{I}{2}] \Rightarrow I + \frac{I}{4} = \frac{- e^{- 2 x} \sin x}{2} - \frac{e^{- 2 x} \cos x}{4} \Rightarrow \frac{5 I}{4} = \frac{- e^{- 2 x} (2 \sin x + \cos x)}{4} ∴ I = \frac{e^{- 2 x}}{5} (- 2 \sin x - \cos x) + C$

Page No 18.149:

Question 12:

$\int x^{2} e^{x^{3}} \cos x^{3} d x$

Answer:

$Given integral is, \int x^{2} e^{x^{3}} \cos (x^{3}) d x Let x^{3} = t \Rightarrow 3 x^{2} d x = d t \Rightarrow x^{2} d x = \frac{d t}{3} Integral becomes, \frac{1}{3} \int e^{t} \cos t d t = \frac{1}{3} I . . . . . (1) Where, I = \int e^{t} \cos t d t I = \int e^{t} \cos t d t Considering \cos t as first and e^{t} as second function I = \cos t e^{t} - \int - \sin t e^{t} d t \Rightarrow I = e^{t} \cos t + \int \sin t e^{t} d t Again considering \sin t as first and e^{t} as second function I = e^{t} \cos t + \sin t e^{t} - \int \cos t e^{t} d t \Rightarrow I = e^{t} \cos t + \sin t e^{t} - I \Rightarrow 2 I = e^{t} (\sin t + \cos t) \Rightarrow I = \frac{e^{t}}{2} (\sin t + \cos t) ∴ \int x^{2} e^{x^{3}} \cos (x^{3}) d x = \frac{1}{3} [\frac{e^{t}}{2} (\sin t + \cos t)] + C [From (1)] = \frac{e^{x^{3}}}{6} (\sin x^{3} + \cos x^{3}) + C$

Page No 18.15:

Question 15:

$\int \sqrt{x} (3 - 5 x) d x$

Answer:

$\int \sqrt{x} (3 - 5 x) d x = \int x^{\frac{1}{2}} (3 - 5 x) d x = \int (3 x^{\frac{1}{2}} - 5 x^{\frac{3}{2}}) d x = 3 [\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] - 5 [\frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + C = 2 x^{\frac{3}{2}} - 2 x^{\frac{5}{2}} + C$

Page No 18.15:

Question 16:

$\int \frac{(x + 1) (x - 2)}{\sqrt{x}} d x$

Answer:

$\int \frac{(x + 1) (x - 2)}{\sqrt{x}} d x = \int (\frac{x^{2} - 2 x + x - 2}{\sqrt{x}}) d x = \int (\frac{x^{2} - x - 2}{\sqrt{x}}) d x = \int (x^{\frac{3}{2}} - x^{\frac{1}{2}} - 2 x^{- \frac{1}{2}}) d x = [\frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] - [\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] - 2 [\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = \frac{2}{5} x^{\frac{5}{2}} - \frac{2}{3} x^{\frac{3}{2}} - 4 x^{\frac{1}{2}} + C$

Page No 18.15:

Question 17:

$\int \frac{x^{5} + x^{- 2} + 2}{x^{2}} d x$

Answer:

$\int (\frac{x^{5} + x^{- 2} + 2}{x^{2}}) d x = \int (\frac{x^{5}}{x^{2}} + \frac{x^{- 2}}{x^{2}} + \frac{2}{x^{2}}) d x = \int (x^{3} + x^{- 4} + 2 x^{- 2}) d x = \frac{x^{3 + 1}}{3 + 1} + \frac{x^{- 4 + 1}}{- 4 + 1} + 2 \frac{x^{- 2 + 1}}{- 2 + 1} + C = \frac{x^{4}}{4} - \frac{1}{3 x^{3}} - \frac{2}{x} + C$

Page No 18.15:

Question 18:

$\int {(3 x + 4)}^{2} d x$

Answer:

$\int {(3 x + 4)}^{2} d x = \int (9 x^{2} + 2 \times 3 x \times 4 + 16) d x = 9 \int x^{2} d x + 24 \int x d x + 16 \int d x = 9 [\frac{x^{3}}{3}] + 24 [\frac{x^{2}}{2}] + 16 x + C = 3 x^{3} + 12 x^{2} + 16 x + C$

Page No 18.15:

Question 19:

$\int \frac{2 x^{4} + 7 x^{3} + 6 x^{2}}{x^{2} + 2 x} d x$

Answer:

$\int (\frac{2 x^{4} + 7 x^{3} + 6 x^{2}}{x^{2} + 2 x}) d x = \int \frac{x^{2} (2 x^{2} + 7 x + 6)}{x (x + 2)} d x = \int \frac{x [2 x^{2} + 4 x + 3 x + 6]}{x + 2} d x = \int \frac{x (2 x (x + 2) + 3 (x + 2))}{(x + 2)} d x = \int \frac{x (2 x + 3) (x + 2)}{(x + 2)} d x = \int (2 x^{2} + 3 x) d x = 2 \int x^{2} d x + 3 \int x d x = 2 [\frac{x^{3}}{3}] + 3 [\frac{x^{2}}{2}] + C = \frac{2}{3} x^{3} + \frac{3}{2} x^{2} + C$

Page No 18.15:

Question 20:

$\int \frac{5 x^{4} + 12 x^{3} + 7 x^{2}}{x^{2} + x} d x$

Answer:

$\int (\frac{5 x^{4} + 12 x^{3} + 7 x^{2}}{x^{2} + x}) d x = \int \frac{x^{2} (5 x^{2} + 12 x + 7)}{x (x + 1)} d x = \int x \frac{(5 x^{2} + 5 x + 7 x + 7)}{(x + 1)} d x = \int \frac{x (5 x (x + 1) + 7 (x + 1))}{(x + 1)} d x = \int x \frac{(5 x + 7) (x + 1)}{(x + 1)} d x = \int (5 x^{2} + 7 x) d x = \frac{5 x^{3}}{3} + \frac{7 x^{2}}{2} + C$

Page No 18.15:

Question 21:

$\int \frac{\sin^{2} x}{1 + \cos x} d x$

Answer:

$\int \frac{\sin^{2} x}{1 + \cos x} d x = \int \frac{(1 - \cos^{2} x)}{(1 + \cos x)} d x = \int \frac{(1 - \cos x) (1 + \cos x)}{(1 + \cos x)} d x = \int (1 - \cos x) d x = x - \sin x + C$

Page No 18.15:

Question 22:

$\int (\sec^{2} x + {cosec}^{2} x) d x$

Answer:

$\int (\sec^{2} x + {cosec}^{2} x) d x = \int \sec^{2} x d x + \int {cosec}^{2} x d x = \tan x - \cot x + C$

Page No 18.15:

Question 23:

$\int \frac{\sin^{3} x - \cos^{3} x}{\sin^{2} x \cos^{2} x} d x$

Answer:

$\int (\frac{\sin^{3} x - \cos^{3} x}{\sin^{2} x \cdot \cos^{2} x}) d x = \int \frac{\sin^{3} x}{\sin^{2} x \cdot \cos^{2} x} d x - \int \frac{\cos^{3} x}{\sin^{2} x \cdot \cos^{2} x} d x = \int \frac{\sin x}{\cos^{2} x} d x - \int \frac{\cos x}{\sin^{2} x} d x = \int \frac{\sin x}{\cos x} \times \frac{1}{\cos x} d x - \int \frac{\cos x}{\sin x} \times \frac{1}{\sin x} d x = \int \sec x \tan x d x - \int cosec x \cot x d x = \sec x - (- cosec x) + C = \sec x + cosec x + C$

Page No 18.15:

Question 24:

$\int \frac{5 \cos^{3} x + 6 \sin^{3} x}{2 \sin^{2} x \cos^{2} x} d x$

Answer:

$\int (\frac{5 \cos^{3} x + 6 \sin^{3} x}{2 \sin^{2} x \cos^{2} x}) d x = \int (\frac{5 \cos^{3} x}{2 \sin^{2} x \cos^{2} x} + \frac{6 \sin^{3} x}{2 \sin^{2} x \cos^{2} x}) d x = \int (\frac{5}{2} \frac{\cos x}{\sin^{2} x} + 3 \frac{\sin x}{\cos^{2} x}) d x = \frac{5}{2} \int (\frac{\cos x}{\sin x} \times \frac{1}{\sin x}) d x + 3 \int \frac{\sin x}{\cos x} \times \frac{1}{\cos x} d x = \frac{5}{2} \int (cosec x \cot x) d x + 3 \int \sec x \tan x d x = \frac{5}{2} (- cosec x) + 3 \sec x + C = - \frac{5}{2} cosec x + 3 \sec x + C$

Page No 18.15:

Question 25:

$\int {(\tan x + \cot x)}^{2} d x$

Answer:

$\int {(\tan x + \cot x)}^{2} = \int (\tan^{2} x + \cot^{2} x + 2 \tan x \cot x) d x = \int (\tan^{2} x + \cot^{2} x + 2) d x = \int [(\sec^{2} x - 1) + ({cosec}^{2} x - 1) + 2] d x = \int (\sec^{2} x + {cosec}^{2} x) d x = \tan x - \cot x + C$

Page No 18.15:

Question 26:

$\int \frac{1 - \cos 2 x}{1 + \cos 2 x} d x$

Answer:

$\int (\frac{1 - \cos 2 x}{1 + \cos 2 x}) d x = \int \frac{2 \sin^{2} x}{2 \cos^{2} x} d x [∵ 1 - \cos 2 θ = 2 \sin^{2} θ & 1 + \cos 2 θ = 2 \cos^{2} θ] = \int \tan^{2} x d x = \int (\sec^{2} x - 1) d x = \int \sec^{2} x d x - \int d x = \tan x - x + C$

Page No 18.15:

Question 27:

$\int \frac{\cos x}{1 - \cos x} d x or \int \frac{\cot x}{cosec x - \cot x} d x$

Answer:

$\int \frac{\cot x}{cosec x - \cot x} d x = \int \frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x} - \frac{\cos x}{\sin x}} d x = \int (\frac{\cos x}{1 - \cos x}) \times \frac{(1 + \cos x)}{(1 + \cos x)} d x = \int (\frac{\cos x + \cos^{2} x}{1 - \cos^{2} x}) d x = \int (\frac{\cos x + \cos^{2} x}{\sin^{2} x}) d x = \int (\frac{\cos x}{\sin x} \times \frac{1}{\sin x} + \frac{\cos^{2} x}{\sin^{2} x}) d x = \int [(\cot x cosec x) + \cot^{2} x] d x = \int [cosec x \cot x + {cosec}^{2} x - 1] d x = - cosec x - \cot x - x + C$

Page No 18.15:

Question 28:

$\int \frac{\cos^{2} x - \sin^{2} x}{\sqrt{1} + \cos 4 x} d x$

Answer:

$\int (\frac{\cos^{2} x - \sin^{2} x}{\sqrt{1 + \cos 4 x}}) d x = \int \frac{\cos (2 x)}{\sqrt{2 \cos^{2} (2 x)}} d x [∴ 1 + \cos A = 2 \cos^{2} (\frac{A}{2}) & \cos^{2} A - \sin^{2} A = \cos 2 A] = \frac{1}{\sqrt{2}} \int (\frac{\cos 2 x}{\cos 2 x}) d x = \frac{1}{\sqrt{2}} [x] + C = \frac{x}{\sqrt{2}} + C$

Page No 18.15:

Question 29:

$\int \frac{1}{1 - \cos x} d x$

Answer:

$\int \frac{d x}{1 - \cos x} = \int \frac{d x}{1 - \cos x} \times \frac{1 + \cos x}{1 + \cos x} = \int (\frac{1 + \cos x}{1 - \cos^{2} x}) d x = \int (\frac{1 + \cos x}{\sin^{2} x}) d x = \int (\frac{1}{\sin^{2} x} + \frac{\cos x}{\sin x} \times \frac{1}{\sin x}) d x = \int ({cosec}^{2} x + cosec x \cot x) d x = - \cot x - cosec x + C$

Page No 18.15:

Question 30:

$\int \frac{1}{1 - \sin x} d x$

Answer:

$\int \frac{d x}{1 - \sin x} = \int \frac{(1 + \sin x)}{(1 - \sin x) \times (1 + \sin x)} d x = \int (\frac{1 + \sin x}{1 - \sin^{2} x}) d x = \int (\frac{1 + \sin x}{\cos^{2} x}) d x = \int (\frac{1}{\cos^{2} x} + \frac{\sin x}{\cos x} \times \frac{1}{\cos x}) d x = \int (\sec^{2} x + \sec x \tan x) d x = \tan x + \sec x + C$

Page No 18.15:

Question 31:

$\int \frac{\tan x}{\sec x + \tan x} d x$

Answer:

$\int \frac{\tan x}{\sec x + \tan x} d x = \int \frac{\tan x}{(\sec x + \tan x)} \times (\frac{\sec x - \tan x}{\sec x - \tan x}) d x = \int \frac{\tan x (\sec x - \tan x)}{(\sec^{2} x - \tan^{2} x)} d x = \int (\frac{\sec x \tan x - \tan^{2} x}{1}) d x = \int \sec x \tan x d x - \int (s e c^{2} x - 1) d x = \sec x - \tan x + x + C$

Page No 18.15:

Question 32:

$\int \frac{cosec x}{cosec x - \cot x} d x$

Answer:

$\int (\frac{cosec x}{cosec x - \cot x}) d x = \int \frac{cosec x (cosec x + \cot x)}{(cosec x - \cot x) (cosec x + \cot x)} d x = \int \frac{cosec x (cosec x + \cot x)}{({cosec}^{2} x - \cot^{2} x)} d x = \int ({cosec}^{2} x + cosec x \cot x) d x = - \cot x - cosec x + C$

Page No 18.15:

Question 33:

$\int \frac{1}{1 + \cos 2 x} d x$

Answer:

$\int \frac{d x}{1 + \cos (2 x)} [∴ 1 + \cos θ = 2 \cos^{2} (\frac{θ}{2})] = \int \frac{d x}{2 \cos^{2} x} = \frac{1}{2} \int \sec^{2} x d x = \frac{1}{2} \tan x + C$

Page No 18.15:

Question 34:

$\int \frac{1}{1 - \cos 2 x} d x$

Answer:

$\int \frac{d x}{1 - \cos (2 x)} [∴ 1 - \cos A = 2 \sin^{2} (\frac{A}{2})] = \int \frac{d x}{2 \sin^{2} x} = \frac{1}{2} \int {cosec}^{2} x d x = \frac{1}{2} [- \cot x] + C = - \frac{1}{2} \cot x + C$

Page No 18.15:

Question 35:

$\int \tan^{- 1} (\frac{\sin 2 x}{1 + \cos 2 x}) d x$

Answer:

$\int \tan^{- 1} [\frac{\sin (2 x)}{1 + \cos 2 x}] d x = \int \tan^{- 1} [\frac{2 \sin x \cos x}{2 \cos^{2} x}] d x [∴ \sin 2 x = 2 \sin x \cos x & 1 + \cos 2 x = 2 \cos^{2} x] = \int \tan^{- 1} [\tan x] = \int \tan^{- 1} [\tan x] = \int x d x = \frac{x^{2}}{2} + C$

Page No 18.15:

Question 36:

$\int \cos^{- 1} (\sin x) d x$

Answer:

$\int \cos^{- 1} (\sin x) d x = \int \cos^{- 1} (\cos (\frac{π}{2} - x)) d x [∴ \sin x = \cos (\frac{π}{2} - x)] = \int (\frac{π}{2} - x) d x = \frac{π x}{2} - \frac{x^{2}}{2} + C$

Page No 18.15:

Question 37:

$\int \cot^{- 1} (\frac{\sin 2 x}{1 - \cos 2 x}) d x$

Answer:

$\int \cot^{- 1} (\frac{\sin 2 x}{1 - \cos 2 x}) d x = \int \cot^{- 1} (\frac{2 \sin x \cos x}{2 \sin^{2} x}) d x [∴ \sin 2 x = 2 \sin x \cos x & 1 - \cos 2 x = 2 \sin^{2} x] = \int \cot^{- 1} (\cot x) d x = \int x d x = \frac{x^{2}}{2} + C$

Page No 18.15:

Question 38:

$\int \sin^{- 1} (\frac{2 \tan x}{1 + \tan^{2} x}) d x$

Answer:

$\int \sin^{- 1} (\frac{2 \tan x}{1 + \tan^{2} x}) d x = \int \sin^{- 1} (\sin 2 x) d x [∴ \sin 2 x = \frac{2 \tan x}{1 + \tan^{2} x}] = 2 \int x d x = 2 (\frac{x^{2}}{2}) + C = x^{2} + C$

Page No 18.15:

Question 39:

$\int \frac{(x^{3} + 8) (x - 1)}{x^{2} - 2 x + 4} d x$

Answer:

$\int \frac{(x^{3} + 8) (x - 1)}{(x^{2} - 2 x + 4)} d x = \int \frac{(x^{3} + 2^{3}) (x - 1)}{(x^{2} - 2 x + 4)} d x = \int \frac{(x + 2) (x^{2} - 2 x + 4) (x - 1)}{(x^{2} - 2 x + 4)} d x [∴ a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})] = \int (x + 2) (x - 1) d x = \int (x^{2} - x + 2 x - 2) d x = \int (x^{2} + x - 2) d x = \int x^{2} d x + \int x d x - 2 \int 1 d x = \frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x + C$

Page No 18.15:

Question 40:

$\int {(a \tan x + b \cot x)}^{2} d x$

Answer:

$\int {(a \tan x + b \cot x)}^{2} d x = \int (a^{2} \tan^{2} x + b^{2} \cot^{2} x + 2 a b \tan x \cot x) d x = a^{2} \int \tan^{2} x d x + b^{2} \int \cot^{2} x d x + 2 a b \int d x = a^{2} \int (\sec^{2} x - 1) d x + b^{2} \int ({cosec}^{2} x - 1) d x + 2 a b \int d x = a^{2} [\tan x - x] + b^{2} [- \cot x - x] + 2 a b x + C = a^{2} \tan x - b^{2} \cot x - (a^{2} + b^{2} - 2 a b) x + C$

Page No 18.15:

Question 41:

$\int \frac{x^{3} - 3 x^{2} + 5 x - 7 + x^{2} a^{x}}{2 x^{2}} d x$

Answer:

$\int (\frac{x^{3} - 3 x^{2} + 5 x - 7 + x^{2} a^{x}}{2 x^{2}}) d x = \int (\frac{x^{3}}{2 x^{2}} - \frac{3 x^{2}}{2 x^{2}} + \frac{5 x}{2 x^{2}} - \frac{7}{2 x^{2}} + \frac{x^{2} a^{x}}{2 x^{2}}) d x = \int (\frac{x}{2} - \frac{3}{2} + \frac{5}{2 x} - \frac{7}{2} x^{- 2} + \frac{a^{x}}{2}) d x = \frac{1}{2} \int x d x - \frac{3}{2} \int d x + \frac{5}{2} \int \frac{d x}{x} - \frac{7}{2} \int x^{- 2} d x + \frac{1}{2} \int a^{x} d x = \frac{1}{2} [\frac{x^{2}}{2}] - \frac{3}{2} x + \frac{5}{2} \ln |x| - \frac{7}{2} [\frac{x^{- 2 + 1}}{- 2 + 1}] + \frac{1}{2} [\frac{a^{x}}{\ln a}] + C = \frac{x^{2}}{4} - \frac{3}{2} x + \frac{5}{2} \ln |x| + \frac{7}{2 x} + \frac{a^{x}}{2 \ln a} + C = \frac{1}{2} [\frac{x^{2}}{2} - 3 x + 5 \ln |x| + \frac{7}{x} + \frac{a^{x}}{\ln a}] + C$

Page No 18.15:

Question 42:

$\int \frac{\cos x}{1 + \cos x} d x$

Answer:

$\int \frac{\cos x}{1 + \cos x} d x = \int \frac{\cos x (1 - \cos x)}{(1 + \cos x) (1 - \cos x)} d x = \int \frac{\cos x - \cos^{2} x}{1 - \cos^{2} x} d x = \int \frac{\cos x - \cos^{2} x}{\sin^{2} x} d x = \int \frac{\cos x}{\sin^{2} x} - \frac{\cos^{2} x}{\sin^{2} x} d x = \int (\cot x cosec x - \cot^{2} x) d x = \int (\cot x cosec x - co \sec^{2} x + 1) d x = \int \cot x cosec x d x - \int co \sec^{2} x d x + \int 1 d x = - cosec x + \cot x + x + C$

Page No 18.15:

Question 43:

$\int \frac{1 - \cos x}{1 + \cos x} d x$

Answer:

$\int (\frac{1 - \cos x}{1 + \cos x}) d x = \int \frac{{(1 - \cos x)}^{2}}{1 - \cos^{2} x} d x = \int \frac{1 + \cos^{2} x - 2 \cos x}{\sin^{2} x} d x = \int (\frac{1}{\sin^{2} x} + \frac{\cos^{2} x}{\sin^{2} x} - \frac{2 \cos x}{\sin^{2} x}) d x = \int ({cosec}^{2} x + \cot^{2} x - 2 \cot x . cosec x) d x = \int ({cosec}^{2} x + {cosec}^{2} x - 1 - 2 \cot x . cosec x) d x = \int (2 {cosec}^{2} x - 1 - 2 \cot x . cosec x) d x = \int 2 {cosec}^{2} x d x - \int 1 d x - \int 2 \cot x . cosec x d x = - 2 \cot x - x + 2 cosec x + C = 2 (cosec x - \cot x) - x + C$

Page No 18.15:

Question 45:

If f' (x) = x − $\frac{1}{x^{2}}$ and f(1) = $\frac{1}{2},$ find f(x)

Answer:

$f' (x) = x - \frac{1}{x^{2}} f' (x) = x - x^{- 2} \int f' (x) d x = \int (x - x^{- 2}) d x f (x) = \frac{x^{2}}{2} - \frac{x^{- 2 + 1}}{- 2 + 1} + C = \frac{x^{2}}{2} + \frac{1}{x} + C f (1) = \frac{1}{2} (Given) \Rightarrow \frac{1^{2}}{2} + \frac{1}{1} + C = \frac{1}{2} \Rightarrow C = - 1 ∴ f (x) = \frac{x^{2}}{2} + \frac{1}{x} - 1$

Page No 18.15:

Question 46:

If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)

Answer:

$f' (x) = x + b, f (1) = 5, f (2) = 13 f' (x) = x + b \int f' (x) d x = \int (x + b) d x f (x) = \frac{x^{2}}{2} + b x + C . . . . (i) f (1) = 5, f (2) = 13 (Given) Puting x = 1 in (i) f (1) = \frac{1^{2}}{2} + b 1 + C 5 = \frac{1}{2} + b + C . . . (i i) Puting x = 2 in (i) f (2) = \frac{2^{2}}{2} + b 2 + C 13 = \frac{4}{2} + 2 b + C 13 = 2 + 2 b + C . . . (i i i) Solving (i i) and (i i i) we get, b = \frac{13}{2} and C = - 2 Thus, f (x) = \frac{x^{2}}{2} + \frac{13}{2} x - 2$

Page No 18.15:

Question 47:

If f' (x) = 8x³ − 2x, f(2) = 8, find f(x)

Answer:

$f' (x) = 8 x^{3} - 2 x f (2) = 8 f' (x) = 8 x^{3} - 2 x \int f' (x) d x = \int (8 x^{3} - 2 x) d x = 8 \int x^{3} d x - 2 \int x d x f (x) = 8 [\frac{x^{4}}{4}] - 2 \times \frac{x^{2}}{2} + C f (x) = 2 x^{4} - x^{2} + C f (2) = 8 (Given) f (2) = 2 \times 2^{4} - 2^{2} + C 8 = 32 - 4 + C C = - 20 ∴ f (x) = 2 x^{4} - x^{2} - 20$

Page No 18.15:

Question 48:

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f $(\frac{π}{2})$ = 5, find f(x)

Answer:

$f' (x) = a \sin x + b \cos x f' (0) = 4, f (0) = 3 f (\frac{π}{2}) = 5 f' (x) = a \sin x + b \cos x \int f' (x) d x = \int (a \sin x + b \cos x) d x f (x) = - a \cos x + b \sin x + C . . . (i) Now puting x = 0 in equation (i) f (0) = - a \cos 0 + b \sin 0 + C 3 = - a \times 1 + b \times 0 + C 3 = - a + C . . . (i i) Now puting x = \frac{π}{2} in equation (i) f (\frac{π}{2}) = - a \cos \frac{π}{2} + b \sin \frac{π}{2} + C 5 = - a \cos \frac{π}{2} + b \sin (\frac{π}{2}) + C 5 = - a \times 0 + b \times 1 + C 5 = b + C . . . (i i i) We also have f' (0) = 4 f' (x) = a \sin x + b \cos x f' (0) = a \sin 0 + b \cos 0 4 = a \times 0 + b \times 1 4 = b . . . (i v) solving (i i), (i i i) and (i v) we get, b = 4 C = 1 a = - 2 ∴ f (x) = 2 \cos x + 4 \sin x + 1$

Page No 18.15:

Question 49:

Write the primitive or anti-derivative of $f (x) = \sqrt{x} + \frac{1}{\sqrt{x}} .$

Answer:

$f (x) = \sqrt{x} + \frac{1}{\sqrt{x}}$
Integrating both sides
$\int f (x) d x = \int (\sqrt{x} + \frac{1}{\sqrt{x}}) d x = \int (x^{\frac{1}{2}} + x^{- \frac{1}{2}}) d x = [\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + [\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = \frac{2}{3} x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + C$

Page No 18.154:

Question 1:

$\int \sqrt{3 + 2 x - x^{2}} d x$

Answer:

$\int \sqrt{3 + 2 x - x^{2}} d x = \int \sqrt{3 - (x^{2} - 2 x)} d x = \int \sqrt{3 - (x^{2} - 2 x + 1 - 1)} d x = \int \sqrt{4 - {(x - 1)}^{2}} d x = \int \sqrt{2^{2} - {(x - 1)}^{2}} d x [∵ \int \sqrt{a^{2} - x^{2}} d x = \frac{1}{2} x \sqrt{a^{2} - x^{2}} + \frac{1}{2} a^{2} \sin^{- 1} \frac{x}{a} + C] = (\frac{x - 1}{2}) \sqrt{2^{2} - {(x - 1)}^{2}} + \frac{2^{2}}{2} \sin^{- 1} (\frac{x - 1}{2}) + C = \frac{x - 1}{2} \sqrt{3 + 2 x - x^{2}} + \sin^{- 1} (\frac{x - 1}{2}) + C$

Page No 18.154:

Question 2:

$\int \sqrt{x^{2} + x + 1} d x$

Answer:

$\int \sqrt{x^{2} + x + 1} d x = \int \sqrt{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} d x = \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = (\frac{x + \frac{1}{2}}{2}) \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} + \frac{3}{8} \log |(x + \frac{1}{2}) + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C [∵ \int \sqrt{x^{2} + a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} + a^{2}} + \frac{1}{2} a^{2} \ln |x + \sqrt{x^{2} + a^{2}}| + C] = (\frac{2 x + 1}{4}) \sqrt{x^{2} + x + 1} + \frac{3}{8} \log |(2 x + 1) + \frac{1}{2} + \sqrt{x^{2} + x + 1}| + C$

Page No 18.154:

Question 3:

$\int \sqrt{x - x^{2}} d x$

Answer:

$\int \sqrt{x - x^{2}} d x = \int \sqrt{- (x^{2} - x)} d x = \int \sqrt{- \{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}\}} d x = \int \sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{1}{2})}^{2}} d x = (\frac{x - \frac{1}{2}}{2}) \sqrt{x - x^{2}} + \frac{1}{8} \sin^{- 1} (\frac{x - \frac{1}{2}}{\frac{1}{2}}) + C [∵ \int \sqrt{a^{2} - x^{2}} d x = \frac{1}{2} x \sqrt{a^{2} - x^{2}} + \frac{1}{2} a^{2} \sin^{- 1} \frac{x}{a} + C] = (\frac{2 x - 1}{4}) \sqrt{x - x^{2}} + \frac{1}{8} \sin^{- 1} (2 x - 1) + C$

Page No 18.154:

Question 4:

$\int \sqrt{1 + x - 2 x^{2}} d x$

Answer:

$\int \sqrt{1 + x - 2 x^{2}} d x = \int \sqrt{2 (\frac{1}{2} + \frac{x}{2} - x^{2})} d x = \sqrt{2} \int \sqrt{\frac{1}{2} - (x^{2} - \frac{x}{2})} d x = \sqrt{2} \int \sqrt{\frac{1}{2} - (x^{2} - \frac{x}{2} + \frac{1}{4^{2}} - \frac{1}{4^{2}})} d x = \sqrt{2} \int \sqrt{\frac{1}{2} + \frac{1}{16} - {(x - \frac{1}{4})}^{2}} d x = \sqrt{2} \int \sqrt{{(\frac{3}{4})}^{2} - {(x - \frac{1}{4})}^{2}} d x = \sqrt{2} [(\frac{x - \frac{1}{4}}{2}) \sqrt{{(\frac{3}{4})}^{2} - {(x - \frac{1}{4})}^{2}} + \frac{9}{32} \sin^{- 1} (\frac{x - \frac{1}{4}}{\frac{3}{4}})] + C [∵ \int \sqrt{a^{2} - x^{2}} d x = \frac{1}{2} x \sqrt{a^{2} - x^{2}} + \frac{1}{2} a^{2} \sin^{- 1} \frac{x}{a} + C] = (\frac{4 x - 1}{8}) \sqrt{1 + x - 2 x^{2}} + \frac{9 \sqrt{2}}{32} \sin^{- 1} (\frac{4 x - 1}{3}) + C$

Page No 18.154:

Question 5:

$\int \cos x \sqrt{4 - \sin^{2} x} d x$

Answer:

$Let I = \int \cos x \sqrt{4 - \sin^{2} x} d x Putting \sin x = t \Rightarrow \cos x d x = d t \int \sqrt{2^{2} - t^{2}} d t = \frac{t}{2} \sqrt{2^{2} - t^{2}} + \frac{2^{2}}{2} \sin^{- 1} (\frac{t}{2}) + C [∵ \int \sqrt{a^{2} - x^{2}} d x = \frac{1}{2} x \sqrt{a^{2} - x^{2}} + \frac{1}{2} a^{2} \sin^{- 1} (\frac{x}{a}) + C] = \frac{\sin x}{2} \sqrt{4 - \sin^{2} x} + 2 \sin^{- 1} (\frac{\sin x}{2}) + C [∵ t = \sin x]$

Page No 18.154:

Question 6:

$\int e^{x} \sqrt{e^{2 x} + 1} d x$

Answer:

$Let I = \int e^{x} \sqrt{e^{2 x} + 1} d x Putting e^{x} = t \Rightarrow e^{x} d x = d t ∴ I = \int \sqrt{t^{2} + 1} d t = \frac{t}{2} \sqrt{t^{2} + 1} + \frac{1^{2}}{2} \ln |t + \sqrt{t^{2} + 1}| + C [∵ \int \sqrt{x^{2} + a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} + a^{2}} + \frac{1}{2} \ln |x + \sqrt{x^{2} + a^{2}}| + C] = \frac{e^{x}}{2} \sqrt{e^{2 x} + 1} + \frac{1}{2} \ln |e^{x} + \sqrt{e^{2 x} + 1}| + C (∵ t = e^{x})$

Page No 18.154:

Question 7:

$\int \sqrt{9 - x^{2}} d x$

Answer:

$\int \sqrt{9 - x^{2}} d x = \int \sqrt{3^{2} - x^{2}} d x = \frac{x}{2} \sqrt{3^{2} - x^{2}} + \frac{3^{2}}{2} \sin^{- 1} (\frac{x}{3}) + C [∵ \int \sqrt{a^{2} - x^{2}} dx = \frac{1}{2} x \sqrt{a^{2} - x^{2}} + \frac{1}{2} a^{2} \sin^{- 1} (\frac{x}{a}) + C] = \frac{x}{2} \sqrt{9 - x^{2}} + \frac{9}{2} \sin^{- 1} (\frac{x}{3}) + C$

Page No 18.154:

Question 8:

$\int \sqrt{16 x^{2} + 25} d x$

Answer:

$\int \sqrt{16 x^{2} + 25} d x = \int \sqrt{16 (x^{2} + \frac{25}{16})} d x = 4 \int \sqrt{x^{2} + {(\frac{5}{4})}^{2}} d x = 4 [\frac{x}{2} \sqrt{x^{2} + {(\frac{5}{4})}^{2}} + \frac{{(\frac{5}{4})}^{2}}{2} \ln |x + \sqrt{x^{2} + {(\frac{5}{4})}^{2}}|] + C [∵ \int \sqrt{x^{2} + a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} + a^{2}} + \frac{1}{2} a^{2} \ln |x + \sqrt{x^{2} + a^{2}}| + C] = 2 x \sqrt{x^{2} + \frac{25}{16}} + \frac{25}{8} \ln |x + \sqrt{x^{2} + \frac{25}{16}}| + C$

Page No 18.154:

Question 9:

$\int \sqrt{4 x^{2} - 5} d x$

Answer:

$\int \sqrt{4 x^{2} - 5} d x = \int \sqrt{4 (x^{2} - \frac{5}{4})} d x = 2 \int \sqrt{x^{2} - {(\frac{\sqrt{5}}{2})}^{2}} d x = 2 [\frac{x}{2} \sqrt{x^{2} - \frac{5}{4}} - \frac{5}{8} \ln |x + \sqrt{x^{2} - \frac{5}{4}}|] + C [∵ \int \sqrt{x^{2} - a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} - a^{2}} - \frac{1}{2} a^{2} \ln |x + \sqrt{x^{2} - a^{2}}| + C] = x \sqrt{x^{2} - \frac{5}{4}} - \frac{5}{4} \ln |x + \sqrt{x^{2} - \frac{5}{4}}| + C$

Page No 18.154:

Question 10:

$\int \sqrt{2 x^{2} + 3 x + 4} d x$

Answer:

$\int \sqrt{2 x^{2} + 3 x + 4} d x = \sqrt{2} \int \sqrt{x^{2} + \frac{3}{2} x + 2} d x = \sqrt{2} \int \sqrt{x^{2} + \frac{3}{2} x + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{2} + 2} d x = \sqrt{2} \int \sqrt{{(x + \frac{3}{4})}^{2} - \frac{9}{16} + 2} d x = \sqrt{2} \int \sqrt{{(x + \frac{3}{4})}^{2} + {(\frac{\sqrt{23}}{4})}^{2}} d x = \sqrt{2} [\frac{x + \frac{3}{4}}{2} \sqrt{{(x + \frac{3}{4})}^{2} + {(\frac{\sqrt{23}}{4})}^{2}} + \frac{23}{32} \ln |x + \frac{3}{4} + \sqrt{x^{2} + \frac{3}{2} x + 2}|] + C [∵ \int \sqrt{x^{2} + a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} + a^{2}} - \frac{1}{2} a^{2} \ln |x + \sqrt{x^{2} + a^{2}}| + C] = \sqrt{2} [(\frac{4 x + 3}{8}) \sqrt{x^{2} + \frac{3}{2} x + 2} + \frac{23}{32} \ln |x + \frac{3}{4} + \sqrt{x^{2} + \frac{3}{2} x + 2}|] + C = (\frac{4 x + 3}{8}) \sqrt{2 x^{2} + 3 x + 4} + \frac{23 \sqrt{2}}{32} \ln |x + \frac{3}{4} + \sqrt{x^{2} + \frac{3}{2} x + 2}| + C$

Page No 18.154:

Question 11:

$\int \sqrt{3 - 2 x - 2 x^{2}} d x$

Answer:

$Let I = \int \sqrt{3 - 2 x - 2 x^{2}} d x = \int \sqrt{3 - (2 x^{2} + 2 x)} d x = \int \sqrt{3 - 2 (x^{2} + x)} d x = \int \sqrt{3 - 2 (x^{2} + x + \frac{1}{4} - \frac{1}{4})} d x = \int \sqrt{3 - 2 {(x + \frac{1}{2})}^{2} + \frac{1}{2}} d x = \int \sqrt{\frac{7}{2} - 2 {(x + \frac{1}{2})}^{2}} d x = \sqrt{2} \int \sqrt{\frac{7}{4} - {(x + \frac{1}{2})}^{2}} d x = \sqrt{2} \int \sqrt{{(\frac{\sqrt{7}}{2})}^{2} - {(x + \frac{1}{2})}^{2}} d x = \sqrt{2} \times (\frac{x + \frac{1}{2}}{2}) \sqrt{{(\frac{\sqrt{7}}{2})}^{2} - {(x + \frac{1}{2})}^{2}} + \sqrt{2} \times \frac{7}{4 \times 2} \sin^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{7}}{2}}) + C = \frac{2 x + 1}{4} \sqrt{3 - 2 x - 2 x^{2}} + \frac{7}{4 \sqrt{2}} \sin^{- 1} (\frac{2 x + 1}{\sqrt{7}}) + C$

Page No 18.154:

Question 12:

$\int x \sqrt{x^{4} + 1} d x$

Answer:

$Let I = \int x \sqrt{x^{4} + 1} d x = \int x \sqrt{{(x^{2})}^{2} + 1} d x Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \sqrt{t^{2} + 1} d t = \frac{1}{2} \int \sqrt{t^{2} + 1^{2}} d t = \frac{1}{2} [\frac{t}{2} \sqrt{t^{2} + 1} + \frac{1^{2}}{2} \log |t + \sqrt{t^{2} + 1}|] + C = \frac{1}{2} [\frac{x^{2}}{2} \sqrt{x^{4} + 1} + \frac{1}{2} \log |x^{2} + \sqrt{x^{4} + 1}|] + C = \frac{x^{2}}{4} \sqrt{x^{4} + 1} + \frac{1}{4} \log |x^{2} + \sqrt{x^{4} + 1}| + C$

Page No 18.154:

Question 13:

$\int x^{2} \sqrt{a^{6} - x^{6}} d x$

Answer:

$Let I = \int x^{2} \sqrt{a^{6} - x^{6}} d x = \int x^{2} \sqrt{{(a^{3})}^{2} - {(x^{3})}^{2}} d x Putting x^{3} = t \Rightarrow 3 x^{2} d x = d t \Rightarrow x^{2} d x = \frac{d t}{3} ∴ I = \frac{1}{3} \int \sqrt{{(a^{3})}^{2} - t^{2}} d t = \frac{1}{3} [\frac{t}{2} \sqrt{{(a^{3})}^{2} - t^{2}} + \frac{{(a^{3})}^{2}}{2} \sin^{- 1} (\frac{t}{a^{3}})] + C = \frac{x^{3}}{6} \sqrt{a^{6} - x^{6}} + \frac{a^{6}}{6} \sin^{- 1} (\frac{x^{3}}{a^{3}}) + C$

Page No 18.154:

Question 14:

$\int \frac{\sqrt{16 + {(\log x)}^{2}}}{x} d x$

Answer:

$Let I = \int \sqrt{\frac{16 + {(\log x)}^{2}}{x}} d x Putting \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = \int \sqrt{16 + t^{2}} d t = \int \sqrt{4^{2} + t^{2}} d t = \frac{t}{2} \sqrt{4^{2} + t^{2}} + \frac{4^{2}}{2} \log |t + \sqrt{4^{2} + t^{2}}| + C = \frac{\log x}{2} \sqrt{16 + {(\log x)}^{2}} + 8 \log |\log x + \sqrt{16 + {(\log x)}^{2}}| + C$

Page No 18.155:

Question 15:

$\int \sqrt{2 a x - x^{2}} d x$

Answer:

$Let I = \int \sqrt{2 a x - x^{2}} d x = \int \sqrt{a^{2} + 2 a x - x^{2} - a^{2}} d x = \int \sqrt{a^{2} - (x^{2} - 2 a x + a^{2})} d x = \int \sqrt{a^{2} - {(x - a)}^{2}} d x = (\frac{x - a}{2}) \sqrt{2 a x - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x - a}{a}) + C$

Page No 18.155:

Question 16:

$\int \sqrt{3 - x^{2}} d x$

Answer:

$Let I = \int \sqrt{3 - x^{2}} d x = \int \sqrt{{(\sqrt{3})}^{2} - x^{2}} d x = \frac{x}{2} \sqrt{{(\sqrt{3})}^{2} - x^{2}} + \frac{{(\sqrt{3})}^{2}}{2} \sin^{- 1} (\frac{x}{\sqrt{3}}) + C = \frac{x}{2} \sqrt{3 - x^{2}} + \frac{3}{2} \sin^{- 1} (\frac{x}{\sqrt{3}}) + C$

Page No 18.155:

Question 17:

$\int \sqrt{x^{2} - 2 x} d x$

Answer:

$I = \int \sqrt{x^{2} - 2 x} d x$

$\Rightarrow I = \int \sqrt{x^{2} - 2 x + 1 - 1} d x \Rightarrow I = \int \sqrt{(x - 1)^{2} - 1^{2}} d x ∵ \int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \ln (|x + \sqrt{x^{2} - a^{2}}|) + c ∴ I = \frac{(x - 1)}{2} \sqrt{(x - 1)^{2} - 1} - \frac{1}{2} \ln |(x - 1) + \sqrt{x^{2} - 2 x}| + c$

Page No 18.155:

Question 18:

$\int \sqrt{2 x - x^{2}} d x$

Answer:

$I = \int \sqrt{2 x - x^{2}} d x = \int \sqrt{x (2 - x)} d x$

Let $x = 1 + \sin u$

$or, d x = \cos u d u \Rightarrow I = \int \sqrt{(1 + \sin u) (1 - \sin u)} \cos u d u \Rightarrow I = \int \cos^{2} u d u \Rightarrow I = \frac{1}{2} \int (\cos 2 u + 1) d u$

$\Rightarrow I = \frac{1}{2} (\frac{1}{2} \sin 2 u + u) + c \Rightarrow I = \frac{1}{2} (\sin u \cos u + u) + c \Rightarrow I = \frac{1}{2} (\sin u \sqrt{1 - \sin^{2} u} + u) + c ∴ I = \frac{1}{2} (x - 1) \sqrt{2 x - x^{2}} + \frac{1}{2} \sin^{- 1} (x - 1) + c [∵ u = \sin^{- 1} (x - 1)]$

Page No 18.158:

Question 1:

$\int (x + 1) \sqrt{x^{2} - x + 1} d x$

Answer:

$Let I = \int (x + 1) \sqrt{x^{2} - x + 1} d x Also, x + 1 = λ \frac{d}{d x} (x^{2} - x + 1) + μ \Rightarrow x + 1 = λ (2 x - 1) + μ \Rightarrow x + 1 = λ (2 x - 1) + μ \Rightarrow x + 1 = (2 λ) x + μ - λ Equating the coefficient of like terms 2 λ = 1 \Rightarrow λ = \frac{1}{2} And μ - λ = 1 \Rightarrow μ - \frac{1}{2} = 1 \Rightarrow μ = \frac{3}{2} ∴ I = \int [\frac{1}{2} (2 x - 1) + \frac{3}{2}] \sqrt{x^{2} - x + 1} d x = \frac{1}{2} \int (2 x - 1) \sqrt{x^{2} - x + 1} d x + \frac{3}{2} \int \sqrt{x^{2} - x + 1} d x = \frac{1}{2} \int (2 x - 1) \sqrt{x^{2} - x + 1} d x + \frac{3}{2} \int \sqrt{x^{2} - x + \frac{1}{4} - \frac{1}{4} + 1} d x = \frac{1}{2} \int (2 x - 1) \sqrt{x^{2} - x + 1} d x + \frac{3}{2} \int \sqrt{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} d x = \frac{1}{2} \int (2 x - 1) \sqrt{x^{2} - x + 1} d x + \frac{3}{2} \int \sqrt{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x Let x^{2} - x + 1 = t \Rightarrow (2 x - 1) d x = d t ∴ I = \frac{1}{2} \int \sqrt{t} d t + \frac{3}{2} \int \sqrt{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x = \frac{1}{2} \times (\frac{t^{\frac{3}{2}}}{\frac{3}{2}}) + \frac{3}{2} [\frac{x - \frac{1}{2}}{2} \sqrt{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} + \frac{3}{8} \log |x - \frac{1}{2} + \sqrt{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}|] + C = \frac{1}{3} {(x^{2} - x + 1)}^{\frac{3}{2}} + \frac{3}{8} (2 x - 1) \sqrt{x^{2} - x + 1} + \frac{9}{16} \log |x - \frac{1}{2} + \sqrt{x^{2} - x + 1}| + C$

Page No 18.158:

Question 2:

$\int (x + 1) \sqrt{2 x^{2} + 3} d x$

Answer:

$Let I = \int (x + 1) \sqrt{2 x^{2} + 3} d x Also, x + 1 = λ \frac{d}{d x} (2 x^{2} + 3) + μ \Rightarrow x + 1 = λ (4 x) + μ Equating coefficient of like terms 4 λ = 1 \Rightarrow λ = \frac{1}{4} and μ = 1 ∴ I = \int [\frac{1}{4} (4 x) + 1] \sqrt{2 x^{2} + 3} d x = \frac{1}{4} \int (4 x) \sqrt{2 x^{2} + 3} d x + \int \sqrt{2 x^{2} + 3} d x = \frac{1}{4} \int (4 x) \sqrt{2 x^{2} + 3} d x + \int \sqrt{2 (x^{2} + \frac{3}{2})} d x = \frac{1}{4} \int (4 x) \sqrt{2 x^{2} + 3} d x + \sqrt{2} \int \sqrt{x^{2} + {(\frac{\sqrt{3}}{\sqrt{2}})}^{2}} d x Let 2 x^{2} + 3 = t \Rightarrow 4 x d x = d t ∴ I = \frac{1}{4} \int \sqrt{t} d t + \sqrt{2} [\frac{x}{2} \sqrt{x^{2} + \frac{3}{2}} + \frac{3}{4} \log |x + \sqrt{x^{2} + \frac{3}{2}}|] = \frac{1}{4} [\frac{t^{\frac{3}{2}}}{\frac{3}{2}}] + \sqrt{2} \frac{x}{2} \frac{\sqrt{2 x^{2} + 3}}{\sqrt{2}} + \frac{3 \sqrt{2}}{4} \log |x + \frac{\sqrt{2 x^{2} + 3}}{\sqrt{2}}| + C = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3 \sqrt{2}}{4} \log |\frac{\sqrt{2} x + \sqrt{2 x^{2} + 3}}{\sqrt{2}}| + C = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3 \sqrt{2}}{4} \log |\sqrt{2} x + \sqrt{2 x^{2} + 3}| - \frac{3 \sqrt{2}}{4} \log \sqrt{2} + C = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3 \sqrt{2}}{4} \log |\sqrt{2} x + \sqrt{2 x^{2} + 3}| + C' Where C' = C - \frac{3 \sqrt{2}}{4} \log \sqrt{2}$

Page No 18.159:

Question 3:

$\int (2 x - 5) \sqrt{2 + 3 x - x^{2}} d x$

Answer:

$Let I = \int (2 x - 5) \sqrt{2 + 3 x - x^{2}} d x Also, 2 x - 5 = λ \frac{d}{d x} (2 + 3 x - x^{2}) + μ \Rightarrow 2 x - 5 = λ (- 2 x + 3) + μ \Rightarrow 2 x - 5 = (- 2 λ) x + 3 λ + μ Equating coeffieicents of like terms - 2 λ = 2 \Rightarrow λ = - 1 And 3 λ + μ = - 5 \Rightarrow 3 (- 1) + μ = - 5 \Rightarrow μ = - 5 + 3 \Rightarrow μ = - 2 ∴ 2 x - 5 = - 1 (- 2 x + 3) - 2 Hence, I = \int [- (- 2 x + 3) - 2] \sqrt{2 + 3 x - x^{2}} d x = - \int (- 2 x + 3) \sqrt{2 + 3 x - x^{2}} d x - 2 \int \sqrt{2 + 3 x - x^{2}} d x = - I_{1} - 2 I_{2} . . . . . (1) I_{1} = \int (- 2 x + 3) \sqrt{2 + 3 x - x^{2}} d x Let 2 + 3 x - x^{2} = t \Rightarrow (- 2 x + 3) d x = d t ∴ I_{1} = \int t^{\frac{1}{2}} d t = \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} = \frac{2}{3} t^{\frac{3}{2}} = \frac{2}{3} {(2 + 3 x - x^{2})}^{\frac{3}{2}} . . . . . (2) And I_{2} = \int \sqrt{2 + 3 x - x^{2}} d x I_{2} = \int \sqrt{2 - (x^{2} - 3 x)} d x = \int \sqrt{2 - [x^{2} - 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2}]} d x = \int \sqrt{2 + \frac{9}{4} - {(x - \frac{3}{2})}^{2}} d x = \int \sqrt{{(\frac{\sqrt{17}}{2})}^{2} - {(x - \frac{3}{2})}^{2}} d x = \frac{x - \frac{3}{2}}{2} \sqrt{{(\frac{\sqrt{17}}{2})}^{2} - {(x - \frac{3}{2})}^{2}} + \frac{{(\frac{\sqrt{17}}{2})}^{2}}{2} \sin^{- 1} (\frac{x - \frac{3}{2}}{\frac{\sqrt{17}}{2}}) = \frac{2 x - 3}{4} \sqrt{2 + 3 x - x^{2}} + \frac{17}{8} \sin^{- 1} (\frac{2 x - 3}{\sqrt{17}}) . . . . . (3) From eq (1), (2) and (3) we have I = - \frac{2}{3} {(2 + 3 x - x^{2})}^{\frac{3}{2}} - \frac{(2 x - 3)}{2} \sqrt{2 + 3 x - x^{2}} - \frac{17}{4} \sin^{- 1} (\frac{2 x - 3}{\sqrt{17}}) + C$

Page No 18.159:

Question 4:

$\int (x + 2) \sqrt{x^{2} + x + 1} d x$

Answer:

$Let I = \int (x + 2) \sqrt{x^{2} + x + 1} d x Also, x + 2 = λ \frac{d}{d x} (x^{2} + x + 1) + μ \Rightarrow x + 2 = λ (2 x + 1) + μ \Rightarrow x + 2 = (2 λ) x + λ + μ Equating coefficient of like terms 2 λ = 1 \Rightarrow λ = \frac{1}{2} And λ + μ = 2 \Rightarrow \frac{1}{2} + μ = 2 \Rightarrow μ = \frac{3}{2} ∴ I = \int [(\frac{1}{2} (2 x + 1) + \frac{3}{2}) \sqrt{x^{2} + x + 1}] d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{3}{2} \int \sqrt{x^{2} + x + 1} d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{3}{2} \int \sqrt{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{3}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x Let x^{2} + x + 1 = t \Rightarrow (2 x + 1) d x = d t Then, I = \frac{1}{2} \int \sqrt{t} d t + \frac{3}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x = \frac{1}{2} \int t^{\frac{1}{2}} d t + \frac{3}{2} [\frac{x + \frac{1}{2}}{2} \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} + \frac{3}{8} \log |(x + \frac{1}{2}) + \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}|] + C = \frac{1}{2} [\frac{t \frac{3}{2}}{\frac{3}{2}}] + \frac{3}{8} (2 x + 1) \sqrt{x^{2} + x + 1} + \frac{9}{16} \log |(x + \frac{1}{2}) + \sqrt{x^{2} + x + 1}| + C = \frac{1}{3} {(x^{2} + x + 1)}^{\frac{3}{2}} + \frac{3}{8} (2 x + 1) \sqrt{x^{2} + x + 1} + \frac{9}{16} \log |(x + \frac{1}{2}) + \sqrt{x^{2} + x + 1}| + C$

Page No 18.159:

Question 5:

$\int (4 x + 1) \sqrt{x^{2} - x - 2} d x$

Answer:

$Let I = \int (4 x + 1) \sqrt{x^{2} - x - 2} d x Also, 4 x + 1 = λ \frac{d}{d x} (x^{2} - x - 2) + μ \Rightarrow 4 x + 1 = λ (2 x - 1) + μ \Rightarrow 4 x + 1 = (2 λ) x + μ - λ Equating coefficient of like terms 2 λ = 4 \Rightarrow λ = 2 And μ - λ = 1 \Rightarrow μ - 2 = 1 \Rightarrow μ = 3 ∴ I = \int [2 (2 x - 1) + 3] \sqrt{x^{2} - x - 2} d x = 2 \int (2 x - 1) \sqrt{x^{2} - x - 2} d x + 3 \int \sqrt{x^{2} - x - 2} d x = 2 \int (2 x - 1) \sqrt{x^{2} - x - 2} d x + 3 \int \sqrt{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} - 2} d x = 2 \int (2 x + 1) \sqrt{x^{2} - x - 2} d x + 3 \int \sqrt{{(x - \frac{1}{2})}^{2} - 2 - \frac{1}{4}} d x = \int (2 x - 1) \sqrt{x^{2} - x - 2} d x + 3 \int \sqrt{{(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}} d x Let x^{2} - x - 2 = t \Rightarrow (2 x - 1) d x = d t ∴ I = 2 \int \sqrt{t} d t + 3 [(\frac{x - \frac{1}{2}}{2}) \sqrt{{(x - \frac{1}{2})}^{2} - {(\frac{3}{2})}^{2}} - \frac{{(\frac{3}{2})}^{2}}{2} \log |(x - \frac{1}{2}) + \sqrt{x^{2} - x - 2}|] = 2 [\frac{t^{\frac{3}{2}}}{\frac{3}{2}}] + \frac{3}{4} (2 x - 1) \sqrt{x^{2} - x - 2} - \frac{27}{8} \log |(x - \frac{1}{2}) + \sqrt{x^{2} - x - 2}| + C = \frac{4}{3} {(x^{2} - x - 2)}^{\frac{3}{2}} + \frac{3}{4} (2 x - 1) \sqrt{x^{2} - x - 2} - \frac{27}{8} \log |(x - \frac{1}{2}) + \sqrt{x^{2} - x - 2}| + C$

Page No 18.159:

Question 6:

$\int (x - 2) \sqrt{2 x^{2} - 6 x + 5} d x$

Answer:

$Let I = \int (x - 2) \sqrt{2 x^{2} - 6 x + 5} d x Also, x - 2 = λ \frac{d}{d x} (2 x^{2} - 6 x + 5) + μ \Rightarrow x - 2 = (4 λ) x + μ - 6 λ Equating the coefficient of like terms 4 λ = 1 \Rightarrow λ = \frac{1}{4} And μ - 6 λ = - 2 \Rightarrow μ - 6 \times \frac{1}{4} = - 2 \Rightarrow μ = - 2 + \frac{3}{2} = - \frac{1}{2} ∴ I = \int [\frac{1}{4} (4 x - 6) - \frac{1}{2}] \sqrt{2 x^{2} - 6 x + 5} d x = \frac{1}{4} \int (4 x - 6) \sqrt{2 x^{2} - 6 x + 5} d x - \frac{1}{2} \int \sqrt{2 x^{2} - 6 x + 5} d x Let 2 x^{2} - 6 x + 5 = t \Rightarrow (4 x - 6) d x = d t ∴ I = \frac{1}{4} \int t^{\frac{1}{2}} d t - \frac{1}{2} \int \sqrt{2 (x^{2} - 3 x + \frac{5}{2})} d x = \frac{1}{4} \int t^{\frac{1}{2}} - \frac{\sqrt{2}}{2} \int \sqrt{x^{2} - 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{5}{2}} d x = \frac{1}{4} [\frac{t^{\frac{3}{2}}}{\frac{3}{2}}] - \frac{1}{\sqrt{2}} \int \sqrt{{(x - \frac{3}{2})}^{2} - \frac{9}{4} + \frac{5}{2}} d x = \frac{1}{6} t^{\frac{3}{2}} - \frac{1}{\sqrt{2}} \int \sqrt{{(x - \frac{3}{2})}^{2} - \frac{9 + 10}{4}} d x = \frac{1}{6} t^{\frac{3}{2}} - \frac{1}{\sqrt{2}} \int \sqrt{{(x - \frac{3}{2})}^{2} + {(\frac{1}{2})}^{2}} d x = \frac{1}{6} {(2 x^{2} - 6 x + 5)}^{\frac{3}{2}} - \frac{1}{\sqrt{2}} [(\frac{x - \frac{3}{2}}{2}) \sqrt{{(x - \frac{3}{2})}^{2} + {(\frac{1}{2})}^{2}} + \frac{1}{8} \log |(x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + \frac{5}{2}}|] + C = \frac{1}{6} {(2 x^{2} - 6 x + 5)}^{\frac{3}{2}} - \frac{1}{\sqrt{2}} [\frac{2 x - 3}{4} \sqrt{x^{2} - 3 x + \frac{5}{2}} + \frac{1}{8} \log |\frac{2 x - 3}{2} + \sqrt{x^{2} - 3 x + \frac{5}{2}}|] + C$

Page No 18.159:

Question 7:

$\int (x + 1) \sqrt{x^{2} + x + 1} d x$

Answer:

$Let I = \int (x + 1) \sqrt{x^{2} + x + 1} d x Also, x + 1 = λ \frac{d}{d x} (x^{2} + x + 1) + μ \Rightarrow x + 1 = λ (2 x + 1) + μ \Rightarrow x + 1 = (2 λ) x + λ + μ Equating coefficient of like terms 2 λ = 1 \Rightarrow λ = \frac{1}{2} And λ + μ = 1 \Rightarrow \frac{1}{2} + μ = 1 ∴ μ = \frac{1}{2} ∴ I = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{1}{2} \int \sqrt{x^{2} + x + 1} d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{1}{2} \int \sqrt{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x Let x^{2} + x + 1 = t \Rightarrow (2 x + 1) d x = d t Then,$
$I = \frac{1}{2} \int \sqrt{t} d t + \frac{1}{2} [\frac{x + \frac{1}{2}}{2} \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} + \frac{3}{8} \log |(x + \frac{1}{2}) + \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}|] + C = \frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}} + \frac{1}{2} [(\frac{2 x + 1}{4}) \sqrt{x^{2} + x + 1} + \frac{3}{8} \log |(x + \frac{1}{2}) + \sqrt{x^{2} + x + 1}|] + C = \frac{1}{3} {(x^{2} + x + 1)}^{\frac{3}{2}} + \frac{1}{2} [(\frac{2 x + 1}{4}) \sqrt{x^{2} + x + 1} + \frac{3}{8} \log |(x + \frac{1}{2}) + \sqrt{x^{2} + x + 1}|] + C$

Page No 18.159:

Question 8:

$\int (2 x + 3) \sqrt{x^{2} + 4 x + 3} d x$

Answer:

$Let I = \int (2 x + 3) \sqrt{x^{2} + 4 x + 3} d x Also, 2 x + 3 = λ \frac{d}{d x} (x^{2} + 4 x + 3) + μ \Rightarrow 2 x + 3 = λ (2 x + 4) + μ \Rightarrow 2 x + 3 = (2 λ) x + 4 λ + μ Equating coefficient of like terms . 2 λ = 2 \Rightarrow λ = 1 And 4 λ + μ = 3 \Rightarrow 4 + μ = 3 \Rightarrow μ = - 1 ∴ I = \int (2 x + 4 - 1) \sqrt{x^{2} + 4 x + 3} d x = \int (2 x + 4) \sqrt{x^{2} + 4 x + 3} d x - \int \sqrt{x^{2} + 4 x + 3} d x = \int (2 x + 4) \sqrt{x^{2} + 4 x + 3} d x - \int \sqrt{x^{2} + 4 x + 4 - 1} d x = \int (2 x + 4) \sqrt{x^{2} + 4 x + 3 d x} - \int \sqrt{{(x + 2)}^{2} - 1^{2}} d x Let x^{2} + 4 x + 3 = t \Rightarrow (2 x + 4) d x = d t Then, I = \int \sqrt{t} d t - \int \sqrt{{(x + 2)}^{2} - 1^{2}} d x = \frac{2}{3} t^{\frac{3}{2}} - [\frac{x + 2}{2} \sqrt{{(x + 2)}^{2} - 1} - \frac{1^{2}}{2} \log |(x + 2) + \sqrt{{(x + 2)}^{2} - 1}|] + C = \frac{2}{3} {(x^{2} + 4 x + 3)}^{\frac{3}{2}} - \frac{1}{2} [(x + 2) \sqrt{x^{2} + 4 x + 3} - \log |(x + 2) + \sqrt{x^{2} + 4 x + 3}|] + C$

Page No 18.159:

Question 9:

$\int (2 x - 5) \sqrt{x^{2} - 4 x + 3} d x$

Answer:

$Let I = \int (2 x - 5) \sqrt{x^{2} - 4 x + 3} d x = \int (2 x - 4 - 1) \sqrt{x^{2} - 4 x + 3} d x = \int (2 x - 4) \sqrt{x^{2} - 4 x + 3} d x - \int \sqrt{x^{2} - 4 x + 3} d x = \int (2 x - 4) \sqrt{x^{2} - 4 x + 3} d x - \int \sqrt{x^{2} - 4 x + 4 - 4 + 3} d x = \int (2 x - 4) \sqrt{x^{2} - 4 x + 3} d x - \int \sqrt{{(x - 2)}^{2} - 1^{2}} d x Let x^{2} - 4 x + 3 = t \Rightarrow (2 x - 4) d x = d t ∴ I = \int \sqrt{t} d t - \int \sqrt{{(x - 2)}^{2} - 1^{2}} d x = \frac{2}{3} t^{\frac{3}{2}} - [\frac{x - 2}{2} \sqrt{{(x - 2)}^{2} - 1^{2}} - \frac{1^{2}}{2} \log |(x - 2) + \sqrt{{(x - 2)}^{2} - 1}|] + C = \frac{2}{3} {(x^{2} - 4 x + 3)}^{\frac{3}{2}} - (\frac{x - 2}{2}) \sqrt{x^{2} - 4 x + 3} + \frac{1}{2} \log |(x - 2) + \sqrt{x^{2} - 4 x + 3}| + C$

Page No 18.159:

Question 10:

$\int x \sqrt{x^{2} + x} d x$

Answer:

$Let I = \int x \sqrt{x^{2} + x} d x Also, x = λ \frac{d}{d x} (x^{2} + x) + μ \Rightarrow x = λ (2 x + 1) + μ \Rightarrow x = (2 λ) x + λ + μ Equating coefficient of like terms 2 λ = 1 \Rightarrow λ = \frac{1}{2} And λ + μ = 0 \Rightarrow μ = - \frac{1}{2} ∴ I = \int [\frac{1}{2} (2 x + 1) - \frac{1}{2}] \sqrt{x^{2} + x} d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x} d x - \frac{1}{2} \int \sqrt{x^{2} + x} d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x} d x - \frac{1}{2} \int \sqrt{x^{2} + x + \frac{1}{4} - \frac{1}{4}} d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x} d x - \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d x Let x^{2} + x = t \Rightarrow (2 x + 1) d x = d t Then, I = \frac{1}{2} \int \sqrt{t} d t - \frac{1}{2} [\frac{x + \frac{1}{2}}{2} \sqrt{x^{2} + x} - \frac{1}{8} \log |(x + \frac{1}{2}) + \sqrt{x^{2} + x}|] + C = \frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}} - (\frac{2 x + 1}{8}) \sqrt{x^{2} + x} + \frac{1}{16} \log |(x + \frac{1}{2}) + \sqrt{x^{2}} + x| + C = \frac{1}{3} {(x^{2} + x)}^{\frac{3}{2}} - (\frac{2 x + 1}{8}) \sqrt{x^{2} + x} + \frac{1}{16} \log |(x + \frac{1}{2}) + \sqrt{x^{2}} + x| + C$

Page No 18.159:

Question 11:

Evaluate the following integrals:

$\int (x - 3) \sqrt{x^{2} + 3 x - 18} d x$

Answer:

$Let I = \int (x - 3) \sqrt{x^{2} + 3 x - 18} d x We express x - 3 = A (\frac{d}{d x} (x^{2} + 3 x - 18)) + B x - 3 = A (2 x + 3) + B Equating the coefficients of x and constants, we get 1 = 2 A and - 3 = 3 A + B or A = \frac{1}{2} and B = - \frac{9}{2} ∴ I = \int (\frac{1}{2} (2 x + 3) - \frac{9}{2}) \sqrt{x^{2} + 3 x - 18} d x = \frac{1}{2} \int (2 x + 3) \sqrt{x^{2} + 3 x - 18} d x - \frac{9}{2} \int \sqrt{x^{2} + 3 x - 18} d x = \frac{1}{2} I_{1} - \frac{9}{2} I_{2} . . . (1) Now, I_{1} = \int (2 x + 3) \sqrt{x^{2} + 3 x - 18} d x Let x^{2} + 3 x - 18 = u On differentiating both sides, we get (2 x + 3) d x = d u ∴ I_{1} = \int \sqrt{u} d u = \frac{2}{3} u^{\frac{3}{2}} + c_{1} = \frac{2}{3} {(x^{2} + 3 x - 18)}^{\frac{3}{2}} + c_{1} . . . (2) And, I_{2} = \int \sqrt{x^{2} + 3 x - 18} d x = \int \sqrt{x^{2} + 3 x + \frac{9}{4} - \frac{9}{4} - 18} d x = \int \sqrt{{(x + \frac{3}{2})}^{2} - {(\frac{9}{2})}^{2}} d x Let (x + \frac{3}{2}) = u On differentiating both sides, we get d x = d u ∴ I_{2} = \int \sqrt{{(u)}^{2} - {(\frac{9}{2})}^{2}} d u = \frac{u}{2} \sqrt{{(u)}^{2} - {(\frac{9}{2})}^{2}} - \frac{1}{2} {(\frac{9}{2})}^{2} \log |u + \sqrt{{(u)}^{2} - {(\frac{9}{2})}^{2}}| + c_{2} = \frac{x + \frac{3}{2}}{2} \sqrt{{(x + \frac{3}{2})}^{2} - {(\frac{9}{2})}^{2}} - \frac{1}{2} {(\frac{9}{2})}^{2} \log |(x + \frac{3}{2}) + \sqrt{{(x + \frac{3}{2})}^{2} - {(\frac{9}{2})}^{2}}| + c_{2} = \frac{2 x + 3}{4} \sqrt{x^{2} + 3 x - 18} - \frac{81}{8} \log |(x + \frac{3}{2}) + \sqrt{x^{2} + 3 x - 18}| + c_{2} . . . (3) From (1), (2) and (3), we get ∴ I = \frac{1}{2} (\frac{2}{3} {(x^{2} + 3 x - 18)}^{\frac{3}{2}} + c_{1}) - \frac{9}{2} (\frac{2 x + 3}{4} \sqrt{x^{2} + 3 x - 18} - \frac{81}{8} \log |(x + \frac{3}{2}) + \sqrt{x^{2} + 3 x - 18}| + c_{2}) = \frac{1}{3} {(x^{2} + 3 x - 18)}^{\frac{3}{2}} - \frac{9}{8} (2 x + 3) \sqrt{x^{2} + 3 x - 18} + \frac{729}{16} \log |(x + \frac{3}{2}) + \sqrt{x^{2} + 3 x - 18}| + c Hence, \int (x - 3) \sqrt{x^{2} + 3 x - 18} d x = \frac{1}{3} {(x^{2} + 3 x - 18)}^{\frac{3}{2}} - \frac{9}{8} (2 x + 3) \sqrt{x^{2} + 3 x - 18} + \frac{729}{16} \log |(x + \frac{3}{2}) + \sqrt{x^{2} + 3 x - 18}| + c$

Page No 18.159:

Question 12:

Evaluate the following integrals:

$\int (x + 3) \sqrt{3 - 4 x - x^{2}} d x$

Answer:

$Let I = \int (x + 3) \sqrt{3 - 4 x - x^{2}} d x We express x + 3 = A (\frac{d}{d x} (3 - 4 x - x^{2})) + B x + 3 = A (- 4 - 2 x) + B Equating the coefficients of x and constants, we get 1 = - 2 A and 3 = - 4 A + B or A = - \frac{1}{2} and B = 1 ∴ I = \int (- \frac{1}{2} (- 4 - 2 x) + 1) \sqrt{3 - 4 x - x^{2}} d x = - \frac{1}{2} \int (- 4 - 2 x) \sqrt{3 - 4 x - x^{2}} d x + \int \sqrt{3 - 4 x - x^{2}} d x = - \frac{1}{2} I_{1} + I_{2} . . . (1) Now, I_{1} = \int (- 4 - 2 x) \sqrt{3 - 4 x - x^{2}} d x Let 3 - 4 x - x^{2} = u On differentiating both sides, we get (- 4 - 2 x) d x = d u ∴ I_{1} = \int \sqrt{u} d u = \frac{2}{3} u^{\frac{3}{2}} + c_{1} = \frac{2}{3} {(3 - 4 x - x^{2})}^{\frac{3}{2}} + c_{1} . . . (2) And, I_{2} = \int \sqrt{3 - 4 x - x^{2}} d x = \int \sqrt{3 + 4 - 4 - 4 x - x^{2}} d x = \int \sqrt{{(\sqrt{7})}^{2} - {(x + 2)}^{2}} d x Let (x + 2) = u On differentiating both sides, we get d x = d u ∴ I_{2} = \int \sqrt{{(\sqrt{7})}^{2} - {(u)}^{2}} d u = \frac{u}{2} \sqrt{{(\sqrt{7})}^{2} - {(u)}^{2}} + \frac{1}{2} {(\sqrt{7})}^{2} \sin^{- 1} (\frac{u}{\sqrt{7}}) + c_{2} = \frac{x + 2}{2} \sqrt{7 - {(x + 2)}^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + c_{2} = \frac{x + 2}{2} \sqrt{3 - 4 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + c_{2} . . . (3) From (1), (2) and (3), we get ∴ I = - \frac{1}{2} (\frac{2}{3} {(3 - 4 x - x^{2})}^{\frac{3}{2}} + c_{1}) + (\frac{x + 2}{2} \sqrt{3 - 4 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + c_{2}) = - \frac{1}{3} {(3 - 4 x - x^{2})}^{\frac{3}{2}} + \frac{x + 2}{2} \sqrt{3 - 4 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + c Hence, \int (x + 3) \sqrt{3 - 4 x - x^{2}} d x = - \frac{1}{3} {(3 - 4 x - x^{2})}^{\frac{3}{2}} + \frac{x + 2}{2} \sqrt{3 - 4 x - x^{2}} + \frac{7}{2} \sin^{- 1} (\frac{x + 2}{\sqrt{7}}) + c$

Page No 18.159:

Question 13:

$\int (3 x + 1) \sqrt{4 - 3 x - 2 x^{2}} d x$

Answer:

$I = \int (3 x + 1) \sqrt{4 - 3 x - 2 x^{2}} d x Let (3 x + 1) = A \frac{d}{d x} (4 - 3 x - 2 x^{2}) + B \Rightarrow (3 x + 1) = A (- 3 - 4 x) + B \Rightarrow (3 x + 1) = - 4 A x + (B - 3 A) \Rightarrow 3 = - 4 A and (B - 3 A) = 1 \Rightarrow A = - \frac{3}{4} and B = - \frac{5}{4}$

$\Rightarrow (3 x + 1) = - \frac{3}{4} (- 3 - 4 x) - \frac{5}{4} \Rightarrow I = - \frac{3}{4} \int (- 3 - 4 x) \sqrt{4 - 3 x - 2 x^{2}} d x - \frac{5}{4} \int \sqrt{4 - 3 x - 2 x^{2}} d x Let I = - \frac{3}{4} I_{1} - \frac{5}{4} I_{2} . . . (i) Now, I_{1} = \int (- 3 - 4 x) \sqrt{4 - 3 x - 2 x^{2}} d x Let (4 - 3 x - 2 x^{2}) = t, or, (- 3 - 4 x) d x = d t \Rightarrow I_{1} = \int \sqrt{t} d t = \frac{2}{3} t^{\frac{3}{2}} + c_{1} \Rightarrow I_{1} = \frac{2}{3} {(4 - 3 x - 2 x^{2})}^{\frac{3}{2}} + c_{1}$

$I_{2} = \int \sqrt{4 - 3 x - 2 x^{2}} d x = \int \sqrt{2 (2 - \frac{3}{2} x - x^{2})} d x = \sqrt{2} \int \sqrt{(\frac{17}{4} - \frac{9}{4} - \frac{3}{2} x - x^{2})} d x = \sqrt{2} \int \sqrt{[{(\frac{\sqrt{17}}{2})}^{2} - (\frac{9}{4} + \frac{3}{2} x + x^{2})]} d x = \sqrt{2} \int \sqrt{[{(\frac{\sqrt{17}}{2})}^{2} - {(x + \frac{3}{2})}^{2}]} d x = \sqrt{2} \sin (\frac{x + \frac{3}{2}}{\frac{\sqrt{17}}{2}}) + c_{2} = \sqrt{2} \sin (\frac{2 x + 3}{\sqrt{17}}) + c_{2}$

Using (i), we get

$I = - \frac{3}{4} \times \frac{2}{3} {(4 - 3 x - 2 x^{2})}^{\frac{3}{2}} - \frac{5}{4} \times \sqrt{2} \sin (\frac{2 x + 3}{\sqrt{17}}) + C ∴ I = - \frac{1}{2} {(4 - 3 x - 2 x^{2})}^{\frac{3}{2}} - \frac{5 \sqrt{2}}{4} \sin (\frac{2 x + 3}{\sqrt{17}}) + C$

Page No 18.159:

Question 14:

$\int (2 x + 5) \sqrt{10 - 4 x - 3 x^{2}} d x$

Answer:

$I = \int (2 x + 5) \sqrt{10 - 4 x - 3 x^{2}} d x Let (2 x + 5) = A \frac{d}{d x} (10 - 4 x - 3 x^{2}) + B \Rightarrow (2 x + 5) = A (- 4 - 6 x) + B \Rightarrow (2 x + 5) = - 6 A x + (B - 4 A) \Rightarrow 2 = - 6 A and (B - 4 A) = 5 \Rightarrow A = - \frac{1}{3} and B = \frac{11}{3}$

$\Rightarrow (2 x + 5) = - \frac{1}{3} (- 4 - 6 x) + \frac{11}{3} \Rightarrow I = - \frac{1}{3} \int (- 4 - 6 x) \sqrt{10 - 4 x - 3 x^{2}} d x + \frac{11}{3} \int \sqrt{10 - 4 x - 3 x^{2}} d x Let I = - \frac{1}{3} I_{1} + \frac{11}{3} I_{2} . . . (i) Now, I_{1} = \int (- 4 - 6 x) \sqrt{10 - 4 x - 3 x^{2}} d x Let (10 - 4 x - 3 x^{2}) = t, or, (- 4 - 6 x) d x = d t \Rightarrow I_{1} = \int \sqrt{t} d t = \frac{2}{3} t^{\frac{3}{2}} + c_{1} \Rightarrow I_{1} = \frac{2}{3} {(10 - 4 x - 3 x^{2})}^{\frac{3}{2}} + c_{1}$

$I_{2} = \int \sqrt{(10 - 4 x - 3 x^{2})} d x = \int \sqrt{3 (\frac{10}{3} - \frac{4}{3} x - x^{2})} d x = \sqrt{3} \int \sqrt{(\frac{26}{9} - \frac{4}{9} - \frac{4}{3} x - x^{2})} d x = \sqrt{3} \int \sqrt{[{(\frac{\sqrt{26}}{3})}^{2} - (\frac{4}{9} + \frac{4}{3} x + x^{2})]} d x = \sqrt{3} \int \sqrt{[{(\frac{\sqrt{26}}{3})}^{2} - {(x + \frac{2}{3})}^{2}]} d x = \sqrt{3} \sin (\frac{x + \frac{2}{3}}{\frac{\sqrt{26}}{3}}) + c_{2} = \sqrt{3} \sin (\frac{3 x + 2}{\sqrt{26}}) + c_{2}$

Using (i), we get

$I = - \frac{1}{3} \times \frac{2}{3} {(10 - 4 x - 3 x^{2})}^{\frac{3}{2}} + \frac{11}{3} \times \sqrt{3} \sin (\frac{3 x + 2}{\sqrt{26}}) + C ∴ I = - \frac{2}{9} {(10 - 4 x - 3 x^{2})}^{\frac{3}{2}} + \frac{11 \sqrt{3}}{3} \sin (\frac{3 x + 2}{\sqrt{26}}) + C$

Page No 18.176:

Question 1:

$\int \frac{2 x + 1}{(x + 1) (x - 2)} d x$

Answer:

$\int \frac{(2 x + 1)}{(x + 1) (x - 2)} d x Let \frac{2 x + 1}{(x + 1) (x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2} . . . . (1) \Rightarrow \frac{2 x + 1}{(x + 1) (x - 2)} = \frac{A (x - 2) + B (x + 1)}{(x + 1) (x - 2)} Then, (2 x + 1) = A (x - 2) + B (x + 1) . . . . (2) Putting (x - 2) = 0 or, x = 2 in eq (2) \Rightarrow 2 \times 2 + 1 = A \times 0 + B (2 + 1) \Rightarrow B = \frac{5}{3} Putting (x + 1) = 0 or, x = - 1 in eq (2) 2 \times - 1 + 1 + A (- 1 - 2) + B \times 0 \Rightarrow - 1 = A (- 3) \Rightarrow A = \frac{1}{3} Substituting the values of A and B in eq (1), we get ∴ \frac{2 x + 1}{(x + 1) (x - 2)} = \frac{1}{3} (x + 1) + \frac{5}{3} (x - 2) \int \frac{(2 x + 1) d x}{(x + 1) (x - 2)} = \frac{1}{3} \int \frac{1}{x + 1} d x + \frac{5}{3} \int \frac{1}{x - 2} d x = \frac{1}{3} \ln |x + 1| + \frac{5}{3} \ln |x - 2| + C$

Page No 18.176:

Question 2:

$\int \frac{1}{x (x - 2) (x - 4)} d x$

Answer:

$\int \frac{1}{x (x - 2) (x - 4)} d x Let \frac{1}{x (x - 2) (x - 4)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x - 4} \Rightarrow \frac{1}{x (x - 2) (x - 4)} = \frac{A (x - 2) (x - 4) + B (x) (x - 4) + C x \cdot (x - 2)}{x (x - 2) (x - 4)} \Rightarrow 1 = A (x - 2) (x - 4) + B (x) \cdot (x - 4) + C x . (x - 2) . . . (1) Putting x = 0 in eq (1) \Rightarrow 1 = A (0 - 2) (0 - 4) + B \times 0 + C \times 0 \Rightarrow \frac{1}{8} = A Putting (x - 2) = 0 or x = 2 in eq (1) \Rightarrow 1 = A \times 0 + B (2) (2 - 4) + C \times 2 \times 0 \Rightarrow B = - \frac{1}{4} Putting (x - 4) = 0 or x = 4 in eq (1) \Rightarrow 1 = A \times 0 + B \times 0 + C \cdot 4 (4 - 2) \Rightarrow C = \frac{1}{8} ∴ \frac{1}{x (x - 2) (x - 4)} = \frac{1}{8 x} - \frac{1}{4 (x - 2)} + \frac{1}{8 (x - 4)} \Rightarrow \int \frac{d x}{x (x - 2) (x - 4)} = \frac{1}{8} \int \frac{1}{x} d x - \frac{1}{4} \int \frac{1}{x - 2} d x + \frac{1}{8} \int \frac{1}{x - 4} d x = \frac{1}{8} \ln |x| - \frac{1}{4} \ln |x - 2| + \frac{1}{8} \ln |x - 4| + C = \frac{1}{8} (\ln |x| + \ln |x - 4| - 2 \ln |x - 2|) + C = \frac{1}{8} [\ln |\frac{x (x - 4)}{{(x - 2)}^{2}}|] + C$

Page No 18.176:

Question 3:

$\int \frac{x^{2} + x - 1}{x^{2} + x - 6} d x$

Answer:

$\int (\frac{x^{2} + x - 1}{x^{2} + x - 6}) d x = \int (\frac{x^{2} + x - 6 + 6 - 1}{x^{2} + x - 6}) d x = \int (\frac{x^{2} + x - 6}{x^{2} + x - 6}) d x + 5 \int \frac{1}{x^{2} + x - 6} d x = \int d x + 5 \int \frac{1}{x^{2} + 3 x - 2 x - 6} d x = \int d x + 5 \int \frac{1}{x (x + 3) - 2 (x + 3)} d x = \int d x + 5 \int \frac{1}{(x - 2) (x + 3)} d x . . . (1) Let \frac{1}{(x - 2) (x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3} \Rightarrow \frac{1}{(x - 2) (x + 3)} = \frac{A (x + 3) + B (x - 2)}{(x - 2) (x + 3)} \Rightarrow 1 = A (x + 3) + B (x - 2) . . . . (2) \begin{array}{l} Putting x + 3 = 0 or x = - 3 in eq (2) \\ \Rightarrow 1 = A \times 0 + B (- 3 - 2) \\ \Rightarrow B = - \frac{1}{5} \end{array} \begin{array}{l} Putting x - 2 = 0 or x = 2 in eq (2) \\ \Rightarrow 1 = A (2 + 3) + B \times 0 \\ \Rightarrow A = \frac{1}{5} \end{array}$

$∴ \frac{1}{(x - 2) (x + 3)} = \frac{1}{5} (x - 2) - \frac{1}{5} (x + 3) \Rightarrow \int \frac{1}{(x - 2) (x + 3)} d x = \frac{1}{5} \int \frac{d x}{x - 2} - \frac{1}{5} \int \frac{d x}{x + 3} = \frac{1}{5} \ln |x - 2| - \frac{1}{5} \ln |x + 3| + C = \frac{1}{5} \ln |\frac{x - 2}{x + 3}| + C . . . (3) From eq (1) & eq (3) ∴ \int (\frac{x^{2} + x - 1}{x^{2} + x - 6}) d x = x + \frac{5}{5} \ln |\frac{x - 2}{x + 3}| + C = x + \ln |x - 2| - \ln |x + 3| + C$

Page No 18.176:

Question 4:

$\int \frac{3 + 4 x - x^{2}}{(x + 2) (x - 1)} d x$

Answer:

$\int (\frac{3 + 4 x - x^{2}}{x^{2} - x + 2 x - 2}) d x = \int (\frac{- x^{2} + 4 x + 3}{x^{2} + x - 2}) d x \frac{- x^{2} + 4 x + 3}{x^{2} + x - 2} = - 1 + \frac{5 x + 1}{x^{2} + x - 2} . . . (1) ∴ \frac{5 x + 1}{x^{2} + x - 2} = \frac{5 x + 1}{x^{2} + 2 x - x - 2} = \frac{5 x + 1}{x (x + 2) - 1 (x + 2)} Let \frac{5 x + 1}{(x - 1) (x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2} \Rightarrow \frac{5 x + 1}{(x - 1) (x + 2)} = \frac{A (x + 2) + B (x - 1)}{(x - 1) (x + 2)} \Rightarrow 5 x + 1 = A (x + 2) + B (x - 1) . . . (2) Putting x + 2 = 0 or x = - 2 in eq (2) \Rightarrow 5 x - 2 + 1 = A \times 0 + B (- 2 - 1) \Rightarrow B = 3 Putting x - 1 = 0 or x = 1 in eq (2) \Rightarrow 5 \times 1 + 1 = A (3) + B \times 0 \Rightarrow A = 2 ∴ \frac{5 x + 1}{(x - 1) (x + 2)} = \frac{2}{x - 1} + \frac{3}{x + 2} . . . (3) From (1) & (3) \frac{- x^{2} + 4 x + 3}{x^{2} + x - 2} = - 1 + \frac{2}{x - 1} + \frac{3}{x + 2} \Rightarrow \int \frac{- x^{2} + 4 x + 3}{x^{2} + x - 2} d x = \int - 1 d x + \int \frac{2}{x - 1} d x + \int \frac{3}{x + 2} d x = - x + 2 \ln (x - 1) + 3 \ln (x + 2) + C$

Page No 18.176:

Question 5:

$\int \frac{x^{2} + 1}{x^{2} - 1} d x$

Answer:

$\int (\frac{x^{2} + 1}{x^{2} - 1}) d x = \int (\frac{x^{2} - 1 + 2}{x^{2} - 1}) d x = \int d x + 2 \int \frac{1}{x^{2} - 1^{2}} d x = \int d x + 2 \int \frac{1}{(x - 1) (x + 1)} d x . . . (1) ∴ \frac{1}{(x - 1) (x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} \Rightarrow \frac{1}{(x - 1) (x + 1)} = \frac{A (x + 1) + B (x - 1)}{(x - 1) (x + 1)} \Rightarrow 1 = A (x + 1) B (x - 1) . . . . (2) Putting x + 1 = 0 or x = - 1 in eq (2) \Rightarrow 1 = A \times 0 + B (- 1 - 1) \Rightarrow B = \frac{- 1}{2} Putting x - 1 = 0 or x = 1 in eq (2) \Rightarrow 1 = A (1 + 1) + B \times 0 \Rightarrow A = \frac{1}{2} ∴ \frac{1}{(x - 1) (x + 1)} = \frac{1}{2 (x - 1)} - \frac{1}{2 (x + 1)} . . . (3) From eq (1) & eq (3) \int (\frac{x^{2} + 1}{x^{2} - 1}) d x = \int d x + 2 \int [\frac{1}{2 (x - 1)} - \frac{1}{2 (x + 1)}] d x = \int d x + \int \frac{d x}{x - 1} - \int \frac{d x}{x + 1} = x + \ln |x - 1| = - \ln |x + 1| + C = x + \ln |\frac{x - 1}{x + 1}| + C$

Page No 18.176:

Question 6:

$\int \frac{x^{2}}{(x - 1) (x - 2) (x - 3)} d x$

Answer:

$\int \frac{x^{2}}{(x - 1) (x - 2) (x - 3)} d x Let \frac{x^{2}}{(x - 1) (x - 2) (x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3} \Rightarrow \frac{x^{2}}{(x - 1) (x - 2) (x - 3)} = \frac{A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2)}{(x - 1) (x - 2) (x - 3)} \Rightarrow x^{2} = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2) . . . . (1) Putting x - 1 = 0 or x = 1 in eq (1) \Rightarrow 1 = A (1 - 2) (1 - 3) \Rightarrow 1 = A (- 1) (- 2) \Rightarrow A = \frac{1}{2} Putting x - 2 = 0 or x = 2 in eq (1) \Rightarrow 4 = B (2 - 1) (2 - 3) \Rightarrow B = - 4 Putting x - 3 = 0 or x = 3 in eq (1) \Rightarrow 9 = C (3 - 1) (3 - 2) \Rightarrow C = \frac{9}{2} ∴ \frac{x^{2}}{(x - 1) (x - 2) (x - 3)} = \frac{1}{2 (x - 1)} - \frac{4}{x - 2} + \frac{9}{2 (x - 3)} \int \frac{x^{2}}{(x - 1) (x - 2) (x - 3)} d x = \frac{1}{2} \int \frac{1}{x - 1} d x - 4 \int \frac{1}{x - 2} d x + \frac{9}{2} \int \frac{1}{x - 3} d x = \frac{1}{2} \ln |x - 1| - 4 \ln |x - 2| + \frac{9}{2} \ln |x - 3| + C$

Page No 18.176:

Question 7:

$\int \frac{5 x}{(x + 1) (x^{2} - 4)} d x$

Answer:

$\int \frac{5 x}{(x + 1) (x^{2} - 4)} d x Let \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{x + 2} \Rightarrow \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{A (x - 2) (x + 2) + B (x + 1) (x + 2) + C (x + 1) (x - 2)}{(x + 1) (x - 2) (x + 2)} \Rightarrow 5 x = A (x - 2) (x + 2) + B (x + 1) (x + 2) + C (x + 1) (x - 2) . . . . (1) Putting x - 2 = 0 or x = 2 in eq (1) \Rightarrow 5 \times 2 = B (2 + 1) (2 + 2) \Rightarrow B = \frac{10}{3 \times 4} = \frac{5}{6} Putting x + 2 = 0 or x = - 2 in eq (1) \Rightarrow 5 \times - 2 = C (- 2 + 1) (- 2 - 2) \Rightarrow \frac{- 10}{- 1 \times - 4} = C \Rightarrow C = \frac{- 5}{2} Putting x + 1 = 0 or x = - 1 in eq (1) \Rightarrow - 5 = A (- 1 - 2) (- 1 + 2) \Rightarrow \frac{- 5}{- 3} = A \Rightarrow A = \frac{5}{3} ∴ \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{5}{3} \times \frac{1}{x + 1} + \frac{5}{6 (x - 2)} - \frac{5}{2 (x + 2)} \Rightarrow \frac{5 x}{(x + 1) (x - 2) (x + 2)} = \frac{5}{6} \times \frac{2}{x + 1} + \frac{5}{6 (x - 2)} - \frac{5}{6} (\frac{3}{x + 2}) ∴ \int \frac{5 x}{(x + 1) (x - 2) (x + 2)} d x = \frac{5}{6} \int \frac{2}{x + 1} d x + \frac{5}{6} \int \frac{1}{x - 2} d x - \frac{5}{6} \int \frac{3}{x + 2} d x = \frac{5}{6} [2 \ln |x + 1| + \ln |x - 2| - 3 \ln |x + 2|] + C = \frac{5}{6} [\ln {|x + 1|}^{2} + \ln |x - 2| - \ln {|x + 2|}^{3}] + C = \frac{5}{6} \ln |\frac{{(x + 1)}^{2} (x - 2)}{{(x + 2)}^{3}}| + C$

Page No 18.176:

Question 8:

$\int \frac{x^{2} + 1}{x (x^{2} - 1)} d x$

Answer:

$\int \frac{(x^{2} + 1)}{x (x^{2} - 1)} d x = \int \frac{(x^{2} + 1)}{x (x - 1) (x + 1)} d x Let \frac{x^{2} + 1}{x (x - 1) (x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1} \Rightarrow \frac{x^{2} + 1}{x (x - 1) (x + 1)} = \frac{A (x - 1) (x + 1) + B (x) (x + 1) + C (x) (x - 1)}{x (x - 1) (x + 1)} \Rightarrow x^{2} + 1 = A (x - 1) (x + 1) + B (x) (x + 1) + C (x) (x - 1) . . . . (1) Putting x - 1 = 0 or x = 1 in eq (1) \Rightarrow 1 + 1 = A \times 0 + B (1) (1 + 1) + C \times 0 \Rightarrow B = 1 Putting x = 0 in eq (1) \Rightarrow 0 + 1 = A (0 - 1) (0 + 1) \Rightarrow A = - 1 Putting x + 1 = 0 or x = - 1 in eq (1) \Rightarrow {(- 1)}^{2} + 1 = A \times 0 + B \times 0 + C (- 1) (- 1 - 1) \Rightarrow 2 = C \times 2 \Rightarrow C = 1 ∴ \frac{x^{2} + 1}{x (x^{2} - 1)} = \frac{- 1}{x} + \frac{1}{x - 1} + \frac{1}{x + 1} \Rightarrow \int \frac{(x^{2} + 1)}{x (x^{2} - 1)} d x = - \int \frac{1}{x} d x + \int \frac{1}{x - 1} d x + \int \frac{1}{x + 1} d x = - \ln |x| + \ln |x - 1| + \ln |x + 1| + C = - \ln |x| + \ln |x^{2} - 1| + C = \ln |\frac{x^{2} - 1}{x}| + C$

Page No 18.176:

Question 9:

$\int \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)} d x$

Answer:

$\int \frac{2 x - 3}{(x^{2} - 1) (2 x + 3)} d x = \int \frac{(2 x - 3)}{(x - 1) (x + 1) (2 x + 3)} d x Let \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{2 x + 3} \Rightarrow \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} = \frac{A (x + 1) (2 x + 3) + B (x + 1) (2 x + 3) + C (x^{2} - 1)}{(x - 1) (x + 1) (2 x + 3)} \Rightarrow 2 x - 3 = A (x + 1) (2 x + 3) + B (x - 1) (2 x + 3) + C (x + 1) (x - 1) . . . (1) Putting x + 1 = 0 or x = - 1 in eq (1) \Rightarrow - 2 - 3 = B (- 1 - 1) (- 2 + 3) \Rightarrow - 5 = B (- 2) (1) \Rightarrow B = \frac{5}{2} Putting x - 1 = 0 or x = 1 in eq (1) \Rightarrow 2 - 3 = A (1 + 1) (2 + 3) \Rightarrow - 1 = A (2) (5) \Rightarrow A = \frac{- 1}{10} Putting 2 x + 3 = 0 or x = \frac{- 3}{2} in eq (1) \Rightarrow 2 \times - \frac{3}{2} - 3 = A \times 0 + B \times 0 + C (- \frac{3}{2} + 1) (\frac{- 3}{2} - 1) \Rightarrow - 6 = C (- \frac{1}{2}) (\frac{- 5}{2}) \Rightarrow C = - \frac{24}{5} ∴ \frac{2 x - 3}{(x - 1) (x + 1) (2 x + 3)} = \frac{- 1}{10 (x - 1)} + \frac{5}{2 (x + 1)} - \frac{24}{5 (2 x + 3)} \Rightarrow \int \frac{(2 x - 3)}{(x - 1) (x + 1) (2 x + 3)} d x = \frac{- 1}{10} \int \frac{1}{x - 1} d x + \frac{5}{2} \int \frac{1}{x + 1} d x - \frac{24}{5} \int \frac{1}{2 x + 3} d x = \frac{- 1}{10} \ln |x - 1| + \frac{5}{2} \ln |x + 1| - \frac{24}{5} \ln \frac{|2 x + 3|}{3} + C = - \frac{1}{10} \ln |x - 1| + \frac{5}{2} \ln |x + 1| - \frac{12}{5} \ln |2 x + 3| + C$

Page No 18.176:

Question 10:

$\int \frac{x^{3}}{(x - 1) (x - 2) (x - 3)} d x$

Answer:

$\int \frac{x^{3}}{(x - 1) (x - 2) (x - 3)} d x = \int \frac{x^{3}}{(x - 1) (x^{2} - 5 x + 6)} d x = \int \frac{x^{3}}{x^{3} - 5 x^{2} + 6 x - x^{2} + 5 x - 6} d x = \int \frac{x^{3}}{x^{3} - 6 x^{2} + 11 x - 6} d x ∴ \frac{x^{3}}{x^{3} - 6 x^{2} + 11 x - 6} = 1 + \frac{6 x^{2} + 11 x + 6}{x^{2} - 6 x^{2} + 11 x - 6} \Rightarrow \frac{x^{3}}{x^{3} - 6 x^{2} + 11 x - 6} = 1 + \frac{6 x^{2} - 11 x + 6}{(x - 1) (x - 2) (x - 3)} ∴ \int \frac{x^{3}}{(x - 1) (x - 2) (x - 3)} d x = \int d x + \int \frac{(6 x^{2} - 11 x + 6)}{(x - 1) (x - 2) (x - 3)} d x . . . (1)$

$Let \frac{6 x^{2} - 11 x + 6}{(x - 1) (x - 2) (x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3} \Rightarrow \frac{6 x^{2} - 11 x + 6}{(x - 1) (x - 2) (x - 3)} = \frac{A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2)}{(x - 1) (x - 2) (x - 3)} \Rightarrow 6 x^{2} - 11 x + 6 = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2) . . . (2) Putting x - 2 = 0 or x = 2 in eq (2) \Rightarrow 6 \times 4 - 22 + 6 = B (2 - 1) (2 - 3) \Rightarrow 8 = B (- 1) \Rightarrow B = - 8 Putting x - 3 = 0 or x = 3 in eq (2) \Rightarrow 6 \times 3^{2} - 11 \times 3 + 6 = C (3 - 1) (3 - 2) \Rightarrow 27 = C (2) (1) \Rightarrow C = \frac{27}{2} Putting x - 1 = 0 or x = 1 in eq (2) \Rightarrow 6 \times 1 - 11 + 6 = A (1 - 2) (1 - 3) \Rightarrow 1 = A (- 1) (- 2) \Rightarrow A = \frac{1}{2} ∴ \frac{6 x^{2} - 11 x + 6}{(x - 1) (x - 2) (x - 3)} = \frac{1}{2 (x - 1)} - \frac{8}{x - 2} + \frac{27}{2 (x - 3)} . . . (3) From eq (2) and eq (3) ∴ \int \frac{x^{3} d x}{(x - 1) (x - 2) (x - 3)} = \int d x + \frac{1}{2} \int \frac{1}{x - 1} d x - 8 \int \frac{1}{x - 2} d x + \frac{27}{2} \int \frac{1}{x - 3} d x = x + \frac{1}{2} \ln |x - 1| - 8 \ln |x - 2| + \frac{27}{2} \ln |x - 3| + C$

Page No 18.176:

Question 11:

$\int \frac{\sin 2 x}{(1 + \sin x) (2 + \sin x)} d x$

Answer:

$We have, I = \int \frac{\sin 2 x d x}{(1 + \sin x) (2 + \sin x)} = \int \frac{2 \sin x \cos x d x}{(1 + \sin x) (2 + \sin x)} Putting \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \frac{2 t d t}{(1 + t) (2 + t)} = 2 \int \frac{t d t}{(1 + t) (2 + t)} Let \frac{t}{(1 + t) (2 + t)} = \frac{A}{1 + t} + \frac{B}{2 + t} \Rightarrow \frac{t}{(1 + t) (2 + t)} = \frac{A (2 + t) + B (1 + t)}{(1 + t) (2 + t)} \Rightarrow t = A (2 + t) + B (1 + t) Putting 2 + t = 0 \Rightarrow t = - 2 - 2 = A \times 0 + B (- 2 + 1) \Rightarrow - 2 = B (- 1) \Rightarrow B = 2 let t + 1 = 0 t = - 1 \Rightarrow - 1 = A (- 1 + 2) + B \times 0 A = - 1 ∴ I = 2 \int (\frac{- 1}{t + 1} + \frac{2}{t + 2}) d t = 2 [- \log |t + 1| + 2 \log |t + 2|] + C = 4 \log |t + 2| - 2 \log |t + 1| + C = \log |\frac{{(t + 2)}^{4}}{{(t + 1)}^{2}}| + C = \log |\frac{{(\sin x + 2)}^{4}}{{(\sin x + 1)}^{2}}| + C$

Page No 18.176:

Question 12:

$\int \frac{2 x}{(x^{2} + 1) (x^{2} + 3)} d x$

Answer:

$We have, I = \int \frac{2 x d x}{(x^{2} + 1) (x^{2} + 3)} Putting x^{2} = t \Rightarrow 2 x d x = d t ∴ I = \int \frac{d t}{(t + 1) (t + 3)} Let \frac{1}{(t + 1) (t + 3)} = \frac{A}{t + 1} + \frac{B}{t + 3} \Rightarrow \frac{1}{(t + 1) (t + 3)} = \frac{A (t + 3) + B (t + 1)}{(t + 1) (t + 3)} \Rightarrow 1 = A (t + 3) + B (t + 1) Putting t + 3 = 0 \Rightarrow t = - 3 1 = A \times 0 + B (- 3 + 1) \Rightarrow B = - \frac{1}{2} Putting t + 1 = 0 \Rightarrow t = - 1 1 = A (- 1 + 3) + B (- 1 + 1) \Rightarrow 1 = A \times 2 + B \times 0 \Rightarrow A = \frac{1}{2} Then, I = \frac{1}{2} \int \frac{d t}{t + 1} - \frac{1}{2} \int \frac{d t}{t + 3} = \frac{1}{2} \log |t + 1| - \frac{1}{2} \log |t + 3| + C = \frac{1}{2} \log |\frac{t + 1}{t + 3}| + C = \frac{1}{2} \log |\frac{x^{2} + 1}{x^{2} + 3}| + C$

Page No 18.176:

Question 13:

$\int \frac{1}{x \log x (2 + \log x)} d x$

Answer:

$We have, I = \int \frac{d x}{x \log x (2 + \log x)} Putting \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = \int \frac{d t}{t (t + 2)} Let \frac{1}{t (t + 2)} = \frac{A}{t} + \frac{B}{t + 2} \Rightarrow \frac{1}{t (t + 2)} = \frac{A (t + 2) + B t}{t (t + 2)} \Rightarrow 1 = A (t + 2) + B t Putting t + 2 = 0 \Rightarrow t = - 2 1 = A \times 0 + B (- 2) \Rightarrow B = - \frac{1}{2} Putting t = 0 1 = A (0 + 2) + B \times 0 \Rightarrow A = \frac{1}{2} Then, I = \frac{1}{2} \int \frac{d t}{t} - \frac{1}{2} \int \frac{d t}{t + 2} = \frac{1}{2} [\log |t| - \log |t + 2|] + C = \frac{1}{2} \log |\frac{t}{t + 2}| + C = \frac{1}{2} \log |\frac{\log x}{\log x + 2}| + C$

Page No 18.176:

Question 14:

Evaluate the following integrals:

$\int \frac{x^{2} + x + 1}{(x^{2} + 1) (x + 2)} d x$

Answer:

$Let I = \int \frac{x^{2} + x + 1}{(x^{2} + 1) (x + 2)} d x We express \frac{x^{2} + x + 1}{(x^{2} + 1) (x + 2)} = \frac{A}{x + 2} + \frac{B x + C}{x^{2} + 1} \Rightarrow x^{2} + x + 1 = A (x^{2} + 1) + (B x + C) (x + 2) Equating the coefficients of x^{2}, x and constants, we get 1 = A + B and 1 = 2 B + C and 1 = A + 2 C or A = \frac{3}{5} and B = \frac{2}{5} and C = \frac{1}{5} ∴ I = \int (\frac{\frac{3}{5}}{x + 2} + \frac{\frac{2}{5} x + \frac{1}{5}}{x^{2} + 1}) d x = \frac{3}{5} \int \frac{1}{x + 2} d x + \frac{2}{5} \int \frac{x}{x^{2} + 1} d x + \frac{1}{5} \int \frac{1}{x^{2} + 1} d x = \frac{3}{5} I_{1} + \frac{2}{5} I_{2} + \frac{1}{5} I_{3} . . . (1) Now, I_{1} = \int \frac{1}{x + 2} d x Let x + 2 = u On differentiating both sides, we get d x = d u ∴ I_{1} = \int \frac{1}{u} d u = \log |u| + c_{1} = \log |x + 2| + c_{1} . . . (2) And, I_{2} = \int \frac{x}{x^{2} + 1} d x Let (x^{2} + 1) = u On differentiating both sides, we get 2 x d x = d u ∴ I_{2} = \frac{1}{2} \int \frac{1}{u} d u = \frac{1}{2} \log |u| + c_{2} = \frac{1}{2} \log |x^{2} + 1| + c_{2} . . . (3) And, I_{3} = \int \frac{1}{x^{2} + 1} d x = \tan^{- 1} x + c_{3} . . . (4) From (1), (2), (3) and (4), we get ∴ I = \frac{3}{5} (\log |x + 2| + c_{1}) + \frac{2}{5} (\frac{1}{2} \log |x^{2} + 1| + c_{2}) + \frac{1}{5} (\tan^{- 1} x + c_{3}) = \frac{3}{5} \log |x + 2| + \frac{1}{5} \log |x^{2} + 1| + \frac{1}{5} \tan^{- 1} x + c Hence, \int \frac{x^{2} + x + 1}{(x^{2} + 1) (x + 2)} d x = \frac{3}{5} \log |x + 2| + \frac{1}{5} \log |x^{2} + 1| + \frac{1}{5} \tan^{- 1} x + c$

Page No 18.176:

Question 15:

$\int \frac{a x^{2} + b x + c}{(x - a) (x - b) (x - c)} d x, where a, b, c are distinct$

Answer:

$We have, I = \int \frac{a x^{2} + b x + c}{(x - a) (x - b) (x - c)} d x Let \frac{a x^{2} + b x + c}{(x - a) (x - b) (x - c)} = \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c} \Rightarrow a x^{2} + b x + c = A (x - b) (x - c) + B (x - c) (x - a) + C (x - a) (x - b) \Rightarrow a x^{2} + b x + c = A [x^{2} - (b + c) x + b c] + B [x^{2} - (c + a) x + c a] + C [x^{2} - (a + b) x + a b] \Rightarrow a x^{2} + b x + c = (A + B + C) x^{2} - [A (b + c) + B (c + a) + C (a + b)] x + A b c + B c a + C a b Equating the coefficients on both sides, we get a = A + B + C . . . . . (1) b = - [A (b + c) + B (c + a) + C (a + b)] . . . . . (2) c = A b c + B c a + C a b . . . . . (3) Solving (1), (2) and (3), we get A = \frac{a^{3} + a b + c}{(a - b) (a - c)} B = \frac{a b^{2} + b^{2} + c}{(b - a) (b - c)} C = \frac{a c^{2} + b c + c}{(c - a) (c - b)} ∴ I = \int [\frac{a^{3} + a b + c}{(a - b) (a - c)} \times \frac{1}{x - a} + \frac{a b^{2} + b^{2} + c}{(b - a) (b - c)} \times \frac{1}{x - b} + \frac{a c^{2} + b c + c}{(c - a) (c - b)} \times \frac{1}{x - c}] d x = \frac{a^{3} + a b + c}{(a - b) (a - c)} \log |x - a| + \frac{a b^{2} + b^{2} + c}{(b - a) (b - c)} \log |x - b| + \frac{a c^{2} + b c + c}{(c - a) (c - b)} \log |x - c| + K$

Page No 18.176:

Question 16:

Evaluate the following integrals:

$\int \frac{x}{(x^{2} + 1) (x - 1)} d x$

Answer:

$Let I = \int \frac{x}{(x^{2} + 1) (x - 1)} d x We express \frac{x}{(x^{2} + 1) (x - 1)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + 1} \Rightarrow x = A (x^{2} + 1) + (B x + C) (x - 1) Equating the coefficients of x^{2}, x and constants, we get 0 = A + B and 1 = - B + C and 0 = A - C or A = \frac{1}{2} and B = - \frac{1}{2} and C = \frac{1}{2} ∴ I = \int (\frac{\frac{1}{2}}{x - 1} + \frac{- \frac{1}{2} x + \frac{1}{2}}{x^{2} + 1}) d x = \frac{1}{2} \int \frac{1}{x - 1} d x - \frac{1}{2} \int \frac{x}{x^{2} + 1} d x + \frac{1}{2} \int \frac{1}{x^{2} + 1} d x = \frac{1}{2} I_{1} - \frac{1}{2} I_{2} + \frac{1}{2} I_{3} . . . (1) Now, I_{1} = \int \frac{1}{x - 1} d x Let x - 1 = u On differentiating both sides, we get d x = d u ∴ I_{1} = \int \frac{1}{u} d u = \log |u| + c_{1} = \log |x - 1| + c_{1} . . . (2) And, I_{2} = \int \frac{x}{x^{2} + 1} d x Let (x^{2} + 1) = u On differentiating both sides, we get 2 x d x = d u ∴ I_{2} = \frac{1}{2} \int \frac{1}{u} d u = \frac{1}{2} \log |u| + c_{2} = \frac{1}{2} \log |x^{2} + 1| + c_{2} . . . (3) And, I_{3} = \int \frac{1}{x^{2} + 1} d x = \tan^{- 1} x + c_{3} . . . (4) From (1), (2), (3) and (4), we get ∴ I = \frac{1}{2} (\log |x - 1| + c_{1}) - \frac{1}{2} (\frac{1}{2} \log |x^{2} + 1| + c_{2}) + \frac{1}{2} (\tan^{- 1} x + c_{3}) = \frac{1}{2} \log |x - 1| - \frac{1}{4} \log |x^{2} + 1| + \frac{1}{2} \tan^{- 1} x + c Hence, \int \frac{x}{(x^{2} + 1) (x - 1)} d x = \frac{1}{2} \log |x - 1| - \frac{1}{4} \log |x^{2} + 1| + \frac{1}{2} \tan^{- 1} x + c$

Page No 18.176:

Question 17:

$\int \frac{1}{(x - 1) (x + 1) (x + 2)} d x$

Answer:

$We have, I = \int \frac{d x}{(x - 1) (x + 1) (x + 2)} Let \frac{1}{(x - 1) (x + 1) (x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 2} \Rightarrow \frac{1}{(x - 1) (x + 1) (x + 2)} = \frac{A (x + 1) (x + 2) + B (x - 1) (x + 2) + C (x - 1) (x + 1)}{(x - 1) (x + 1) (x + 2)} \Rightarrow 1 = A (x + 1) (x + 2) + B (x - 1) (x + 2) + C (x - 1) (x + 1) Putting x - 1 = 0 \Rightarrow x = 1 1 = A (1 + 1) (1 + 2) + B \times 0 + C \times 0 \Rightarrow 1 = A \times 6 \Rightarrow A = \frac{1}{6} Putting x + 1 = 0 \Rightarrow x = - 1 1 = A \times 0 + B (- 2) (1) + C \times 0 \Rightarrow B = - \frac{1}{2} Putting x + 2 = 0 \Rightarrow x = - 2 1 = A \times 0 + B \times 0 + C (- 2 - 1) (- 2 + 1) \Rightarrow 1 = C \times 3 \Rightarrow C = \frac{1}{3} ∴ I = \frac{1}{6} \int \frac{d x}{x - 1} - \frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{3} \int \frac{d x}{x + 2} = \frac{1}{6} \log |x - 1| - \frac{1}{2} \log |x + 1| + \frac{1}{3} \log |x + 2| + C = \frac{1}{6} \log |x - 1| - \frac{3}{6} \log |x + 1| + \frac{2}{6} \log |x + 2| + C = \frac{1}{6} [\log |x - 1| - 3 \log |x + 1| + 2 \log |x + 2|] + C = \frac{1}{6} \log |\frac{(x - 1) {(x + 2)}^{2}}{{(x + 1)}^{3}}| + C$

Page No 18.176:

Question 18:

Evaluate the following integrals:

$\int \frac{x^{2}}{(x^{2} + 4) (x^{2} + 9)} d x$

Answer:

$Let I = \int \frac{x^{2}}{(x^{2} + 4) (x^{2} + 9)} d x We express \frac{x^{2}}{(x^{2} + 4) (x^{2} + 9)} = \frac{A x + B}{x^{2} + 4} + \frac{C x + D}{x^{2} + 9} \Rightarrow x^{2} = (A x + B) (x^{2} + 9) + (C x + D) (x^{2} + 4) Equating the coefficients of x^{3}, x^{2}, x and constants, we get 0 = A + C and 1 = B + D and 0 = 9 A + 4 C and 0 = 9 B + 4 D or A = 0 and B = - \frac{4}{5} and C = 0 and D = \frac{9}{5} ∴ I = \int (\frac{- \frac{4}{5}}{x^{2} + 4} + \frac{\frac{9}{5}}{x^{2} + 9}) d x = - \frac{4}{5} \int \frac{1}{x^{2} + 4} d x + \frac{9}{5} \int \frac{1}{x^{2} + 9} d x = - \frac{4}{5} \times \frac{1}{2} \tan^{- 1} \frac{x}{2} + \frac{9}{5} \times \frac{1}{3} \tan^{- 1} \frac{x}{3} + c = - \frac{2}{5} \tan^{- 1} \frac{x}{2} + \frac{3}{5} \tan^{- 1} \frac{x}{3} + c Hence, \int \frac{x^{2}}{(x^{2} + 4) (x^{2} + 9)} d x = - \frac{2}{5} \tan^{- 1} \frac{x}{2} + \frac{3}{5} \tan^{- 1} \frac{x}{3} + c$

Page No 18.176:

Question 19:

$\int \frac{5 x^{2} - 1}{x (x - 1) (x + 1)} d x$

Answer:

$We have, I = \int \frac{(5 x^{2} - 1) d x}{x (x - 1) (x + 1)} Let \frac{5 x^{2} - 1}{x (x - 1) (x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1} \Rightarrow \frac{5 x^{2} - 1}{x (x - 1) (x + 1)} = \frac{A (x^{2} - 1) + B x \cdot (x + 1) + C \cdot x \cdot (x - 1)}{x (x - 1) (x + 1)} \Rightarrow 5 x^{2} - 1 = A (x^{2} - 1) + B \cdot x (x + 1) + C \cdot x \cdot (x - 1) Putting x = 1 \Rightarrow 5 - 1 = A \times 0 + B (1) (1 + 1) + C \times 0 \Rightarrow 4 = B (2) \Rightarrow B = 2 Putting x = 0 \Rightarrow 5 \times 0 - 1 = A (0 - 1) + B \times 0 + C \times 0 \Rightarrow - 1 = A (- 1) \Rightarrow A = 1 Putting x + 1 = 0 x = - 1 5 - 1 = A \times 0 + B \times 0 + C (- 1) (- 2) \Rightarrow C = 2 ∴ I = \int \frac{d x}{x} + 2 \int \frac{d x}{x - 1} + 2 \int \frac{d x}{x + 1} = \log |x| + 2 \log |x - 1| + 2 \log |x + 1| + C = \log |x| + 2 \{\log |x^{2} - 1|\} + C = \log |x {(x^{2} - 1)}^{2}| + C$

Page No 18.176:

Question 20:

$\int \frac{x^{2} + 6 x - 8}{x^{3} - 4 x} d x$

Answer:

$We have, I = \int (\frac{x^{2} + 6 x - 8}{x^{3} - 4 x}) d x = \int \frac{(x^{2} + 6 x - 8)}{x (x^{2} - 4)} d x = \int \frac{(x^{2} + 6 x - 8)}{x (x - 2) (x + 2)} d x Let \frac{x^{2} + 6 x - 8}{x (x - 2) (x + 2)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2} \Rightarrow \frac{x^{2} + 6 x - 8}{x (x - 2) (x + 2)} = \frac{A (x - 2) (x + 2) + B (x) (x + 2) + C (x) (x - 2)}{x (x - 2) (x + 2)} \Rightarrow x^{2} + 6 x - 8 = A (x^{2} - 4) + B (x^{2} + 2 x) + C (x^{2} - 2 x) Putting x - 2 = 0 \Rightarrow x = 2 4 + 6 \times 2 - 8 = A \times 0 + B (4 + 4) \Rightarrow 8 = B \times 8 \Rightarrow B = 1 Putting x = - 2 4 - 12 - 8 = A \times 0 + B \times 0 + C \times 8 \Rightarrow C = - 2 Putting x = 0 - 8 = A (- 4) + B \times 0 + C \times 0 \Rightarrow A = 2 ∴ I = \int \frac{2}{x} + \int \frac{d x}{x - 2} - 2 \int \frac{d x}{x + 2} = 2 \log |x| + \log |x - 2| - 2 \log |x + 2| + C = \log x^{2} + \log |x - 2| - \log {|x + 2|}^{2} + C = \log |\frac{x^{2} (x - 2)}{{(x + 2)}^{2}}| + C$

Page No 18.176:

Question 21:

$\int \frac{x^{2} + 1}{(2 x + 1) (x^{2} - 1)} d x$

Answer:

$We have, I = \int \frac{(x^{2} + 1) d x}{(2 x + 1) (x^{2} - 1)} = \int \frac{(x^{2} + 1) d x}{(2 x + 1) (x - 1) (x + 1)} Let \frac{(x^{2} + 1)}{(2 x + 1) (x - 1) (x + 1)} = \frac{A}{2 x + 1} + \frac{B}{x - 1} + \frac{C}{x + 1} \Rightarrow \frac{(x^{2} + 1)}{(2 x + 1) (x - 1) (x + 1)} = \frac{A (x^{2} - 1) + B (2 x + 1) (x + 1) + C (2 x + 1) (x - 1)}{(2 x + 1) (x - 1) (x + 1)} \Rightarrow x^{2} + 1 = A (x^{2} - 1) + B (2 x + 1) (x + 1) + C (2 x + 1) (x - 1) Putting x - 1 = 0 \Rightarrow x = 1 1 + 1 = A \times 0 + B (2 + 1) (1 + 1) + C \times 0 \Rightarrow 2 = B (3) (2) \Rightarrow B = \frac{1}{3} Putting x + 1 = 0 \Rightarrow x = - 1 1 + 1 = A \times 0 + B \times 0 + C (- 2 + 1) (- 1 - 1) \Rightarrow 2 = C (- 1) (- 2) \Rightarrow C = 1 Putting 2 x + 1 = 0 \Rightarrow x = - \frac{1}{2} {(- \frac{1}{2})}^{2} + 1 = A (\frac{1}{4} - 1) \Rightarrow \frac{1}{4} + 1 = A (- \frac{3}{4}) \Rightarrow \frac{5}{4} = A (- \frac{3}{4}) A = - \frac{5}{3} ∴ I = - \frac{5}{3} \int \frac{d x}{2 x + 1} + \frac{1}{3} \int \frac{d x}{x - 1} + \int \frac{d x}{x + 1} = - \frac{5}{3} \times \frac{\log |2 x + 1|}{2} + \frac{1}{3} \log |x - 1| + \log |x + 1| + C = - \frac{5}{6} \log |2 x + 1| + \frac{1}{3} \log |x - 1| + \log |x + 1| + C$

Page No 18.177:

Question 22:

$\int \frac{1}{x [6 {(\log x)}^{2} + 7 \log x + 2]} d x$

Answer:

$We have, I = \int \frac{d x}{x \{6 {(\log x)}^{2} + 7 \log x + 2\}} Putting \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = \int \frac{d t}{6 t^{2} + 7 t + 2} = \int \frac{d t}{(3 t + 2) (2 t + 1)} Let \frac{1}{(3 t + 2) (2 t + 1)} = \frac{A}{3 t + 2} + \frac{B}{2 t + 1} \Rightarrow \frac{1}{(3 t + 2) (2 t + 1)} = \frac{A (2 t + 1) + B (3 t + 2)}{(3 t + 2) (2 t + 1)} \Rightarrow 1 = A (2 t + 1) + B (3 t + 2) Putting 2 t + 1 = 0 \Rightarrow t = - \frac{1}{2} 1 = 0 + B (3 \times - \frac{1}{2} + 2) \Rightarrow 1 = B (\frac{1}{2}) \Rightarrow B = 2 Putting 3 t + 2 = 0 \Rightarrow t = - \frac{2}{3} 1 = A (2 \times - \frac{2}{3} + 1) + 0 \Rightarrow 1 = A (- \frac{4}{3} + 1) \Rightarrow 1 = A (- \frac{1}{3}) \Rightarrow A = - 3 ∴ I = \int (- \frac{3}{3 t + 2} + \frac{2}{2 t + 1}) d t = - 3 \frac{\log |3 t + 2|}{3} + 2 \frac{\log |2 t + 1|}{2} + C = - \log |3 t + 2| + \log |2 t + 1| + C = \log |\frac{2 t + 1}{3 t + 2}| + C = \log |\frac{2 \log x + 1}{3 \log x + 2}| + C$

Page No 18.177:

Question 23:

$\int \frac{1}{x (x^{n} + 1)} d x$

Answer:

$We have, I = \int \frac{d x}{x (x^{n} + 1)} = \int \frac{x^{n - 1} d x}{x^{n - 1} x (x^{n} + 1)} = \int \frac{x^{n - 1} d x}{x^{n} (x^{n} + 1)} Putting x^{n} = t \Rightarrow n x^{n - 1} d x = d t \Rightarrow x^{n - 1} d x = \frac{d t}{n} ∴ I = \frac{1}{n} \int \frac{d t}{t (t + 1)} Let \frac{1}{t (t + 1)} = \frac{A}{t} + \frac{B}{t + 1} \Rightarrow \frac{1}{t (t + 1)} = \frac{A (t + 1) + B t}{t (t + 1)} \Rightarrow 1 = A (t + 1) + B t Putting t + 1 = 0 \Rightarrow t = - 1 1 = A \times 0 + B (- 1) \Rightarrow B = - 1 Putting t = 0 1 = A (0 + 1) + B \times 0 \Rightarrow A = 1 Then, I = \frac{1}{n} \int \frac{d t}{t} - \frac{1}{n} \int \frac{d t}{t + 1} = \frac{1}{n} \log |t| - \frac{1}{n} \log |t + 1| + C = \frac{1}{n} \log |\frac{t}{t + 1}| + C = \frac{1}{n} \log |\frac{x^{n}}{x^{n} + 1}| + C$

Page No 18.177:

Question 24:

$\int \frac{x}{(x^{2} - a^{2}) (x^{2} - b^{2})} d x$

Answer:

$We have, I = \int \frac{x d x}{(x^{2} - a^{2}) (x^{2} - b^{2})} Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{(t - a^{2}) (t - b^{2})} Let \frac{1}{(t - a^{2}) (t - b^{2})} = \frac{A}{t - a^{2}} + \frac{B}{t - b^{2}} \Rightarrow \frac{1}{(t - a^{2}) (t - b^{2})} = \frac{A (t - b^{2}) + B (t - a^{2})}{(t - a^{2}) (t - b^{2})} \Rightarrow 1 = A (t - b^{2}) + B (t - a^{2}) Putting t = b^{2} 1 = A \times 0 + B (b^{2} - a^{2}) \Rightarrow B = \frac{1}{b^{2} - a^{2}} Putting t = a^{2} 1 = A (a^{2} - b^{2}) + B \times 0 \Rightarrow A = \frac{1}{a^{2} - b^{2}} I = \frac{1}{2} \int \frac{d t}{(t - a^{2}) (t - b^{2})} = \frac{1}{2 (a^{2} - b^{2})} \int \frac{d t}{t - a^{2}} + \frac{1}{2 (b^{2} - a^{2})} \int \frac{d t}{t - b^{2}} = \frac{1}{2 (a^{2} - b^{2})} \log |t - a^{2}| + \frac{1}{2 (b^{2} - a^{2})} \log |t - b^{2}| + C = \frac{1}{2 (a^{2} - b^{2})} [\log |t - a^{2}| - \log |t - b^{2}|] + C = \frac{1}{2 (a^{2} - b^{2})} [\log |\frac{t - a^{2}}{t - b^{2}}|] + C = \frac{1}{2 (a^{2} - b^{2})} \log |\frac{x^{2} - a^{2}}{x^{2} - b^{2}}| + C$

Page No 18.177:

Question 25:

Evaluate the following integrals:

$\int \frac{x^{2} + 1}{(x^{2} + 4) (x^{2} + 25)} d x$

Answer:

$Let I = \int \frac{x^{2} + 1}{(x^{2} + 4) (x^{2} + 25)} d x We express \frac{x^{2} + 1}{(x^{2} + 4) (x^{2} + 25)} = \frac{A x + B}{x^{2} + 4} + \frac{C x + D}{x^{2} + 25} \Rightarrow x^{2} + 1 = (A x + B) (x^{2} + 25) + (C x + D) (x^{2} + 4) Equating the coefficients of x^{3}, x^{2}, x and constants, we get 0 = A + C and 1 = B + D and 0 = 25 A + 4 C and 1 = 25 B + 4 D or A = 0 and B = - \frac{1}{7} and C = 0 and D = \frac{8}{7} ∴ I = \int (\frac{- \frac{1}{7}}{x^{2} + 4} + \frac{\frac{8}{7}}{x^{2} + 25}) d x = - \frac{1}{7} \int \frac{1}{x^{2} + 4} d x + \frac{8}{7} \int \frac{1}{x^{2} + 25} d x = - \frac{1}{7} \times \frac{1}{2} \tan^{- 1} \frac{x}{2} + \frac{8}{7} \times \frac{1}{5} \tan^{- 1} \frac{x}{5} + c = - \frac{1}{14} \tan^{- 1} \frac{x}{2} + \frac{8}{35} \tan^{- 1} \frac{x}{5} + c Hence, \int \frac{x^{2} + 1}{(x^{2} + 4) (x^{2} + 25)} d x = - \frac{1}{14} \tan^{- 1} \frac{x}{2} + \frac{8}{35} \tan^{- 1} \frac{x}{5} + c$ A

Page No 18.177:

Question 26:

Evaluate the following integrals:

$\int \frac{x^{3} + x + 1}{x^{2} - 1} d x$ $\int \frac{x^{3} + x + 1}{x^{2} - 1} d x$

Answer:

$Let I = \int \frac{x^{3} + x + 1}{x^{2} - 1} d x Here the integrand \frac{x^{3} + x + 1}{x^{2} - 1} is not a proper rational function, so we divide x^{3} + x + 1 by x^{2} - 1 and find that \frac{x^{3} + x + 1}{x^{2} - 1} = x + \frac{2 x + 1}{x^{2} - 1} = x + \frac{2 x + 1}{(x + 1) (x - 1)} Let \frac{2 x + 1}{(x + 1) (x - 1)} = \frac{A}{x + 1} + \frac{B}{x - 1} \Rightarrow 2 x + 1 = A (x - 1) + B (x + 1) Equating the coefficients of x and constants, we get 2 = A + B and 1 = - A + B or A = \frac{1}{2} and B = \frac{3}{2} ∴ I = \int (x + \frac{\frac{1}{2}}{x + 1} + \frac{\frac{3}{2}}{x - 1}) d x = \int x d x + \frac{1}{2} \int \frac{1}{x + 1} d x + \frac{3}{2} \int \frac{1}{x - 1} d x = \frac{x^{2}}{2} + \frac{1}{2} \log |x + 1| + \frac{3}{2} \log |x - 1| + c = \frac{x^{2}}{2} + \frac{1}{2} \log |x + 1| + \frac{3}{2} \log |x - 1| + c Hence, \int \frac{x^{3} + x + 1}{x^{2} - 1} d x = \frac{x^{2}}{2} + \frac{1}{2} \log |x + 1| + \frac{3}{2} \log |x - 1| + c$

Page No 18.177:

Question 27:

Evaluate the following integrals:

$\int \frac{3 x - 2}{{(x + 1)}^{2} (x + 3)} d x$

Answer:

$Let I = \int \frac{3 x - 2}{{(x + 1)}^{2} (x + 3)} d x We express \frac{3 x - 2}{{(x + 1)}^{2} (x + 3)} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{C}{x + 3} \Rightarrow 3 x - 2 = A (x + 1) (x + 3) + B (x + 3) + C {(x + 1)}^{2} Equating the coefficients of x^{2}, x and constants, we get 0 = A + C and 3 = 4 A + B + 2 C and - 2 = 3 A + 3 B + C or A = \frac{11}{4} and B = - \frac{5}{2} and C = - \frac{11}{4} ∴ I = \int (\frac{\frac{11}{4}}{x + 1} + \frac{- \frac{5}{2}}{{(x + 1)}^{2}} + \frac{- \frac{11}{4}}{x + 3}) d x = \frac{11}{4} \int \frac{1}{x + 1} d x - \frac{5}{2} \int \frac{1}{{(x + 1)}^{2}} d x - \frac{11}{4} \int \frac{1}{x + 3} d x = \frac{11}{4} \log |x + 1| + \frac{5}{2 (x + 1)} - \frac{11}{4} \log |x + 3| + c Hence, \int \frac{3 x - 2}{{(x + 1)}^{2} (x + 3)} d x = \frac{11}{4} \log |x + 1| + \frac{5}{2 (x + 1)} - \frac{11}{4} \log |x + 3| + c$

Page No 18.177:

Question 28:

$\int \frac{2 x + 1}{(x + 2) {(x - 3)}^{2}} d x$

Answer:

$We have, I = \int \frac{(2 x + 1) d x}{(x + 2) {(x - 3)}^{2}} Let \frac{2 x + 1}{(x + 2) {(x - 3)}^{2}} = \frac{A}{x + 2} + \frac{B}{x - 3} + \frac{C}{{(x - 3)}^{2}} \Rightarrow \frac{2 x + 1}{(x + 2) {(x - 3)}^{2}} = \frac{A {(x - 3)}^{2} + B (x + 2) (x - 3) + C (x + 2)}{(x + 2) {(x - 3)}^{2}} \Rightarrow 2 x + 1 = A (x^{2} - 6 x + 9) + B (x^{2} - x - 6) + C (x + 2) \Rightarrow 2 x + 1 = (A + B) x^{2} + (- 6 A - B + C) x + (9 A - 6 B + 2 C) Equating the coefficients of like terms A + B = 0 . . . . . (1) - 6 A - B + C = 2 . . . . . (2) 9 A - 6 B + 2 C = 1 . . . . . (3) Solving (1), (2) and (3), we get A = - \frac{3}{25}, B = \frac{3}{25} and C = \frac{7}{5} ∴ \frac{(2 x + 1) d x}{(x + 2) {(x - 3)}^{2}} = - \frac{3}{25 (x + 2)} + \frac{3}{25 (x - 3)} + \frac{7}{5 {(x - 3)}^{2}} \Rightarrow I = - \frac{3}{25} \int \frac{d x}{x + 2} + \frac{3}{25} \int \frac{d x}{x - 3} + \frac{7}{5} \int {(x - 3)}^{- 2} d x = - \frac{3}{25} \log |x + 2| + \frac{3}{25} \log |x - 3| + \frac{7}{5} [\frac{{(x - 3)}^{- 1}}{- 1}] + C = - \frac{3}{25} \log |x + 2| + \frac{3}{25} \log |x - 3| - \frac{7}{5 (x - 3)} + C$

Page No 18.177:

Question 29:

$\int \frac{x^{2} + 1}{{(x - 2)}^{2} (x + 3)} d x$

Answer:

$We have, I = \int \frac{(x^{2} + 1) d x}{{(x - 2)}^{2} (x + 3)} Let \frac{x^{2} + 1}{{(x - 2)}^{2} (x + 3)} = \frac{A}{x - 2} + \frac{B}{{(x - 2)}^{2}} + \frac{C}{x + 3} \Rightarrow \frac{x^{2} + 1}{{(x - 2)}^{2} (x + 3)} = \frac{A (x - 2) (x + 3) + B (x + 3) + C {(x - 2)}^{2}}{{(x - 2)}^{2} (x + 3)} \Rightarrow x^{2} + 1 = A (x^{2} - 2 x + 3 x - 6) + B (x + 3) + C (x^{2} - 4 x + 4) \Rightarrow x^{2} + 1 = A (x^{2} + x - 6) + B (x + 3) + C (x^{2} - 4 x + 4) Equating coefficients of like terms A + C = 1 . . . . . (1) A + B - 4 C = 0 . . . . . (2) - 6 A + 3 B + 4 C = 1 . . . . . (3) Solving (1), (2) and (3), we get A = \frac{3}{5}, B = 1 and C = \frac{2}{5} ∴ I = \frac{3}{5} \int \frac{d x}{x - 2} + \int \frac{d x}{{(x - 2)}^{2}} + \frac{2}{5} \int \frac{d x}{x + 3} = \frac{3}{5} \log |x - 2| + [\frac{{(x - 2)}^{- 2 + 1}}{- 2 + 1}] + \frac{2}{5} \log |x + 3| + C = \frac{3}{5} \log |x - 2| - \frac{1}{x - 2} + \frac{2}{5} \log |x + 3| + C$

Page No 18.177:

Question 30:

$\int \frac{x}{{(x - 1)}^{2} (x + 2)} d x$

Answer:

$We have, I = \int \frac{x d x}{{(x - 1)}^{2} (x + 2)} Let \frac{x}{{(x - 1)}^{2} (x + 2)} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{x + 2} \Rightarrow \frac{x}{{(x - 1)}^{2} (x + 2)} = \frac{A (x - 1) (x + 2) + B (x + 2) + C {(x - 1)}^{2}}{{(x - 1)}^{2} (x + 2)} \Rightarrow x = A (x^{2} + 2 x - x - 2) + B (x + 2) + C (x^{2} - 2 x + 1) \Rightarrow x = A (x^{2} + x - 2) + B (x + 2) + C (x^{2} - 2 x + 1) \Rightarrow x = (A + C) x^{2} + (A + B - 2 C) x + (- 2 A + 2 B + C) Equating coefficients of like terms A + C = 0 . . . . . (1) A + B - 2 C = 1 . . . . . (2) - 2 A + 2 B + C = 0 . . . . . (3) Solving (1), (2) and (3), we get A = \frac{2}{9}, B = \frac{1}{3} and C = - \frac{2}{9} ∴ \frac{x}{{(x - 1)}^{2} (x + 2)} = \frac{2}{9 (x - 1)} + \frac{1}{3 {(x - 1)}^{2}} - \frac{2}{9 (x + 2)} \Rightarrow I = \frac{2}{9} \int \frac{d x}{x - 1} + \frac{1}{3} \int \frac{d x}{{(x - 1)}^{2}} - \frac{2}{9} \int \frac{d x}{x + 2} = \frac{2}{9} \log |x - 1| + \frac{1}{3} \times (\frac{- 1}{x - 1}) - \frac{2}{9} \log |x + 2| + C = \frac{2}{9} \log |\frac{x - 1}{x + 2}| - \frac{1}{3 (x - 1)} + C$

Page No 18.177:

Question 31:

$\int \frac{x^{2}}{(x - 1) {(x + 1)}^{2}} d x$

Answer:

$We have, I = \int \frac{x^{2} d x}{(x - 1) {(x + 1)}^{2}} Let \frac{x^{2}}{(x - 1) {(x + 1)}^{2}} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{{(x + 1)}^{2}} \Rightarrow \frac{x^{2}}{(x - 1) {(x + 1)}^{2}} = \frac{A {(x + 1)}^{2} + B (x + 1) (x - 1) + C (x - 1)}{{(x + 1)}^{2} (x - 1)} \Rightarrow x^{2} = A (x^{2} + 2 x + 1) + B (x^{2} - 1) + C (x - 1) \Rightarrow x^{2} = (A + B) x^{2} + x (2 A + C) + (A - B - C) Equating coefficients of like terms A + B = 1 . . . . . (1) 2 A + C = 0 . . . . . (2) A - B - C = 0 . . . . . (3) Solving (1), (2) and (3), we get A = \frac{1}{4}, B = \frac{3}{4} and C = - \frac{1}{2} ∴ \frac{x^{2}}{(x - 1) {(x + 1)}^{2}} = \frac{1}{4 (x - 1)} + \frac{3}{4 (x + 1)} - \frac{1}{2 {(x + 1)}^{2}} \Rightarrow I = \frac{1}{4} \int \frac{d x}{x - 1} + \frac{3}{4} \int \frac{d x}{x + 1} - \frac{1}{2} \int \frac{d x}{{(x + 1)}^{2}} = \frac{1}{4} \log |x - 1| + \frac{3}{4} \log |x + 1| - \frac{1}{2} \times \frac{- 1}{x + 1} + C = \frac{1}{4} \log |x - 1| + \frac{3}{4} \log |x + 1| + \frac{1}{2 (x + 1)} + C$

Page No 18.177:

Question 32:

$\int \frac{x^{2} + x - 1}{{(x + 1)}^{2} (x + 2)} d x$

Answer:

$We have, I = \int \frac{(x^{2} + x - 1) d x}{{(x + 1)}^{2} (x + 2)} Let \frac{x^{2} + x - 1}{{(x + 1)}^{2} (x + 2)} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{C}{x + 2} \Rightarrow \frac{x^{2} + x - 1}{{(x + 1)}^{2} (x + 1)} = \frac{A (x + 1) (x + 2) + B (x + 2) + C {(x + 1)}^{2}}{{(x + 1)}^{2} (x + 2)} \Rightarrow x^{2} + x - 1 = A (x^{2} + 3 x + 2) + B (x + 2) + C (x^{2} + 2 x + 1) Equating coefficients of like terms A + C = 1 . . . . . (1) 3 A + B + 2 C = 1 . . . . . (2) 2 A + 2 B + C = - 1 . . . . . (3) Solving (1), (2) and (3), we get A = 0 B = - 1 C = 1 ∴ \frac{x^{2} + x - 1}{{(x + 1)}^{2} (x + 2)} = \frac{- 1}{{(x + 1)}^{2}} + \frac{1}{x + 2} \Rightarrow I = \int \frac{- d x}{{(x + 1)}^{2}} + \int \frac{d x}{x + 2} = - \int {(x + 1)}^{- 2} d x + \int \frac{d x}{x + 2} = - [\frac{{(x + 1)}^{- 2 + 1}}{- 2 + 1}] + \log |x + 2| + C = \frac{1}{(x + 1)} + \log |x + 2| + C$

Page No 18.177:

Question 33:

$\int \frac{2 x^{2} + 7 x - 3}{x^{2} (2 x + 1)} d x$

Answer:

$We have, I = \int \frac{(2 x^{2} + 7 x - 3) d x}{x^{2} (2 x + 1)} Let \frac{2 x^{2} + 7 x - 3}{x^{2} (2 x + 1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{2 x + 1} \Rightarrow \frac{2 x^{2} + 7 x - 3}{x^{2} (2 x + 1)} = \frac{A (x) (2 x + 1) + B (2 x + 1) + C x^{2}}{x^{2} (2 x + 1)} \Rightarrow 2 x^{2} + 7 x - 3 = A (2 x^{2} + x) + B (2 x + 1) + C x^{2} \Rightarrow 2 x^{2} + 7 x - 3 = (2 A + C) x^{2} + (A + 2 B) x + B Equating coefficients of like terms 2 A + C = 2 . . . . . (1) A + 2 B = 7 . . . . . (2) B = - 3 . . . . . (3) Solving (1), (2) and (3), we get A = 13 B = - 3 C = - 24 ∴ \frac{2 x^{2} + 7 x - 3}{x^{2} (2 x + 1)} = \frac{13}{x} - \frac{3}{x^{2}} - \frac{24}{2 x + 1} \Rightarrow I = 13 \int \frac{d x}{x} - 3 \int x^{- 2} d x - 24 \int \frac{d x}{2 x + 1} = 13 \log |x| + \frac{3}{x} - 24 \frac{\log |2 x + 1|}{2} + C = 13 \log |x| + \frac{3}{x} - 12 \log |2 x + 1| + C$

Page No 18.177:

Question 34:

$\int \frac{5 x^{2} + 20 x + 6}{x^{3} + 2 x^{2} + x} d x$

Answer:

$We have, I = \int \frac{5 x^{2} + 20 x + 6}{x^{3} + 2 x^{2} + x} = \int \frac{(5 x^{2} + 20 x + 6) d x}{x (x^{2} + 2 x + 1)} = \int \frac{(5 x^{2} + 20 x + 6) d x}{x {(x + 1)}^{2}} Let \frac{5 x^{2} + 20 x + 6}{x {(x + 1)}^{2}} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{{(x + 1)}^{2}} \Rightarrow \frac{5 x^{2} + 20 x + 6}{x {(x + 1)}^{2}} = \frac{A {(x + 1)}^{2} + B (x) (x + 1) + C (x)}{x {(x + 1)}^{2}} \Rightarrow 5 x^{2} + 20 x + 6 = A (x^{2} + 2 x + 1) + B (x^{2} + x) + C x \Rightarrow 5 x^{2} + 20 x + 6 = (A + B) x^{2} + (2 A + B + C) x + A Equating coefficients of like terms A + B = 5 . . . . . (1) 2 A + B + C = 20 . . . . . (2) A = 6 . . . . . (3) Solving (1), (2) and (3), we get A = 6 B = - 1 C = 9 ∴ \frac{5 x^{2} + 20 x + 6}{x {(x + 1)}^{2}} = \frac{6}{x} - \frac{1}{x + 1} + \frac{9}{{(x + 1)}^{2}} \Rightarrow I = 6 \int \frac{d x}{x} - \int \frac{d x}{x + 1} + 9 \int \frac{d x}{{(x + 1)}^{2}} = 6 \log |x| - \log |x + 1| - \frac{9}{x + 1} + C$

Page No 18.177:

Question 35:

$\int \frac{18}{(x + 2) (x^{2} + 4)} d x$

Answer:

$We have, I = \int \frac{18}{(x + 2) (x^{2} + 4)} d x Let \frac{18}{(x + 2) (x^{2} + 4)} = \frac{A}{x + 2} + \frac{B x + C}{x^{2} + 4} \Rightarrow \frac{18}{(x + 2) (x^{2} + 4)} = \frac{A (x^{2} + 4) + (B x + C) (x + 2)}{(x + 2) (x^{2} + 4)} \Rightarrow 18 = A x^{2} + 4 A + B x^{2} + 2 B x + C x + 2 C \Rightarrow 18 = (A + B) x^{2} + x (2 B + C) + 4 A + 2 C Equating coefficients of like terms A + B = 0 . . . . . (1) 2 B + C = 0 . . . . . (2) 4 A + 2 C = 18 . . . . . (3) Solving (1), (2) and (3), we get A = \frac{9}{4} B = - \frac{9}{4} C = \frac{9}{2} ∴ \frac{18}{(x + 2) (x^{2} + 4)} = \frac{9}{4 (x + 2)} + \frac{- \frac{9}{4} x + \frac{9}{2}}{x^{2} + 4} \Rightarrow \frac{18}{(x + 2) (x^{2} + 4)} = \frac{9}{4 (x + 2)} - \frac{9}{4} (\frac{x}{x^{2} + 4}) + \frac{9}{2 (x^{2} + 4)} \Rightarrow \int \frac{18 d x}{(x + 2) (x^{2} + 4)} = \frac{9}{4} \int \frac{d x}{x + 2} - \frac{9}{4} \int \frac{x d x}{x^{2} + 4} + \frac{9}{2} \int \frac{d x}{x^{2} + 2^{2}} let x^{2} + 4 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{9}{4} \int \frac{d x}{x + 2} - \frac{9}{8} \int \frac{d t}{t} + \frac{9}{2} \int \frac{d x}{x^{2} + 2^{2}} = \frac{9}{4} \log |x + 2| - \frac{9}{8} \log |t| + \frac{9}{2} \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + C' = \frac{9}{4} \log |x + 2| - \frac{9}{8} \log |x^{2} + 4| + \frac{9}{4} \tan^{- 1} (\frac{x}{2}) + C'$

Page No 18.177:

Question 36:

$\int \frac{5}{(x^{2} + 1) (x + 2)} d x$

Answer:

$We have, I = \int \frac{5 d x}{(x^{2} + 1) (x + 2)} Let \frac{5}{(x + 2) (x^{2} + 1)} = \frac{A}{x + 2} + \frac{B x + C}{x^{2} + 1} \Rightarrow \frac{5}{(x + 2) (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (x + 2)}{(x + 2) (x^{2} + 1)} \Rightarrow 5 = A (x^{2} + 1) + B x^{2} + 2 B x + C x + 2 C \Rightarrow 5 = (A + B) x^{2} + (2 B + C) x + (A + 2 C) Equating coefficients of like terms A + B = 0 . . . . . (1) 2 B + C = 0 . . . . . (2) A + 2 C = 5 . . . . . (3) Solving (1), (2) and (3), we get A = 1 B = - 1 C = 2 ∴ \frac{5}{(x + 2) (x^{2} + 1)} = \frac{1}{x + 2} + (\frac{- x + 2}{x^{2} + 1}) \Rightarrow \int \frac{5 d x}{(x + 2) (x^{2} + 1)} = \int \frac{d x}{x + 2} - \int \frac{x d x}{x^{2} + 1} + 2 \int \frac{d x}{x^{2} + 1} let x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \int \frac{d x}{x + 2} - \frac{1}{2} \int \frac{d t}{t} + 2 \int \frac{d x}{x^{2} + 1^{2}} = \log |x + 2| - \frac{1}{2} \log |t| + 2 \tan^{- 1} x + C' = \log |x + 2| - \frac{1}{2} \log |x^{2} + 1| + 2 \tan^{- 1} x + C'$

Page No 18.177:

Question 37:

$\int \frac{x}{(x + 1) (x^{2} + 1)} d x$

Answer:

$We have, I = \int \frac{x d x}{(x + 1) (x^{2} + 1)} Let \frac{x}{(x + 1) (x^{2} + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} + 1} \Rightarrow \frac{x}{(x + 1) (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (x + 1)}{(x + 1) (x^{2} + 1)} \Rightarrow x = A (x^{2} + 1) + B x^{2} + B x + C x + C \Rightarrow x = (A + B) x^{2} + (B + C) x + (A + C) Equating coefficients of like terms A + B = 0 . . . . . (1) B + C = 1 . . . . . (2) A + C = 0 . . . . . (3) Solving (1), (2) and (3), we get A = - \frac{1}{2} B = \frac{1}{2} C = \frac{1}{2} ∴ \frac{x}{(x + 1) (x^{2} + 1)} = - \frac{1}{2 (x + 1)} + \frac{\frac{x}{2} + \frac{1}{2}}{x^{2} + 1} \Rightarrow \int \frac{x d x}{(x + 1) (x^{2} + 1)} = - \frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{2} \int \frac{x d x}{x^{2} + 1} + \frac{1}{2} \int \frac{d x}{x^{2} + 1} let x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = - \frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{4} \int \frac{d t}{t} + \frac{1}{2} \int \frac{d x}{x^{2} + 1^{2}} = - \frac{1}{2} \log |x + 1| + \frac{1}{4} \log |t| + \frac{1}{2} \tan^{- 1} x + C' = - \frac{1}{2} \log |x + 1| + \frac{1}{4} \log |x^{2} + 1| + \frac{1}{2} \tan^{- 1} x + C'$

Page No 18.177:

Question 38:

$\int \frac{1}{1 + x + x^{2} + x^{3}} d x$

Answer:

$We have, I = \int \frac{d x}{1 + x + x^{2} + x^{3}} = \int \frac{d x}{(1 + x) + x^{2} (1 + x)} = \int \frac{d x}{(x + 1) (x^{2} + 1)} Let \frac{1}{(x + 1) (x^{2} + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} + 1} \Rightarrow \frac{1}{(x + 1) (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (x + 1)}{(x + 1) (x^{2} + 1)} \Rightarrow 1 = A (x^{2} + 1) + B x^{2} + B x + C x + C \Rightarrow 1 = (A + B) x^{2} + (B + C) x + (A + C) Equating coefficients of like terms A + B = 0 . . . . . (1) B + C = 0 . . . . . (2) A + C = 1 . . . . . (3) Solving (1), (2) and (3), we get A = \frac{1}{2} B = - \frac{1}{2} C = \frac{1}{2} ∴ I = \frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{2} \int (\frac{- x + 1}{x^{2} + 1}) d x = \frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{2} \int \frac{x d x}{x^{2} + 1} + \frac{1}{2} \int \frac{d x}{x^{2} + 1^{2}} let x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{4} \int \frac{d t}{t} + \frac{1}{2} \int \frac{d x}{x^{2} + 1^{2}} = \frac{1}{2} \log |x + 1| - \frac{1}{4} \log |t| + \frac{1}{2} \tan^{- 1} (x) + C' = \frac{1}{2} \log |x + 1| - \frac{1}{4} \log |x^{2} + 1| + \frac{1}{2} \tan^{- 1} (x) + C'$

Page No 18.177:

Question 39:

$\int \frac{1}{{(x + 1)}^{2} (x^{2} + 1)} d x$

Answer:

$We have, I = \int \frac{d x}{{(x + 1)}^{2} (x^{2} + 1)} Let \frac{1}{{(x + 1)}^{2} (x^{2} + 1)} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{C x + D}{x^{2} + 1} \Rightarrow \frac{1}{{(x + 1)}^{2} (x^{2} + 1)} = \frac{A (x + 1) (x^{2} + 1) + B (x^{2} + 1) + (C x + D) {(x + 1)}^{2}}{{(x + 1)}^{2} (x^{2} + 1)} \Rightarrow 1 = A (x^{3} + x + x^{2} + 1) + B (x^{2} + 1) + (C x + D) (x^{2} + 2 x + 1) \Rightarrow 1 = A (x^{3} + x^{2} + x + 1) + B (x^{2} + 1) + C x^{3} + 2 C x^{2} + C x + D x^{2} + 2 D x + D \Rightarrow 1 = (A + C) x^{3} + (A + B + 2 C + D) x^{2} + (A + C + 2 D) x + A + B + D Equating coefficients of like terms A + C = 0 . . . . . (1) A + B + 2 C + D = 0 . . . . . (2) A + C + 2 D = 0 . . . . . (3) A + B + D = 1 . . . . . (4) A = \frac{1}{2}, B = \frac{1}{2}, C = - \frac{1}{2} and D = 0 ∴ \frac{1}{{(x + 1)}^{2} (x^{2} + 1)} = \frac{1}{2 (x + 1)} + \frac{1}{2 {(x + 1)}^{2}} - \frac{1}{2} \times \frac{x}{x^{2} + 1} \Rightarrow \int \frac{d x}{{(x + 1)}^{2} (x^{2} + 1)} = \frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{2} \int \frac{d x}{{(x + 1)}^{2}} - \frac{1}{2} \int \frac{x d x}{x^{2} + 1} Putting x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d x}{x + 1} + \frac{1}{2} \int \frac{d x}{{(x + 1)}^{2}} - \frac{1}{4} \int \frac{d t}{t} = \frac{1}{2} \log |x + 1| - \frac{1}{2 (x + 1)} - \frac{1}{4} \log |t| + C' = \frac{1}{2} \log |x + 1| - \frac{1}{2 (x + 1)} - \frac{1}{4} \log |x^{2} + 1| + C'$

Page No 18.177:

Question 40:

$\int \frac{2 x}{x^{3} - 1} d x$

Answer:

$We have, I = \int \frac{2 x d x}{x^{3} - 1} = \int \frac{2 x d x}{(x - 1) (x^{2} + x + 1)} Let \frac{2 x}{(x - 1) (x^{2} + x + 1)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + x + 1} \Rightarrow \frac{2 x}{(x - 1) (x^{2} + x + 1)} = \frac{A (x^{2} + x + 1) + (B x + C) (x - 1)}{(x - 1) (x^{2} + x + 1)} \Rightarrow 2 x = A (x^{2} + x + 1) + B x^{2} - B x + C x - C \Rightarrow 2 x = (A + B) x^{2} + (A - B + C) x + A - C Equating coefficients of like terms A + B = 0 . . . . . (1) A - B + C = 2 . . . . . (2) A - C = 0 . . . . . (3) Solving (1), (2) and (3), we get A = \frac{2}{3} B = - \frac{2}{3} C = \frac{2}{3} ∴ I = \frac{2}{3} \int \frac{d x}{x - 1} + \frac{2}{3} \int (\frac{- x + 1}{x^{2} + x + 1}) d x Let - x + 1 = a \frac{d}{d x} (x^{2} + x + 1) + b \Rightarrow - x + 1 = a (2 x + 1) + b \Rightarrow - x + 1 = (2 a) x + a + b Equating coefficients of like terms 2 a = - 1 \Rightarrow a = - \frac{1}{2} And a + b = 1 \Rightarrow - \frac{1}{2} + b = 1 \Rightarrow b = \frac{3}{2} ∴ I = \frac{2}{3} \int \frac{d x}{x - 1} + \frac{2}{3} \int [\frac{- \frac{1}{2} (2 x + 1) + \frac{3}{2}}{(x^{2} + x + 1)}] d x = \frac{2}{3} \int \frac{d x}{x - 1} - \frac{1}{3} \int (\frac{2 x + 1}{x^{2} + x + 1}) d x + \int \frac{d x}{x^{2} + x + 1} = \frac{2}{3} \int \frac{d x}{x - 1} - \frac{1}{3} \int (\frac{2 x + 1}{x^{2} + x + 1}) d x + \int \frac{d x}{x^{2} + x + \frac{1}{4} - \frac{1}{4} + 1} = \frac{2}{3} \int \frac{d x}{x - 1} - \frac{1}{3} \int \frac{(2 x + 1) d x}{x^{2} + x + 1} + \int \frac{d x}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} let x^{2} + x + 1 = t \Rightarrow (2 x + 1) d x = d t Then, I = \frac{2}{3} \int \frac{d x}{x - 1} - \frac{1}{3} \int \frac{d t}{t} + \int \frac{d x}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{2}{3} \log |x - 1| - \frac{1}{3} \log |t| + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C' = \frac{2}{3} \log |x - 1| - \frac{1}{3} \log |x^{2} + x + 1| + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + C'$

Page No 18.177:

Question 41:

$\int \frac{d x}{(x^{2} + 1) (x^{2} + 4)}$

Answer:

$We have, I = \int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} Putting x^{2} = t Then, \frac{1}{(x^{2} + 1) (x^{2} + 4)} = \frac{1}{(t + 1) (t + 4)} Let \frac{1}{(t + 1) (t + 4)} = \frac{A}{t + 1} + \frac{B}{t + 4} \Rightarrow 1 = A (t + 4) + B (t + 1) Putting t + 4 = 0 \Rightarrow t = - 4 ∴ 1 = A \times 0 + B (- 3) \Rightarrow B = - \frac{1}{3} Putting t + 1 = 0 \Rightarrow t = - 1 ∴ 1 = A (- 1 + 4) + B \times 0 \Rightarrow A = \frac{1}{3} ∴ \frac{1}{(t + 1) (t + 4)} = \frac{1}{3 (t + 1)} - \frac{1}{3 (t + 4)} \Rightarrow \frac{1}{(x^{2} + 1) (x^{2} + 4)} = \frac{1}{3 (x^{2} + 1)} - \frac{1}{3 (x^{2} + 2^{2})} \Rightarrow \int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} = \frac{1}{3} \int \frac{d x}{x^{2} + 1^{2}} - \frac{1}{3} \int \frac{d x}{x^{2} + 2^{2}} = \frac{1}{3} \tan^{- 1} x - \frac{1}{3} \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + C = \frac{1}{3} \tan^{- 1} x - \frac{1}{6} \tan^{- 1} (\frac{x}{2}) + C$

Page No 18.177:

Question 42:

$\int \frac{x^{2}}{(x^{2} + 1) (3 x^{2} + 4)} d x$

Answer:

$We have, I = \int \frac{x^{2} d x}{(x^{2} + 1) (3 x^{2} + 4)} Putting x^{2} = t Then, \frac{x^{2}}{(x^{2} + 1) (3 x^{2} + 4)} = \frac{t}{(t + 1) (3 t + 4)} Let \frac{t}{(t + 1) (3 t + 4)} = \frac{A}{t + 1} + \frac{B}{3 t + 4} \Rightarrow \frac{t}{(t + 1) (3 t + 4)} = \frac{A (3 t + 4) + B (t + 1)}{(t + 1) (3 t + 4)} \Rightarrow t = A (3 t + 4) + B (t + 1) Putting t + 1 = 0 \Rightarrow t = - 1 ∴ - 1 = A (- 3 + 4) + 0 \Rightarrow A = - 1 Putting 3 t + 4 = 0 \Rightarrow t = - \frac{4}{3} ∴ - \frac{4}{3} = 0 + B (- \frac{4}{3} + 1) \Rightarrow - \frac{4}{3} = B \times (- \frac{1}{3}) \Rightarrow B = 4 ∴ \frac{t}{(t + 1) (3 t + 4)} = - \frac{1}{t + 1} + \frac{4}{3 t + 4} \Rightarrow \frac{x^{2}}{(x^{2} + 1) (3 x^{2} + 4)} = \frac{- 1}{x^{2} + 1} + \frac{4}{3 x^{2} + 4} \Rightarrow \frac{x^{2}}{(x^{2} + 1) (3 x^{2} + 4)} = \frac{- 1}{x^{2} + 1} + \frac{4}{3 (x^{2} + \frac{4}{3})} \Rightarrow \int \frac{x^{2} d x}{(x^{2} + 1) (3 x^{2} + 4)} = - \int \frac{d x}{x^{2} + 1} + \frac{4}{3} \int \frac{d x}{x^{2} + {(\frac{2}{\sqrt{3}})}^{2}} = - \tan^{- 1} (x) + \frac{4}{3} \times \frac{\sqrt{3}}{2} \tan^{- 1} (\frac{\sqrt{3} x}{2}) + C = - \tan^{- 1} (x) + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} x}{2}) + C$

Page No 18.177:

Question 43:

$\int \frac{3 x + 5}{x^{3} - x^{2} - x + 1} d x$

Answer:

$We have, I = \int \frac{(3 x + 5) d x}{x^{3} - x^{2} - x + 1} = \int \frac{(3 x + 5) d x}{x^{2} (x - 1) - 1 (x - 1)} = \int \frac{(3 x + 5) d x}{(x^{2} - 1) (x - 1)} = \int \frac{(3 x + 5) d x}{(x - 1) (x + 1) (x - 1)} = \int \frac{(3 x + 5) d x}{{(x - 1)}^{2} (x + 1)} Let \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{{(x - 1)}^{2}} \Rightarrow \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} = \frac{A {(x - 1)}^{2} + B (x + 1) (x - 1) + C (x + 1)}{(x + 1) {(x - 1)}^{2}} \Rightarrow 3 x + 5 = A (x^{2} - 2 x + 1) + B (x^{2} - 1) + C x + C \Rightarrow 3 x + 5 = (A + B) x^{2} + (- 2 A + C) x + (A - B + C) Equating coefficient of like terms A + B = 0 . . . . . (1) - 2 A + C = 3 . . . . . (2) A - B + C = 5 . . . . . (3) Solving these three equations we get A = \frac{1}{2} B = - \frac{1}{2} C = 4 ∴ \frac{3 x + 5}{{(x - 1)}^{2} (x + 1)} = \frac{1}{2 (x + 1)} - \frac{1}{2 (x - 1)} + \frac{4}{{(x - 1)}^{2}} \Rightarrow I = \frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{2} \int \frac{d x}{x - 1} + 4 \int {(x - 1)}^{- 2} d x = \frac{1}{2} \log |x + 1| - \frac{1}{2} \log |x - 1| - \frac{4}{(x - 1)} + C' = \frac{1}{2} \log |\frac{x + 1}{x - 1}| - \frac{4}{x - 1} + C'$

Page No 18.177:

Question 44:

$\int \frac{x^{3} - 1}{x^{3} + x} d x$

Answer:

$We have, I = \int \frac{(x^{3} - 1) d x}{x^{3} + x}$
Degree of numerator is equal to degree of denominator.
We divide numerator by denominator.
$1 x^{3} + x x^{3} - 1 x^{3} + x - - - x - 1 ∴ \frac{x^{3} - 1}{x^{3} + x} = 1 - \frac{(x + 1)}{x^{3} + x} \Rightarrow \frac{x^{3} - 1}{x^{3} + x} = 1 - \frac{(x + 1)}{x (x^{2} + 1)} . . . . . (1) Let \frac{x + 1}{x (x^{2} + 1)} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1} \Rightarrow \frac{x + 1}{x (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (x)}{x (x^{2} + 1)} \Rightarrow x + 1 = A x^{2} + A + B x^{2} + C x \Rightarrow x + 1 = (A + B) x^{2} + C x + A$
Equating coefficient of like terms
A + B = 0
C = 1
A = 1
B = –1
$∴ \frac{x + 1}{x (x^{2} + 1)} = \frac{1}{x} + \frac{- x + 1}{x^{2} + 1} . . . . . (2) Using (1) & (2) \int \frac{(x^{3} - 1) d x}{(x^{3} + x)} = \int (1 - \frac{1}{x} + \frac{x}{x^{2} + 1} - \frac{1}{x^{2} + 1}) d x = \int d x - \int \frac{d x}{x} + \int \frac{x d x}{x^{2} + 1} - \int \frac{d x}{x^{2} + 1} Putting x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \int d x - \int \frac{d x}{x} + \frac{1}{2} \int \frac{d t}{t} - \int \frac{d x}{x^{2} + 1} = x - \log |x| + \frac{1}{2} \log |t| - \tan^{- 1} x + C = x - \log |x| + \frac{1}{2} \log |x^{2} + 1| - \tan^{- 1} x + C$

Page No 18.177:

Question 45:

$\int \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} d x$

Answer:

$We have, I = \int \frac{(x^{2} + x + 1)}{{(x + 1)}^{2} (x + 2)} d x Let \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{C}{x + 2} \Rightarrow \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} = \frac{A (x + 1) (x + 2) + B (x + 2) + C {(x + 1)}^{2}}{{(x + 1)}^{2} (x + 2)} \Rightarrow x^{2} + x + 1 = A (x^{2} + x + 2 x + 2) + B x + 2 B + C (x^{2} + 2 x + 1) \Rightarrow x^{2} + x + 1 = (A + C) x^{2} + (3 A + B + 2 C) x + (2 A + 2 B + C) Equating coefficient of like terms . A + C = 1 . . . . . (1) 3 A + B + 2 C = 1 . . . . . (2) 2 A + 2 B + C = 1 . . . . . (3) Solving these three equations we get A = - 2 B = 1 C = 3 Hence, \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} = \frac{- 2}{x + 1} + \frac{1}{{(x + 1)}^{2}} + \frac{3}{x + 2} ∴ I = - 2 \int \frac{d x}{x + 1} + \int \frac{d}{{(x + 1)}^{2}} + 3 \int \frac{d x}{x + 2} = - 2 \log |x + 1| - \frac{1}{x + 1} + 3 \log |x + 2| + C$

Page No 18.177:

Question 46:

$\int \frac{1}{x (x^{4} + 1)} d x$

Answer:

$We have, I = \int \frac{d x}{x (x^{4} + 1)} = \int \frac{x^{3} d x}{x^{4} (x^{4} + 1)} Putting x^{4} = t \Rightarrow 4 x^{3} d x = d t \Rightarrow x^{3} d x = \frac{d t}{4} ∴ I = \frac{1}{4} \int \frac{d t}{t (t + 1)} Let \frac{1}{t (t + 1)} = \frac{A}{t} + \frac{B}{t + 1} \Rightarrow \frac{1}{t (t + 1)} = \frac{A (t + 1) + B t}{t (t + 1)} \Rightarrow 1 = A (t + 1) + B t Putting t + 1 = 0 \Rightarrow t = - 1 ∴ 1 = A \times 0 + B (- 1) \Rightarrow B = - 1 Putting t = 0 ∴ 1 = A (1) + B \times 0 \Rightarrow A = 1 ∴ I = \frac{1}{4} \int \frac{d t}{t} - \frac{1}{4} \int \frac{d t}{t + 1} = \frac{1}{4} \log |t| - \frac{1}{4} \log |t + 1| + C = \frac{1}{4} \log |\frac{t}{t + 1}| + C = \frac{1}{4} \log |\frac{x^{4}}{x^{4} + 1}| + C$

Page No 18.177:

Question 47:

Evaluate the following integrals:

$\int \frac{1}{x (x^{3} + 8)} d x$

Answer:

$Let I = \int \frac{1}{x (x^{3} + 8)} d x We express \frac{1}{x (x^{3} + 8)} = = \frac{A}{x} + \frac{B x^{2} + C x + D}{x^{3} + 8} \Rightarrow 1 = A (x^{3} + 8) + (B x^{2} + C x + D) (x) Equating the coefficients of x^{3}, x^{2}, x and constants, we get 0 = A + B and 0 = C and 0 = D and 1 = 8 A or A = \frac{1}{8} and B = - \frac{1}{8} and C = 0 and D = 0 ∴ I = \int (\frac{\frac{1}{8}}{x} + \frac{- \frac{1}{8} x^{2}}{(x^{3} + 8)}) d x = \frac{1}{8} \int \frac{1}{x} d x - \frac{1}{24} \int \frac{3 x^{2}}{x^{3} + 8} d x = \frac{1}{8} \log |x| - \frac{1}{24} \log |x^{3} + 8| + c Hence, \int \frac{1}{x (x^{3} + 8)} d x = \frac{1}{8} \log |x| - \frac{1}{24} \log |x^{3} + 8| + c$

Page No 18.177:

Question 48:

$\int \frac{3}{(1 - x) (1 + x^{2})} d x$

Answer:

$We have, I = \int \frac{3 d x}{(1 - x) (1 + x^{2})} = 3 \int \frac{d x}{(1 - x) (1 + x^{2})} Let \frac{1}{(1 - x) (1 + x^{2})} = \frac{A}{1 - x} + \frac{B x + C}{x^{2} + 1} \Rightarrow \frac{1}{(1 - x) (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (1 - x)}{(1 - x) (x^{2} + 1)} \Rightarrow 1 = A x^{2} + A + B x - B x^{2} + C - C x \Rightarrow 1 = (A - B) x^{2} + (B - C) x + A + C Equating coefficients of like terms . A - B = 0 . . . . . (1) B - C = 0 . . . . . (2) A + C = 1 . . . . . (3) Solving (1), (2) and (3), we get A = \frac{1}{2}, B = \frac{1}{2}, C = \frac{1}{2} ∴ \frac{1}{(1 - x) (x^{2} + 1)} = \frac{1}{2 (1 - x)} + \frac{\frac{x}{2} + \frac{1}{2}}{x^{2} + 1} \int \frac{3 d x}{(1 - x) (x^{2} + 1)} = \frac{3}{2} \int \frac{d x}{1 - x} + \frac{3}{2} \int \frac{x d x}{x^{2} + 1} + \frac{3}{2} \int \frac{d x}{x^{2} + 1} Putting x^{2} + 1 = t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{3}{2} \int \frac{d x}{1 - x} + \frac{3}{4} \int \frac{d t}{t} + \frac{3}{2} \int \frac{d x}{x^{2} + 1} = \frac{3}{2} \frac{\log |1 - x|}{- 1} + \frac{3}{4} \log |t| + \frac{3}{2} \times \tan^{- 1} x + C = \frac{- 3}{2} \log |1 - x| + \frac{3}{4} \log |1 + x^{2}| + \frac{3}{2} \tan^{- 1} x + C = \frac{- 3}{4} \times 2 \log |1 - x| + \frac{3}{4} \log |1 + x^{2}| + \frac{3}{4} (2 \tan^{- 1} x) + C = \frac{3}{4} [\log |1 + x^{2}| - \log |{(1 - x)}^{2}|] + \frac{3}{4} (2 \tan^{- 1} x) + C = \frac{3}{4} [\log |\frac{1 + x^{2}}{{(1 - x)}^{2}}| + 2 \tan^{- 1} (x)] + C$

Page No 18.177:

Question 49:

$\int \frac{\cos x}{{(1 - \sin x)}^{3} (2 + \sin x)} d x$

Answer:

$We have, I = \int \frac{\cos x}{{(1 - \sin x)}^{3} (2 + \sin x)} d x Let, \sin x = t \Rightarrow \cos x d x = d t Now, integration becomes, I = \int \frac{d t}{{(1 - t)}^{3} (2 + t)} = - \int \frac{d t}{{(t - 1)}^{3} (t + 2)} Let, \frac{1}{{(t - 1)}^{3} (t + 2)} = \frac{A}{(t - 1)} + \frac{B}{{(t - 1)}^{2}} + \frac{C}{{(t - 1)}^{3}} + \frac{D}{(t + 2)} . . . . . (1) \Rightarrow 1 = A {(t - 1)}^{2} (t + 2) + B (t - 1) (t + 2) + C (t + 2) + D {(t - 1)}^{3} . . . . . (2)$

$Putting t = 1 in (2), we get 1 = 3 C \Rightarrow C = \frac{1}{3} Putting t = - 2 in (2), we get 1 = D {(- 2 - 1)}^{3} \Rightarrow 1 = - 27 D \Rightarrow D = \frac{- 1}{27} Putting t = 0 in (2), we get 1 = 2 A - 2 B + 2 C - D \Rightarrow 1 = 2 A - 2 B + \frac{2}{3} + \frac{1}{27} \Rightarrow 2 A - 2 B = \frac{8}{27} \Rightarrow A - B = \frac{4}{27} Putting t = 2 in (2), we get 1 = 4 A + 4 B + 4 C + D \Rightarrow 1 = 4 A + 4 B + \frac{4}{3} - \frac{1}{27} \Rightarrow A + B = - \frac{2}{27} Now, A - B = \frac{4}{27} and A + B = - \frac{2}{27} \Rightarrow A = \frac{1}{27} and B = \frac{- 1}{9}$
$Substituting the values of A, B, C and D in (1), we get \frac{1}{{(t - 1)}^{3} (t + 2)} = \frac{1}{27 (t - 1)} - \frac{1}{9 {(t - 1)}^{2}} + \frac{1}{3 {(t - 1)}^{3}} + \frac{- 1}{27 (t + 2)} Now, integration becomes I = - \int [\frac{1}{27 (t - 1)} - \frac{1}{9 {(t - 1)}^{2}} + \frac{1}{3 {(t - 1)}^{3}} + \frac{- 1}{27 (t + 2)}] d t = - [\frac{1}{27} \log |t - 1| + \frac{1}{9 (t - 1)} - \frac{1}{6 {(t - 1)}^{2}} - \frac{1}{27} \log |t + 2|] + C = - \frac{1}{27} \log |\sin x - 1| - \frac{1}{9 (\sin x - 1)} + \frac{1}{6 {(\sin x - 1)}^{2}} + \frac{1}{27} \log |\sin x + 2| + C = - \frac{1}{27} \log |1 - \sin x| + \frac{1}{9 (1 - \sin x)} + \frac{1}{6 {(1 - \sin x)}^{2}} + \frac{1}{27} \log |2 + \sin x| + C$

Page No 18.177:

Question 50:

Evaluate the following integrals:

$\int \frac{2 x^{2} + 1}{x^{2} (x^{2} + 4)} d x$

Answer:

$Let I = \int \frac{2 x^{2} + 1}{x^{2} (x^{2} + 4)} d x We express \frac{2 x^{2} + 1}{x^{2} (x^{2} + 4)} = \frac{A}{x^{2}} + \frac{B}{x^{2} + 4} \Rightarrow 2 x^{2} + 1 = A (x^{2} + 4) + B (x^{2}) Equating the coefficients of x^{2} and constants, we get 2 = A + B and 1 = 4 A or A = \frac{1}{4} and B = \frac{7}{4} ∴ I = \int (\frac{\frac{1}{4}}{x^{2}} + \frac{\frac{7}{4}}{x^{2} + 4}) d x = \frac{1}{4} \int \frac{1}{x^{2}} d x + \frac{7}{4} \int \frac{1}{x^{2} + 4} d x = - \frac{1}{4 x} + \frac{7}{8} \tan^{- 1} \frac{x}{2} + c Hence, \int \frac{2 x^{2} + 1}{x^{2} (x^{2} + 4)} d x = - \frac{1}{4 x} + \frac{7}{8} \tan^{- 1} \frac{x}{2} + c$

Page No 18.177:

Question 51:

$\int \frac{\cos x}{(1 - \sin x) (2 - \sin x)} d x$

Answer:

$We have, I = \int \frac{\cos x d x}{(1 - \sin x) (2 - \sin x)} Putting \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \frac{d t}{(1 - t) (2 - t)} = \int \frac{d t}{(t - 1) (t - 2)} Let \frac{1}{(t - 1) (t - 2)} = \frac{A}{t - 1} + \frac{B}{t - 2} \Rightarrow \frac{1}{(t - 1) (t - 2)} = \frac{A (t - 2) + B (t - 1)}{(t - 1) (t - 2)} \Rightarrow 1 = A (t - 2) + B (t - 1) Putting t - 1 = 0 \Rightarrow t = 1 ∴ 1 = A (1 - 2) + B \times 0 \Rightarrow A = - 1 Putting t - 2 = 0 \Rightarrow t = 2 ∴ 1 = A \times 0 + B (2 - 1) \Rightarrow B = 1 ∴ I = \int \frac{- d t}{t - 1} + \int \frac{d t}{t - 2} = - \log |t - 1| + \log |t - 2| + C = \log |\frac{t - 2}{t - 1}| + C = \log |\frac{\sin x - 2}{\sin x - 1}| + C = \log |\frac{2 - \sin x}{1 - \sin x}| + C$

Page No 18.177:

Question 52:

$\int \frac{2 x + 1}{(x - 2) (x - 3)} d x$

Answer:

$We have, I = \int \frac{(2 x + 1) d x}{(x - 2) (x - 3)} Let \frac{2 x + 1}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3} \Rightarrow \frac{2 x + 1}{(x - 2) (x - 3)} = \frac{A (x - 3) + B (x - 2)}{(x - 2) (x - 3)} \Rightarrow 2 x + 1 = A (x - 3) + B (x - 2) Putting x - 3 = 0 \Rightarrow x = 3 ∴ 7 = A \times 0 + B \times (3 - 2) \Rightarrow B = 7 Putting x - 2 = 0 \Rightarrow x = 2 ∴ 5 = A (- 1) \Rightarrow A = - 5 ∴ I = - 5 \int \frac{d x}{x - 2} + 7 \int \frac{d x}{x - 3} = - 5 \log |x - 2| + 7 \log |x - 3| + C = \log {|x - 3|}^{7} - \log {|x - 2|}^{5} + C = \log |\frac{{(x - 3)}^{7}}{{(x - 2)}^{5}}| + C$

Page No 18.177:

Question 53:

$\int \frac{1}{(x^{2} + 1) (x^{2} + 2)} d x$

Answer:

$We have, I = \int \frac{d x}{(x^{2} + 1) (x^{2} + 2)} Putting x^{2} = t Then, \frac{1}{(x^{2} + 1) (x^{2} + 2)} = \frac{1}{(t + 1) (t + 2)} Let \frac{1}{(t + 1) (t + 2)} = \frac{A}{t + 1} + \frac{B}{t + 2} \Rightarrow \frac{1}{(t + 1) (t + 2)} = \frac{A (t + 2) + B (t + 1)}{(t + 1) (t + 2)} \Rightarrow 1 = A (t + 2) + B (t + 1) Putting t + 2 = 0 \Rightarrow t = - 2 ∴ 1 = A \times 0 + B (- 1) \Rightarrow B = - 1 Putting t + 1 = 0 \Rightarrow t = - 1 ∴ 1 = A (- 1 + 2) + B \times 0 \Rightarrow A = 1 ∴ \frac{1}{(t + 1) (t + 2)} = \frac{1}{t + 1} - \frac{1}{t + 2} \Rightarrow \frac{1}{(x^{2} + 1) (x^{2} + 2)} = \frac{1}{x^{2} + 1} - \frac{1}{x^{2} + {(\sqrt{2})}^{2}} ∴ I = \int \frac{d x}{x^{2} + 1^{2}} - \int \frac{d x}{x^{2} + {(\sqrt{2})}^{2}} = \tan^{- 1} (x) - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x}{\sqrt{2}}) + C$

Page No 18.177:

Question 54:

$\int \frac{1}{x (x^{4} - 1)} d x$

Answer:

$We have, I = \int \frac{d x}{x (x^{4} - 1)} = \int \frac{x^{3} d x}{x^{4} (x^{4} - 1)} Putting x^{4} = t \Rightarrow 4 x^{3} d x = d t \Rightarrow x^{3} d x = \frac{d t}{4} ∴ I = \frac{1}{4} \int \frac{d t}{t (t - 1)} Let \frac{1}{t (t - 1)} = \frac{A}{t} + \frac{B}{t - 1} \Rightarrow \frac{1}{t (t - 1)} = \frac{A (t - 1) + B t}{t (t - 1)} \Rightarrow 1 = A (t - 1) + B t Putting t - 1 = 0 \Rightarrow t = 1 ∴ 1 = A \times 0 + B (1) \Rightarrow B = 1 Putting t = 0 ∴ 1 = A (0 - 1) + B \times 0 \Rightarrow A = - 1 ∴ I = - \frac{1}{4} \int \frac{d t}{t} + \frac{1}{4} \int \frac{d t}{t - 1} = - \frac{1}{4} \log |t| + \frac{1}{4} \log |t - 1| + C = \frac{1}{4} \log |\frac{t - 1}{t}| + C = \frac{1}{4} \log |\frac{x^{4} - 1}{x^{4}}| + C$

Page No 18.177:

Question 55:

$\int \frac{1}{x^{4} - 1} d x$

Answer:

$We have, I = \int \frac{d x}{x^{4} - 1} = \int \frac{d x}{(x^{2} - 1) (x^{2} + 1)} = \int \frac{d x}{(x - 1) (x + 1) (x^{2} + 1)} Let \frac{1}{(x - 1) (x + 1) (x^{2} + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C x + D}{x^{2} + 1} \Rightarrow \frac{1}{(x - 1) (x + 1) (x^{2} + 1)} = \frac{A (x^{2} + 1) (x + 1) + B (x - 1) (x^{2} + 1) (C x + D) (x - 1) (x + 1)}{(x - 1) (x + 1) (x^{2} + 1)} \Rightarrow 1 = A (x^{2} + 1) (x + 1) + B (x - 1) (x^{2} + 1) + (C x + D) (x^{2} - 1) \Rightarrow 1 = A (x^{3} + x^{2} + x + 1) + B (x^{3} + x - x^{2} - 1) + (C x^{3} - C x + D x^{2} - D) \Rightarrow 1 = (A + B + C) x^{3} + x^{2} (A - B + D) + x (A + B - C) + A - B - D Equating the coefficients of like terms . A + B + C = 0 . . . . . (1) A - B + D = 0 . . . . . (2) A + B - C = 0 . . . . . (3) A - B - D = 1 . . . . . (4) Solving these four equations we get A = \frac{1}{4}, B = - \frac{1}{4}, C = 0, D = - \frac{1}{2} ∴ \frac{1}{(x - 1) (x + 1) (x^{2} + 1)} = \frac{1}{4 (x - 1)} - \frac{1}{4 (x + 1)} - \frac{1}{2 (x^{2} + 1)} \Rightarrow I = \frac{1}{4} \int \frac{d x}{x - 1} - \frac{1}{4} \int \frac{d x}{x + 1} - \frac{1}{2} \int \frac{d x}{x^{2} + 1} = \frac{1}{4} \log (x - 1) - \frac{1}{4} \log (x + 1) - \frac{1}{2} \tan^{- 1} (x) + C' = \frac{1}{4} \log |\frac{x - 1}{x + 1}| - \frac{1}{2} \tan^{- 1} (x) + C'$

Page No 18.177:

Question 56:

Find $\int \frac{2 x}{(x^{2} + 1) {(x^{2} + 2)}^{2}} d x$

Answer:

$\int \frac{2 x}{(x^{2} + 1) {(x^{2} + 2)}^{2}} d x$

$Let x^{2} = y \Rightarrow 2 x d x = d y \Rightarrow d x = \frac{d y}{2 x}$

$\int \frac{2 x}{(x^{2} + 1) {(x^{2} + 2)}^{2}} d x = \int \frac{d y}{(y + 1) {(y + 2)}^{2}} Let \frac{1}{(y + 1) {(y + 2)}^{2}} = \frac{A}{y + 1} + \frac{B}{y + 2} + \frac{C}{{(y + 2)}^{2}} . . . . . (1) \Rightarrow 1 = A {(y + 2)}^{2} + B (y + 1) (y + 2) + C (y + 1) . . . . . (2) Putting y = - 2 in (2) 1 = C (- 2 + 1) \Rightarrow C = - 1$

$Putting y = - 1 in (2) 1 = A {(- 1 + 2)}^{2} \Rightarrow 1 = A (1) \Rightarrow A = 1$

$Putting y = 0 in (2) 1 = 4 A + B (2) + C \Rightarrow 1 = 4 + 2 B - 1 \Rightarrow 1 = 3 + 2 B \Rightarrow - 2 = 2 B \Rightarrow B = - 1$

Substituting the values of A, B and C in (1)

$\frac{1}{(y + 1) {(y + 2)}^{2}} = \frac{1}{y + 1} - \frac{1}{y + 2} - \frac{1}{{(y + 2)}^{2}} \Rightarrow \int \frac{d y}{(y + 1) {(y + 2)}^{2}} = \int \frac{d y}{y + 1} - \int \frac{d y}{y + 2} - \int \frac{d y}{{(y + 2)}^{2}} = \log |y + 1| - \log |y + 2| + \frac{1}{y + 2} + C$
Hence, $\int \frac{2 x}{(x^{2} + 1) {(x^{2} + 2)}^{2}} d x$ = $= \log |x^{2} + 1| - \log |x^{2} + 2| + \frac{1}{x^{2} + 2} + C$

Page No 18.177:

Question 57:

Evaluate the following integrals:
$\int \frac{x^{2}}{(x - 1) (x^{2} + 1)} d x$

Answer:

I = $\int \frac{x^{2}}{(x - 1) (x^{2} + 1)} d x$

$\frac{x^{2}}{(x - 1) (x^{2} + 1)} = \frac{A}{(x - 1)} + \frac{B x + C}{x^{2} + 1} = \frac{A (x^{2} + 1) + (B x + C) (x - 1)}{(x - 1) (x^{2} + 1)} \Rightarrow \frac{x^{2}}{(x - 1) (x^{2} + 1)} = \frac{(A + B) x^{2} + (C - B) x + (A - C)}{(x - 1) (x^{2} + 1)}$

Comparing coefficients, we get

$A + B = 1; C - B = 0 and A - C = 0 Solving these equations, we get A = B = C = \frac{1}{2} ∴ I = \frac{1}{2} \int \frac{1}{(x - 1)} d x + \frac{1}{2} \int \frac{x}{x^{2} + 1} d x + \frac{1}{2} \int \frac{1}{x^{2} + 1} d x = \frac{1}{2} \ln |x - 1| + \frac{1}{4} \ln |x^{2} + 1| + \frac{1}{2} \tan^{- 1} (x) + c$

Page No 18.177:

Question 58:

Evaluate the following integrals:

$\int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$

Answer:

$Let I = \int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x We express \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{A}{x^{2} + a^{2}} + \frac{B}{x^{2} + b^{2}} \Rightarrow x^{2} = A (x^{2} + b^{2}) + B (x^{2} + a^{2}) Equating the coefficients of x^{2} and constants, we get 1 = A + B and 0 = b^{2} A + a^{2} B or A = - \frac{a^{2}}{b^{2} - a^{2}} and B = \frac{b^{2}}{b^{2} - a^{2}} ∴ I = \int (\frac{- \frac{a^{2}}{b^{2} - a^{2}}}{x^{2} + a^{2}} + \frac{\frac{b^{2}}{b^{2} - a^{2}}}{x^{2} + b^{2}}) d x = - \frac{a^{2}}{b^{2} - a^{2}} \int \frac{1}{x^{2} + a^{2}} d x + \frac{b^{2}}{b^{2} - a^{2}} \int \frac{1}{x^{2} + b^{2}} d x = - \frac{a}{b^{2} - a^{2}} \tan^{- 1} \frac{x}{a} + \frac{b}{b^{2} - a^{2}} \tan^{- 1} \frac{x}{b} + c Hence, \int \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x = - \frac{a}{b^{2} - a^{2}} \tan^{- 1} \frac{x}{a} + \frac{b}{b^{2} - a^{2}} \tan^{- 1} \frac{x}{b} + c$

Page No 18.177:

Question 59:

$\int \frac{1}{\cos x (5 - 4 \sin x)} d x$

Answer:

$We have, I = \int \frac{d x}{\cos x (5 - 4 \sin x)} = \int \frac{\cos x d x}{\cos^{2} x (5 - 4 \sin x)} = \int \frac{\cos x d x}{(1 - \sin^{2} x) (5 - 4 \sin x)} = \int \frac{\cos x d x}{(1 - \sin x) (1 + \sin x) (5 - 4 \sin x)} Putting \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \frac{d t}{(1 - t) (1 + t) (5 - 4 t)} Let \frac{1}{(1 - t) (1 + t) (5 - 4 t)} = \frac{A}{1 - t} + \frac{B}{1 + t} + \frac{C}{5 - 4 t} \Rightarrow \frac{1}{(1 - t) (1 + t) (5 - 4 t)} = \frac{A (1 + t) (5 - 4 t) + B (1 - t) (5 - 4 t) + C (1 - t) (1 + t)}{(1 - t) (1 + t) (5 - 4 t)} \Rightarrow 1 = A (1 + t) (5 - 4 t) + B (1 - t) (5 - 4 t) + C (1 - t) (1 + t) Putting 1 + t = 0 \Rightarrow t = - 1 1 = B (2) (5 + 4) B = \frac{1}{18} Putting 1 - t = 0 \Rightarrow t = 1 1 = A (2) (5 - 4) + B \times 0 + C \times 0 A = \frac{1}{2} Putting 5 - 4 t = 0 \Rightarrow 4 t = 5 \Rightarrow t = \frac{5}{4} 1 = C (1 - \frac{5}{4}) (1 + \frac{5}{4}) \Rightarrow 1 = C (- \frac{1}{4}) (\frac{9}{4}) \Rightarrow C = - \frac{16}{9} ∴ I = \frac{1}{2} \int \frac{d t}{1 - t} + \frac{1}{18} \int \frac{d t}{1 + t} - \frac{16}{9} \int \frac{d t}{5 - 4 t} = \frac{1}{2} \frac{\log |1 - t|}{- 1} + \frac{1}{18} \log |1 + t| - \frac{16}{9} \times \frac{\log |5 - 4 t|}{- 4} + C = \frac{1}{18} \log |1 + t| - \frac{1}{2} \log |1 - t| + \frac{4}{9} \log |5 - 4 t| + C = \frac{1}{18} \log |1 + \sin x| - \frac{1}{2} \log |1 - \sin x| + \frac{4}{9} \log |5 - 4 \sin x| + C$

Page No 18.178:

Question 60:

$\int \frac{1}{\sin x (3 + 2 \cos x)} d x$

Answer:

$We have, I = \int \frac{d x}{\sin x (3 + 2 \cos x)} = \int \frac{\sin x d x}{\sin^{2} x (3 + 2 \cos x)} = \int \frac{\sin x d x}{(1 - \cos^{2} x) (3 + 2 \cos x)} = \int \frac{\sin x d x}{(1 - \cos x) (1 + \cos x) (3 + 2 \cos x)} Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = \int \frac{- d t}{(1 - t) (1 + t) (3 + 2 t)} = \int \frac{d t}{(t - 1) (t + 1) (3 + 2 t)} Let \frac{1}{(t - 1) (t + 1) (3 + 2 t)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{3 + 2 t} \Rightarrow \frac{1}{(t - 1) (t + 1) (3 + 2 t)} = \frac{A (t + 1) (3 + 2 t) + B (t - 1) (3 + 2 t) + C (t + 1) (t - 1)}{(t - 1) (t + 1) (3 + 2 t)} \Rightarrow 1 = A (t + 1) (3 + 2 t) + B (t - 1) (3 + 2 t) + C (t + 1) (t - 1) Putting t + 1 = 0 \Rightarrow t = - 1 1 = A \times 0 + B (- 2) (3 - 2) + C \times 0 \Rightarrow 1 = B (- 2) \Rightarrow B = - \frac{1}{2} Putting t - 1 = 0 \Rightarrow t = 1 1 = A (2) (5) + B \times 0 + C \times 0 \Rightarrow A = \frac{1}{10} Putting 3 + 2 t = 0 \Rightarrow t = - \frac{3}{2} 1 = A \times 0 + B \times 0 + C (- \frac{3}{2} + 1) (- \frac{3}{2} - 1) \Rightarrow 1 = C (- \frac{1}{2}) (- \frac{5}{2}) C = \frac{4}{5} Then, I = \frac{1}{10} \int \frac{d t}{t - 1} - \frac{1}{2} \int \frac{d t}{t + 1} + \frac{4}{5} \int \frac{d t}{3 + 2 t} = \frac{1}{10} \log |t - 1| - \frac{1}{2} \log |t + 1| + \frac{4}{5} \times \frac{\log |3 + 2 t|}{2} + C = \frac{1}{10} \log |t - 1| - \frac{1}{2} \log |t + 1| + \frac{2}{5} \log |3 + 2 t| + C = \frac{1}{10} \log |\cos x - 1| - \frac{1}{2} \log |\cos x + 1| + \frac{2}{5} \log |3 + 2 \cos x| + C$

Page No 18.178:

Question 61:

$\int \frac{1}{\sin x + \sin 2 x} d x$

Answer:

$We have, I = \int \frac{d x}{\sin x + \sin 2 x} = \int \frac{d x}{\sin x + 2 \sin x \cos x} = \int \frac{d x}{\sin x (1 + 2 \cos x)} = \int \frac{\sin x d x}{\sin^{2} x (1 + 2 \cos x)} = \int \frac{\sin x d x}{(1 - \cos^{2} x) (1 + 2 \cos x)} = \int \frac{\sin x d x}{(1 - \cos x) (1 + \cos x) (1 + 2 \cos x)} Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = \int \frac{- d t}{(1 - t) (1 + t) (1 + 2 t)} = \int \frac{d t}{(t - 1) (t + 1) (1 + 2 t)} Let \frac{1}{(t - 1) (t + 1) (1 + 2 t)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{1 + 2 t} \Rightarrow \frac{1}{(t - 1) (t + 1) (1 + 2 t)} = \frac{A (t + 1) (1 + 2 t) + B (t - 1) (1 + 2 t) + C (t - 1) (t + 1)}{(t - 1) (t + 1) (1 + 2 t)} \Rightarrow 1 = A (t + 1) (1 + 2 t) + B (t - 1) (1 + 2 t) + C (t - 1) (t + 1) Putting t + 1 = 0 \Rightarrow t = - 1 1 = B (- 1 - 1) (1 - 2) \Rightarrow 1 = B (- 2) (- 1) \Rightarrow B = \frac{1}{2} Putting t - 1 = 0 \Rightarrow t = 1 1 = A (1 + 1) (1 + 2) \Rightarrow 1 = A (2) (3) \Rightarrow A = \frac{1}{6} Putting 1 + 2 t = 0 t = - \frac{1}{2} \Rightarrow 1 = A \times 0 + B \times 0 + C (- \frac{1}{2} - 1) (- \frac{1}{2} + 1) 1 = C (- \frac{3}{2}) (\frac{1}{2}) C = - \frac{4}{3} Then, I = \frac{1}{6} \int \frac{d t}{t - 1} + \frac{1}{2} \int \frac{d t}{t + 1} - \frac{4}{3} \int \frac{d t}{1 + 2 t} = \frac{1}{6} \log |t - 1| + \frac{1}{2} \log |t + 1| - \frac{4}{3} \times \frac{\log |1 + 2 t|}{2} + C = \frac{1}{6} \log |t - 1| + \frac{1}{2} \log |t + 1| - \frac{2}{3} \log |1 + 2 t| + C = \frac{1}{6} \log |\cos x - 1| + \frac{1}{2} \log |\cos x + 1| - \frac{2}{3} \log |1 + 2 \cos x| + C$

Page No 18.178:

Question 62:

$\int \frac{x + 1}{x (1 + x e^{x})} d x$

Answer:

$We have, I = \int \frac{x + 1}{x (1 + x e^{x})} d x I = \int \frac{e^{x} (x + 1)}{e^{x} x (1 + x e^{x})} d x Put e^{x} = t \Rightarrow e^{x} (x + 1) d x = d t I = \int \frac{d t}{t (1 + t)} . . . . . (1) Let, \frac{1}{t (1 + t)} = \frac{A}{t} + \frac{B}{1 + t} \Rightarrow 1 = A (t + 1) + B t . . . . . (2) Putting t = 0 in (2), we obtain A = 1 Putting t = - 1 in (2), we obtain B = - 1 I = \int (\frac{1}{t} - \frac{1}{1 + t}) d t I = \log |t| - \log |t + 1| + C I = \log |\frac{t}{t + 1}| + C I = \log |\frac{x e^{x}}{x e^{x} + 1}| + C$

Page No 18.178:

Question 63:

$\int \frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)} d x$

Answer:

$We have, I = \int \frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)} Putting x^{2} = t Then, \frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)} = \frac{(t + 1) (t + 2)}{(t + 3) (t + 4)} = \frac{t^{2} + 3 t + 2}{t^{2} + 7 t + 12}$
Degree of numerator is equal to degree of denominator.
We divide numerator by denominator.
$1 t^{2} + 7 t + 12 t^{2} + 3 t + 2 t^{2} + 7 t + 12 - - - - 4 t - 10$
$∴ \frac{t^{2} + 3 t + 2}{t^{2} + 7 t + 12} = 1 - (\frac{4 t + 10}{t^{2} + 7 t + 12}) \Rightarrow \frac{t^{2} + 3 t + 2}{t^{2} + 7 t + 12} = 1 - \frac{4 t + 10}{(t + 3) (t + 4)} . . . . . (1) Let \frac{4 t + 10}{(t + 3) (t + 4)} = \frac{A}{t + 3} + \frac{B}{t + 4} \Rightarrow \frac{4 t + 10}{(t + 3) (t + 4)} = \frac{A (t + 4) + B (t + 3)}{(t + 3) (t + 4)} \Rightarrow 4 t + 10 = A (t + 4) + B (t + 3) Putting t + 4 = 0 \Rightarrow t = - 4 ∴ - 16 + 10 = B (- 1) \Rightarrow B = 6 Putting t + 3 = 0 \Rightarrow t = - 3 ∴ - 12 + 10 = A (- 3 + 4) \Rightarrow A = - 2 ∴ \frac{4 t + 10}{(t + 3) (t + 4)} = \frac{- 2}{t + 3} + \frac{6}{t + 4} . . . . . (2) From (1) & (2) \frac{t^{2} + 3 t + 2}{t^{2} + 7 t + 12} = 1 + \frac{2}{t + 3} - \frac{6}{t + 4} ∴ \int \frac{(x^{2} + 1) (x^{2} + 2) d x}{(x^{2} + 3) (x^{2} + 4)} = \int d x + 2 \int \frac{d x}{x^{2} + {(\sqrt{3})}^{2}} - 6 \int \frac{d x}{x^{2} + 2^{2}} = x + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - \frac{6}{2} \tan^{- 1} (\frac{x}{2}) + C = x + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - 3 \tan^{- 1} (\frac{x}{2}) + C$

Page No 18.178:

Question 64:

$\int \frac{4 x^{4} + 3}{(x^{2} + 2) (x^{2} + 3) (x^{2} + 4)} d x$

Answer:

$We have, I = \int \frac{(4 x^{4} + 3) d x}{(x^{2} + 2) (x^{2} + 3) (x^{2} + 4)} Putting x^{2} = t Then, \frac{4 x^{4} + 3}{(x^{2} + 2) (x^{2} + 3) (x^{2} + 4)} = \frac{4 t^{2} + 3}{(t + 2) (t + 3) (t + 4)} Let \frac{4 t^{2} + 3}{(t + 2) (t + 3) (t + 4)} = \frac{A}{t + 2} + \frac{B}{t + 3} + \frac{C}{t + 4} \Rightarrow \frac{4 t^{2} + 3}{(t + 2) (t + 3) (t + 4)} = \frac{A (t + 3) (t + 4) + B (t + 2) (t + 4) + C (t + 2) (t + 3)}{(t + 2) (t + 3) (t + 4)} \Rightarrow 4 t^{2} + 3 = A (t + 3) (t + 4) + B (t + 2) (t + 4) + C (t + 2) (t + 3) Putting t + 3 = 0 \Rightarrow t = - 3 ∴ 4 \times {(- 3)}^{2} + 3 = B (- 3 + 2) (- 3 + 4) \Rightarrow 39 = B (- 1) \Rightarrow B = - 39 Putting t + 2 = 0 \Rightarrow t = - 2 ∴ 4 {(- 2)}^{2} + 3 = A (- 2 + 3) (- 2 + 4) \Rightarrow 19 = A \times 1 \times 2 \Rightarrow A = \frac{19}{2} Let t + 4 = 0 \Rightarrow t = - 4 ∴ 4 \times {(- 4)}^{2} + 3 = C (- 4 + 2) (- 4 + 3) \Rightarrow 67 = C (- 2) (- 1) \Rightarrow C = \frac{67}{2} ∴ \frac{4 t^{2} + 3}{(t + 2) (t + 3) (t + 4)} = \frac{19}{2 (t + 2)} - \frac{39}{t + 3} + \frac{67}{2 (t + 4)} \Rightarrow \frac{4 x^{4} + 3}{(x^{2} + 2) (x^{2} + 3) (x^{2} + 4)} = \frac{19}{2 (x^{2} + 2)} - \frac{39}{x^{2} + 3} + \frac{67}{2 (x^{2} + 4)} ∴ I = \frac{19}{2} \int \frac{d x}{x^{2} + {(\sqrt{2})}^{2}} - 39 \int \frac{d x}{x^{2} + {(\sqrt{3})}^{2}} - \frac{67}{2} \int \frac{d x}{x^{2} + 2^{2}} = \frac{19}{2} \times \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x}{\sqrt{2}}) - \frac{39}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - \frac{67}{2} \times \frac{1}{2} \tan^{- 1} (\frac{x}{2}) + C = \frac{19}{2 \sqrt{2}} \tan^{- 1} (\frac{x}{\sqrt{2}}) - \frac{39}{\sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) - \frac{67}{4} \tan^{- 1} (\frac{x}{2}) + C$

Page No 18.178:

Question 65:

$\int \frac{x^{4}}{(x - 1) (x^{2} + 1)} d x$

Answer:

$We have, I = \int \frac{x^{4} d x}{(x - 1) (x^{2} + 1)} = \int [\frac{x^{4} - 1 + 1}{(x - 1) (x^{2} + 1)}] d x = \int \frac{(x^{4} - 1) d x}{(x - 1) (x^{2} + 1)} + \int \frac{d x}{(x - 1) (x^{2} + 1)} = \int \frac{(x^{2} - 1) (x^{2} + 1) d x}{(x - 1) (x^{2} + 1)} + \int \frac{d x}{(x - 1) (x^{2} + 1)} = \int \frac{(x - 1) (x + 1) d x}{(x - 1)} + \int \frac{d x}{(x - 1) (x^{2} + 1)} = \int (x + 1) d x + \int \frac{d x}{(x - 1) (x^{2} + 1)} . . . . . (1) Let \frac{1}{(x - 1) (x^{2} + 1)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + 1} \Rightarrow \frac{1}{(x - 1) (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (x - 1)}{(x - 1) (x^{2} + 1)} \Rightarrow 1 = A x^{2} + A + B x^{2} - B x + C x - C \Rightarrow 1 = (A + B) x^{2} + (C - B) x + A - C Equating coefficients of like terms A + B = 0 . . . . . (1) C - B = 0 . . . . . (2) A - C = 1 . . . . . (3) Solving (1), (2) and (3), we get B = - \frac{1}{2}, A = \frac{1}{2}, C = - \frac{1}{2} ∴ \frac{1}{(x - 1) (x^{2} + 1)} = \frac{1}{2 (x - 1)} + \frac{- \frac{x}{2} - \frac{1}{2}}{x^{2} + 1} \Rightarrow \frac{1}{(x - 1) (x^{2} + 1)} = \frac{1}{2 (x - 1)} - \frac{1}{2} (\frac{x}{x^{2} + 1}) - \frac{1}{2 (x^{2} + 1)} . . . . . (2) From (1) & (2) I = \int (x + 1) d x + \frac{1}{2} \int \frac{d x}{x - 1} - \frac{1}{2} \int \frac{x d x}{x^{2} + 1} - \frac{1}{2} \int \frac{d x}{x^{2} + 1} Putting x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \int (x + 1) d x + \frac{1}{2} \int \frac{d x}{x - 1} - \frac{1}{4} \int \frac{d t}{t} - \frac{1}{2} \int \frac{d x}{x^{2} + 1} = \frac{x^{2}}{2} + x + \frac{1}{2} \log |x - 1| - \frac{1}{4} \log |t| - \frac{1}{2} \tan^{- 1} x + C = \frac{x^{2}}{2} + x + \frac{1}{2} \log |x - 1| - \frac{1}{4} \log |x^{2} + 1| - \frac{1}{2} \tan^{- 1} (x) + C$

Page No 18.178:

Question 66:

Evaluate the following integrals:

$\int \frac{x^{2}}{x^{4} - x^{2} - 12} d x$

Answer:

$Let I = \int \frac{x^{2}}{x^{4} - x^{2} - 12} d x We express \frac{x^{2}}{x^{4} - x^{2} - 12} = \frac{x^{2}}{x^{4} - 4 x^{2} + 3 x^{2} - 12} = \frac{x^{2}}{(x^{2} - 4) (x^{2} + 3)} = \frac{A}{x^{2} - 4} + \frac{B}{x^{2} + 3} \Rightarrow x^{2} = A (x^{2} + 3) + B (x^{2} - 4) Equating the coefficients of x^{2} and constants, we get 1 = A + B and 0 = 3 A - 4 B or A = \frac{4}{7} and B = \frac{3}{7} ∴ I = \int (\frac{\frac{4}{7}}{x^{2} - 4} + \frac{\frac{3}{7}}{x^{2} + 3}) d x = \frac{4}{7} \int \frac{1}{x^{2} - 4} d x + \frac{3}{7} \int \frac{1}{x^{2} + 3} d x = \frac{4}{7} \times \frac{1}{4} \log |\frac{x - 2}{x + 2}| + \frac{\sqrt{3}}{7} \tan^{- 1} \frac{x}{\sqrt{3}} + c = \frac{1}{7} \log |\frac{x - 2}{x + 2}| + \frac{\sqrt{3}}{7} \tan^{- 1} \frac{x}{\sqrt{3}} + c Hence, \int \frac{x^{2}}{x^{4} - x^{2} - 12} d x = \frac{1}{7} \log |\frac{x - 2}{x + 2}| + \frac{\sqrt{3}}{7} \tan^{- 1} \frac{x}{\sqrt{3}} + c$

Page No 18.178:

Question 67:

Evaluate the following integrals:

$\int \frac{x^{2}}{1 - x^{4}} d x$

Answer:

$Let I = \int \frac{x^{2}}{1 - x^{4}} d x We express \frac{x^{2}}{1 - x^{4}} = \frac{x^{2}}{(1 - x^{2}) (1 + x^{2})} = \frac{A}{1 - x^{2}} + \frac{B}{1 + x^{2}} \Rightarrow x^{2} = A (1 + x^{2}) + B (1 - x^{2}) Equating the coefficients of x^{2} and constants, we get 1 = A - B and 0 = A + B or A = \frac{1}{2} and B = - \frac{1}{2} ∴ I = \int (\frac{\frac{1}{2}}{1 - x^{2}} + \frac{- \frac{1}{2}}{1 + x^{2}}) d x = \frac{1}{2} \int \frac{1}{1 - x^{2}} d x - \frac{1}{2} \int \frac{1}{1 + x^{2}} d x = \frac{1}{2} \times \frac{1}{2} \log |\frac{1 + x}{1 - x}| - \frac{1}{2} \tan^{- 1} x + c = \frac{1}{4} \log |\frac{1 + x}{1 - x}| - \frac{1}{2} \tan^{- 1} x + c Hence, \int \frac{x^{2}}{1 - x^{4}} d x = \frac{1}{4} \log |\frac{1 + x}{1 - x}| - \frac{1}{2} \tan^{- 1} x + c$

Page No 18.178:

Question 68:

Evaluate the following integrals:

$\int \frac{x^{2}}{x^{4} + x^{2} - 2} d x$

Answer:

$Let I = \int \frac{x^{2}}{x^{4} + x^{2} - 2} d x We express \frac{x^{2}}{x^{4} + x^{2} - 2} = \frac{x^{2}}{x^{4} + 2 x^{2} - x^{2} - 2} = \frac{x^{2}}{(x^{2} + 2) (x^{2} - 1)} = \frac{A}{x^{2} + 2} + \frac{B}{x^{2} - 1} \Rightarrow x^{2} = A (x^{2} - 1) + B (x^{2} + 2) Equating the coefficients of x^{2} and constants, we get 1 = A + B and 0 = - A + 2 B or A = \frac{2}{3} and B = \frac{1}{3} ∴ I = \int (\frac{\frac{2}{3}}{x^{2} + 2} + \frac{\frac{1}{3}}{x^{2} - 1}) d x = \frac{2}{3} \int \frac{1}{x^{2} + 2} d x + \frac{1}{3} \int \frac{1}{x^{2} - 1} d x = \frac{\sqrt{2}}{3} \tan^{- 1} \frac{x}{\sqrt{2}} + \frac{1}{6} \log |\frac{x - 1}{x + 1}| + c Hence, \int \frac{x^{2}}{x^{4} + x^{2} - 2} d x = \frac{\sqrt{2}}{3} \tan^{- 1} \frac{x}{\sqrt{2}} + \frac{1}{6} \log |\frac{x - 1}{x + 1}| + c$

Page No 18.178:

Question 69:

$\int \frac{(x^{2} + 1) (x^{2} + 4)}{(x^{2} + 3) (x^{2} - 5)} d x$

Answer:

I = $\int \frac{(x^{2} + 1) (x^{2} + 4)}{(x^{2} + 3) (x^{2} - 5)} d x$

Since,

$\frac{(x^{2} + 1) (x^{2} + 4)}{(x^{2} + 3) (x^{2} - 5)} = \frac{[(x^{2} + 3) - 2] [(x^{2} - 5) + 9]}{(x^{2} + 3) (x^{2} - 5)} \Rightarrow \frac{(x^{2} + 1) (x^{2} + 4)}{(x^{2} + 3) (x^{2} - 5)} = \frac{(x^{2} + 3) (x^{2} - 5) + 9 (x^{2} + 3) - 2 (x^{2} - 5) - 18}{(x^{2} + 3) (x^{2} - 5)}$

$\Rightarrow \frac{(x^{2} + 1) (x^{2} + 4)}{(x^{2} + 3) (x^{2} - 5)} = 1 + \frac{9}{(x^{2} - 5)} - \frac{2}{(x^{2} + 3)} - \frac{18}{(x^{2} + 3) (x^{2} - 5)}$ ...(i)

Let $I_{1} = \int \frac{1}{(x^{2} + 3) (x^{2} - 5)}$ and $x^{2} = y$

$\Rightarrow \frac{1}{(y + 3) (y - 5)} = \frac{A}{(y + 3)} + \frac{B}{(y - 5)} = \frac{A (y - 5) + B (y + 3)}{(y + 3) (y - 5)} \Rightarrow \frac{1}{(y + 3) (y - 5)} = \frac{(A + B) y - (5 A + 3 B)}{(y + 3) (y - 5)}$

Comparing coefficients, we get

$A + B = 0 and 5 A + 3 B = - 1 By solving the equations, we get A = - \frac{1}{8} and B = \frac{1}{8}$

From (i), we get

$I = \int [1 + \frac{9}{(x^{2} - 5)} - \frac{2}{(x^{2} + 3)} - 18 (\frac{- 1}{8 (x^{2} + 3)} + \frac{1}{8 (x^{2} - 5)})] d x$

$\Rightarrow I = \int [1 + \frac{27}{4 (x^{2} - 5)} + \frac{1}{(x^{2} + 3)}] d x \Rightarrow I = \int 1 d x + \int \frac{27}{4 (x^{2} - 5)} d x + \int \frac{1}{(x^{2} + 3)} d x ∴ I = x + \frac{27}{8 \sqrt{5}} \ln (|\frac{x - \sqrt{5}}{x + \sqrt{5}}|) + \frac{1}{4 \sqrt{3}} \tan^{- 1} (\frac{x}{\sqrt{3}}) + c$

Page No 18.178:

Question 70:

$\int \frac{2 \cos x}{(1 - \sin x) (1 + \sin^{2} x)} d x$

Answer:

$Let \sin x = t \Rightarrow \cos x d x = d t \int \frac{2 d t}{(1 - t) (1 + t^{2})} Using partial fraction \frac{2}{(1 - t) (1 + t^{2})} = \frac{A}{(1 - t)} + \frac{B t + C}{(1 + t^{2})} A + A t^{2} + B t + C - B t^{2} - C t = 2 \Rightarrow A + C = 2 \Rightarrow A = 2 - C . . . . . (1) B t - C t = 0 . . . . . (2) And A t^{2} - B t^{2} = 0 . . . . . (3) Substitute (1) in (3) B = 1 Put B = 1 in (2) C = 1 Pu t C = 1 in (1) Hence, A = 1, B = 1, C = 1 \int \frac{2 d t}{(1 - t) (1 + t^{2})} = \int \frac{d t}{(1 - t)} + \int \frac{(1 + t)}{(1 + t^{2})} d t = \int \frac{d t}{(1 - t)} + \int \frac{d t}{(1 + t^{2})} + \int \frac{t d t}{(1 + t^{2})} = - \ln (1 - t) + \tan^{- 1} t + \frac{1}{2} \ln (1 + t^{2}) = \ln \frac{\sqrt{1 + t^{2}}}{1 - t} + \tan^{- 1} t + C Replacing the value of t = \ln \frac{\sqrt{1 + \sin^{2} x}}{1 - \sin x} + \tan^{- 1} (\sin x) + C$

Page No 18.190:

Question 1:

$\int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x$

Answer:

$We have, I = \int \frac{(x^{2} + 1) d x}{x^{4} + x^{2} + 1} Dividing numerator and denominator by x^{2}, we get I = \int \frac{(1 + \frac{1}{x^{2}}) d x}{x^{2} + 1 + \frac{1}{x^{2}}} = \int \frac{(1 + \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} - 2 + 3} = \int \frac{(1 + \frac{1}{x^{2}}) d x}{{(x - \frac{1}{x})}^{2} + 3} Putting x - \frac{1}{x} = t \Rightarrow (1 + \frac{1}{x^{2}}) d x = d t ∴ I = \int \frac{d t}{t^{2} + 3} = \int \frac{d t}{t^{2} + {(\sqrt{3})}^{2}} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) + C = \frac{1}{\sqrt{3}} \tan^{- 1} [\frac{x - \frac{1}{x}}{\sqrt{3}}] + C = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{3} x}) + C$

Page No 18.190:

Question 2:

$\int \sqrt{\cot θ} d θ$

Answer:

$We have, I = \int \sqrt{\cot θ} d θ Putting \cot θ = t^{2} \Rightarrow - {cosec}^{2} θ d θ = 2 t d t \Rightarrow d θ = - \frac{2 t d t}{\cos e c^{2} θ} \Rightarrow d θ = \frac{- 2 t d t}{1 + c o t^{2} θ} \Rightarrow d θ = \frac{- 2 t d t}{1 + t^{4}} ∴ I = \int t (\frac{- 2 t d t}{1 + t^{4}}) = - \int (\frac{2 t^{2}}{1 + t^{4}}) d t = - \int (\frac{t^{2} + 1 + t^{2} - 1}{t^{4} + 1}) d t = - \int (\frac{t^{2} + 1}{t^{4} + 1}) d t - \int \frac{(t^{2} - 1) d t}{t^{4} + 1} Dividing numerator and denominator by t^{2} I = - \int (\frac{1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}}) d t - \int (\frac{1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}}) d t = - \int \frac{(1 + \frac{1}{t^{2}}) d t}{t^{2} + \frac{1}{t^{2}} - 2 + 2} - \int \frac{(1 - \frac{1}{t^{2}}) d t}{t^{2} + \frac{1}{t^{2}} + 2 - 2} = - \int \frac{(1 + \frac{1}{t^{2}}) d t}{{(t - \frac{1}{t})}^{2} + {(\sqrt{2})}^{2}} - \int \frac{(1 - \frac{1}{t^{2}}) d t}{{(t + \frac{1}{t})}^{2} - {(\sqrt{2})}^{2}} Putting t - \frac{1}{t} = p \Rightarrow (1 + \frac{1}{t^{2}}) d t = d p Putting t + \frac{1}{t} = q \Rightarrow (1 - \frac{1}{t^{2}}) d t = d q I = - \int \frac{d p}{p^{2} + {(\sqrt{2})}^{2}} - \int \frac{d q}{q^{2} - {(\sqrt{2})}^{2}} = - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{p}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} \log |\frac{q - \sqrt{2}}{q + \sqrt{2}}| + C = - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} \log |\frac{t + \frac{1}{t} - \sqrt{2}}{1 + \frac{1}{t} + \sqrt{2}}| + C = - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t^{2} - 1}{\sqrt{2} t}) - \frac{1}{2 \sqrt{2}} \log |\frac{t^{2} + 1 - \sqrt{2} t}{t^{2} + 1 + \sqrt{2} t}| + C = - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\cot θ - 1}{2 \sqrt{\cot θ}}) - \frac{1}{2 \sqrt{2}} \log |\frac{\cot θ + 1 - \sqrt{2 \cot θ}}{\cot θ + 1 + \sqrt{2 \cot θ}}| + C$

Page No 18.190:

Question 3:

$\int \frac{x^{2} + 9}{x^{4} + 81} d x$

Answer:

$We have, I = \int (\frac{x^{2} + 9}{x^{4} + 81}) d x Dividing numerator and denominator by x^{2} I = \int \frac{(1 + \frac{9}{x^{2}}) d x}{x^{2} + \frac{81}{x^{2}}} = \int \frac{(1 + \frac{9}{x^{2}}) d x}{x^{2} + {(\frac{9}{x})}^{2} - 2 \times x \times \frac{9}{x} + 2 \times x \times \frac{9}{x}} = \int \frac{(1 + \frac{9}{x^{2}}) d x}{{(x - \frac{9}{x})}^{2} + {(\sqrt{18})}^{2}} Putting x - \frac{9}{x} = t \Rightarrow (1 + \frac{9}{x^{2}}) d x = d t ∴ I = \int \frac{d t}{t^{2} + {(\sqrt{18})}^{2}} = \int \frac{d t}{t^{2} + {(3 \sqrt{2})}^{2}} = \frac{1}{3 \sqrt{2}} \tan^{- 1} (\frac{t}{3 \sqrt{2}}) + C = \frac{1}{3 \sqrt{2}} \tan^{- 1} (\frac{x - \frac{9}{x}}{3 \sqrt{2}}) + C = \frac{1}{3 \sqrt{2}} \tan^{- 1} (\frac{x^{2} - 9}{3 \sqrt{2} x}) + C$

Page No 18.190:

Question 4:

$\int \frac{1}{x^{4} + x^{2} + 1} d x$

Answer:

$We have, I = \int \frac{d x}{x^{4} + x^{2} + 1} = \frac{1}{2} \int \frac{2 d x}{x^{4} + x^{2} + 1} \Rightarrow \frac{1}{2} \int (\frac{(x^{2} + 1) - (x^{2} - 1)}{x^{4} + x^{2} + 1}) d x \Rightarrow \frac{1}{2} \int (\frac{x^{2} + 1}{x^{4} + x^{2} + 1}) d x - \frac{1}{2} \int (\frac{x^{2} - 1}{x^{4} + x^{2} + 1}) d x Dividing numerator and denominator by x^{2} I = \frac{1}{2} \int (\frac{1 + \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}}) d x - \frac{1}{2} \int (\frac{1 - \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}}) d x = \frac{1}{2} \int (\frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} - 2 + 3}) d x - \frac{1}{2} \int \frac{(1 - \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} + 2 - 1} = \frac{1}{2} \int \frac{(1 + \frac{1}{x^{2}}) d x}{{(x - \frac{1}{x})}^{2} + {(\sqrt{3})}^{2}} - \frac{1}{2} \int \frac{(1 - \frac{1}{x^{2}}) d x}{{(x + \frac{1}{x})}^{2} - 1^{2}} Putting x - \frac{1}{x} = t \Rightarrow (1 + \frac{1}{x^{2}}) d x = d t Putting x + \frac{1}{x} = p \Rightarrow (1 - \frac{1}{x^{2}}) d x = d p ∴ I = \frac{1}{2} \int \frac{d t}{t^{2} + {(\sqrt{3})}^{2}} - \frac{1}{2} \int \frac{d p}{p^{2} - 1^{2}} = \frac{1}{2} \times \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) - \frac{1}{2} \times \frac{1}{2 \times 1} \log |\frac{p - 1}{p + 1}| + C = \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) - \frac{1}{4} \log |\frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1}| + C = \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{x^{2} - 1}{x \sqrt{3}}) - \frac{1}{4} \log |\frac{x^{2} - x + 1}{x^{2} + x + 1}| + C$

Page No 18.190:

Question 5:

$\int \frac{x^{2} - 3 x + 1}{x^{4} + x^{2} + 1} d x$

Answer:

$We have, I = \int (\frac{x^{2} - 3 x + 1}{x^{4} + x^{2} + 1}) d x = \int \frac{(x^{2} + 1) d x}{x^{4} + x^{2} + 1} - 3 \int \frac{x d x}{x^{4} + x^{2} + 1} . . . . . (1) = I_{1} - 3 I_{2} where I_{1} = \int \frac{(x^{2} + 1) d x}{x^{4} + x^{2} + 1}, I_{2} = \int \frac{x d x}{x^{4} + x^{2} + 1} I_{1} = \int (\frac{x^{2} + 1}{x^{4} + x^{2} + 1}) d x Dividing numerator & denominator by x^{2} I_{1} = \int \frac{(1 + \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} + 1} = \int \frac{(1 + \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} - 2 + 3} = \int \frac{(1 + \frac{1}{x^{2}}) d x}{{(x - \frac{1}{x})}^{2} + {(\sqrt{3})}^{2}} Let x - \frac{1}{x} = t \Rightarrow (1 + \frac{1}{x^{2}}) d x = d t ∴ I_{1} = \int \frac{d t}{t^{2} + {(\sqrt{3})}^{2}} I_{1} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) + C_{1} I_{1} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + C_{1} . . . . . (2) I_{2} = \int \frac{x d x}{x^{4} + x^{2} + 1} Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I_{2} = \frac{1}{2} \int \frac{d t}{t^{2} + t + 1} = \frac{1}{2} \int \frac{d t}{t^{2} + t + \frac{1}{4} + \frac{3}{4}} = \frac{1}{2} \int \frac{d t}{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{1}{\frac{\sqrt{3}}{2}} \times \frac{1}{2} [\tan^{- 1} (\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}})] + C_{2} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 t + 1}{\sqrt{3}}) + C_{2} I_{2} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x^{2} + 1}{\sqrt{3}}) + C_{2} . . . (3) From equating (1), (2) and (3) we have I = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + C_{1} - 3 \times [\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x^{2} + 1}{\sqrt{3}}) + C_{2}] = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{3} x}) - \sqrt{3} \tan^{- 1} (\frac{2 x^{2} + 1}{\sqrt{3}}) + C where C = C_{1} + 3 C_{2}$

Page No 18.190:

Question 6:

$\int \frac{x^{2} + 1}{x^{4} - x^{2} + 1} d x$

Answer:

$We have, I = \int (\frac{x^{2} + 1}{x^{4} - x^{2} + 1}) d x Dividing numerator and denominator by x^{2} I = \int \frac{(1 + \frac{1}{x^{2}})}{x^{2} + \frac{1}{x^{2}} - 1} d x = \int \frac{(1 + \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} - 2 + 1} = \int \frac{(1 + \frac{1}{x^{2}}) d x}{{(x - \frac{1}{x})}^{2} + 1} Putting x - \frac{1}{x} = t \Rightarrow (1 + \frac{1}{x^{2}}) d x = d t ∴ I = \int \frac{d t}{t^{2} + 1^{2}} = \tan^{- 1} t + C = \tan^{- 1} (x - \frac{1}{x}) + C = \tan^{- 1} (\frac{x^{2} - 1}{x}) + C$

Page No 18.190:

Question 7:

$\int \frac{x^{2} - 1}{x^{4} + 1} d x$

Answer:

$We have, I = \int (\frac{x^{2} - 1}{x^{4} + 1}) d x Dividing numerator and denominator by x^{2} = \int (\frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}}) d x = \int \frac{(1 - \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} + 2 - 2} = \int \frac{(1 - \frac{1}{x^{2}}) d x}{{(x + \frac{1}{x})}^{2} - {(\sqrt{2})}^{2}} Putting x + \frac{1}{x} = t \Rightarrow (1 - \frac{1}{x^{2}}) d x = d t ∴ I = \int \frac{d t}{t^{2} - {(\sqrt{2})}^{2}} = \frac{1}{2 \sqrt{2}} \log |\frac{t - \sqrt{2}}{t + \sqrt{2}}| + C = \frac{1}{2 \sqrt{2}} \log |\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}| + C = \frac{1}{2 \sqrt{2}} \log |\frac{x^{2} - \sqrt{2} x + 1}{x^{2} + \sqrt{2} x + 1}| + C$

Page No 18.190:

Question 8:

$\int \frac{x^{2} + 1}{x^{4} + 7 x^{2} + 1} d x$

Answer:

$We have, I = \int (\frac{x^{2} + 1}{x^{4} + 7 x^{2} + 1}) d x Dividing numerator and denominator by x^{2} I = \int (\frac{1 + \frac{1}{x^{2}}}{x^{2} + 7 + \frac{1}{x^{2}}}) d x = \int \frac{(1 + \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} - 2 + 9} \Rightarrow \int \frac{(1 + \frac{1}{x^{2}}) d x}{{(x - \frac{1}{x})}^{2} + 3^{2}} Putting x - \frac{1}{x} = t \Rightarrow (1 + \frac{1}{x^{2}}) d x = d t ∴ I = \int \frac{d t}{t^{2} + 3^{2}} = \frac{1}{3} \tan^{- 1} (\frac{t}{3}) + C = \frac{1}{3} \tan^{- 1} (\frac{x - \frac{1}{x}}{3}) + C = \frac{1}{3} \tan^{- 1} (\frac{x^{2} - 1}{3 x}) + C$

Page No 18.190:

Question 9:

$\int \frac{{(x - 1)}^{2}}{x^{4} + x^{2} + 1} d x$

Answer:

$We have, I = \int \frac{{(x - 1)}^{2} d x}{x^{4} + x^{2} + 1} = \int (\frac{x^{2} - 2 x + 1}{x^{4} + x^{2} + 1}) d x = \int (\frac{x^{2} + 1}{x^{4} + x^{2} + 1}) d x - \int \frac{2 x d x}{x^{4} + x^{2} + 1} = I_{1} - I_{2} where, I_{1} = \int \frac{(x^{2} + 1) d x}{x^{4} + x^{2} + 1} I_{2} = \int \frac{2 x d x}{x^{4} + x^{2} + 1} Now, I_{1} = \int (\frac{x^{2} + 1}{x^{4} + x^{2} + 1}) d x Dividing numerator and denominator by x^{2} I_{1} = \int (\frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} + 1}) d x I_{1} = \int \frac{(1 + \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} - 2 + 3} I_{1} = \int \frac{(1 + \frac{1}{x^{2}}) d x}{{(x - \frac{1}{x})}^{2} + {(\sqrt{3})}^{2}} Putting x - \frac{1}{x} = t \Rightarrow (1 + \frac{1}{x^{2}}) d x = d t ∴ I_{1} = \int \frac{d t}{t^{2} + {(\sqrt{3})}^{2}} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) + C_{1} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) + C_{1} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{3} x}) + C_{1} And I_{2} = \int \frac{2 x d x}{x^{4} + x^{2} + 1} Putting x^{2} = t \Rightarrow 2 x d x = d t I_{2} = \int \frac{d t}{t^{2} + t + 1} = \int \frac{d t}{t^{2} + t + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} = \int \frac{d t}{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C_{2} = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 t + 1}{3}) + C_{2} ∴ I = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{3} x}) - \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x^{2} + 1}{\sqrt{3}}) + C$

Page No 18.190:

Question 10:

$\int \frac{1}{x^{4} + 3 x^{2} + 1} d x$

Answer:

$We have, I = \int \frac{d x}{x^{4} + 3 x^{2} + 1} = \frac{1}{2} \int \frac{2 d x}{x^{4} + 3 x^{2} + 1} = \frac{1}{2} \int [\frac{(x^{2} + 1) - (x^{2} - 1)}{x^{4} + 3 x^{2} + 1}] d x = \frac{1}{2} \int (\frac{x^{2} + 1}{x^{4} + 3 x^{2} + 1}) d x - \frac{1}{2} \int \frac{(x^{2} - 1)}{x^{4} + 3 x^{2} + 1} d x Dividing numerator and denominator by x^{2} = \frac{1}{2} \int (\frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} + 3}) d x - \frac{1}{2} \int \frac{(1 - \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} + 3} = \frac{1}{2} \int (\frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} - 2 + 5}) d x - \frac{1}{2} \int \frac{(1 - \frac{1}{x^{2}}) d x}{x^{2} + \frac{1}{x^{2}} + 2 + 1} = \frac{1}{2} \int \frac{(1 + \frac{1}{x^{2}}) d x}{{(x - \frac{1}{x})}^{2} + {(\sqrt{5})}^{2}} - \frac{1}{2} \int \frac{(1 - \frac{1}{x^{2}}) d x}{{(x + \frac{1}{x})}^{2} + 1^{2}} Putting x - \frac{1}{x} = t \Rightarrow (1 + \frac{1}{x^{2}}) d x = d t Putting x + \frac{1}{x} = p \Rightarrow (1 - \frac{1}{x^{2}}) d x = d p ∴ I = \frac{1}{2} \int \frac{d t}{t^{2} + {(\sqrt{5})}^{2}} - \frac{1}{2} \int \frac{d p}{p^{2} + 1^{2}} = \frac{1}{2 \sqrt{5}} \tan^{- 1} (\frac{t}{\sqrt{5}}) - \frac{1}{2} \tan^{- 1} (p) + C = \frac{1}{2 \sqrt{5}} \tan^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{5}}) - \frac{1}{2} \tan^{- 1} (x + \frac{1}{x}) + C = \frac{1}{2 \sqrt{5}} \tan^{- 1} (\frac{x^{2} - 1}{\sqrt{5} x}) - \frac{1}{2} \tan^{- 1} (\frac{x^{2} + 1}{x}) + C$

Page No 18.190:

Question 11:

Evaluate the following integrals:

$\int \frac{1}{\sin^{4} x + \sin^{2} x \cos^{2} x + \cos^{4} x} d x$

Answer:

$Let I = \int \frac{1}{\sin^{4} x + \sin^{2} x \cos^{2} x + \cos^{4} x} d x = \int \frac{1}{{(\sin^{2} x + \cos^{2} x)}^{2} - \sin^{2} x \cos^{2} x} d x = \int \frac{1}{1 - \sin^{2} x \cos^{2} x} d x = \int \frac{\frac{1}{\cos^{4} x}}{\frac{1}{\cos^{4} x} - \frac{\sin^{2} x}{\cos^{2} x}} d x = \int \frac{\sec^{2} x (1 + \tan^{2} x)}{\sec^{4} x - \tan^{2} x} d x Let \tan x = t On differentiating both sides, we get \sec^{2} x d x = d t ∴ I = \int \frac{1 + t^{2}}{{(1 + t^{2})}^{2} - t^{2}} d t = \int \frac{1 + t^{2}}{(t^{4} + t^{2} + 1)} d t = \int \frac{\frac{1}{t^{2}} + 1}{(t^{2} + 1 + \frac{1}{t^{2}})} d t = \int \frac{\frac{1}{t^{2}} + 1}{{(t - \frac{1}{t})}^{2} + 3} d t Let (t - \frac{1}{t}) = u On differentiating both sides, we get (1 + \frac{1}{t^{2}}) d t = d u ∴ I = \int \frac{1}{{(u)}^{2} + 3} d u = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{u}{\sqrt{3}}) + c = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{3}}) + c = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{\tan x - \cot x}{\sqrt{3}}) + c Hence, \int \frac{1}{\sin^{4} x + \sin^{2} x \cos^{2} x + \cos^{4} x} d x = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{\tan x - \cot x}{\sqrt{3}}) + c$

Page No 18.196:

Question 1:

$\int \frac{1}{(x - 1) \sqrt{x + 2}} d x$

Answer:

$We have, I = \int \frac{d x}{(x - 1) \sqrt{x + 2}} Putting x + 2 = t^{2} \Rightarrow d x = 2 t d t ∴ I = \int \frac{2 t d t}{(t^{2} - 2 - 1) t} = \int \frac{2 d t}{t^{2} - {(\sqrt{3})}^{2}} = 2 \times \frac{1}{2 \sqrt{3}} \log |\frac{t - \sqrt{3}}{t + \sqrt{3}}| + C = \frac{1}{\sqrt{3}} \log |\frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}}| + C$

Page No 18.196:

Question 2:

$\int \frac{1}{(x - 1) \sqrt{2 x + 3}} d x$

Answer:

$We have, I = \int \frac{d x}{(x - 1) \sqrt{2 x + 3}} Putting 2 x + 3 = t^{2} \Rightarrow x = \frac{t^{2} - 3}{2} Diff both sides d x = t d t ∴ I = \int \frac{t d t}{[\frac{t^{2} - 3}{2} - 1] t} = \int \frac{2 d t}{t^{2} - 3 - 2} = \frac{2 d t}{t^{2} - 5} = 2 \int \frac{d t}{t^{2} - {(\sqrt{5})}^{2}} = 2 \times \frac{1}{2 \sqrt{5}} \log |\frac{t - \sqrt{5}}{t + \sqrt{5}}| + C = \frac{1}{\sqrt{5}} \log |\frac{\sqrt{2 x + 3} - \sqrt{5}}{\sqrt{2 x + 3} + \sqrt{5}}| + C$

Page No 18.196:

Question 3:

$\int \frac{x + 1}{(x - 1) \sqrt{x + 2}} d x$

Answer:

$We have, I = \int \frac{x + 1}{(x - 1) \sqrt{x + 2}} d x Putting x + 2 = t^{2} \Rightarrow x = t^{2} - 2 Diff both sides d x = 2 t d t I = \int \frac{(t^{2} - 2 + 1) 2 t d t}{(t^{2} - 2 - 1) t} = 2 \int (\frac{t^{2} - 1}{t^{2} - 3}) d t = 2 \int (\frac{t^{2} - 3 + 2}{t^{2} - 3}) d t = 2 \int (\frac{t^{2} - 3}{t^{2} - 3}) d t + 4 \int \frac{d t}{t^{2} - 3} = 2 \int d t + 4 \int \frac{d t}{t^{2} - {(\sqrt{3})}^{2}} = 2 t + 4 \times \frac{1}{2 \sqrt{3}} \log |\frac{t - \sqrt{3}}{t + \sqrt{3}}| + C = 2 \sqrt{x + 2} + \frac{2}{\sqrt{3}} \log |\frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}}| + C$

Page No 18.196:

Question 4:

$\int \frac{x^{2}}{(x - 1) \sqrt{x + 2}} d x$

Answer:

$We have, I = \int \frac{x^{2}}{(x - 1) \sqrt{x + 2}} d x Putting x + 2 = t^{2} x = t^{2} - 2 Diff both sides d x = 2 t d t I = \int \frac{{(t^{2} - 2)}^{2}}{(t^{2} - 2 - 1) t} 2 t d t = 2 \int \frac{{(t^{2} - 2)}^{2} d t}{t^{2} - 3} = 2 \int \frac{(t^{4} - 4 t^{2} + 4)}{t^{2} - 3} d t Dividing numerator by denominator, we get t^{2} - 1 t^{2} - 3 t^{4} - 4 t^{2} + 4 t^{4} - 3 t^{2} - + - t^{2} + 4 - t^{2} + 3 + - 1 ∴ I = 2 \int (t^{2} - 1 + \frac{1}{t^{2} - 3}) d t = 2 \int t^{2} d t - 2 \int d t + 2 \int \frac{d t}{t^{2} - {(\sqrt{3})}^{2}} = 2 [\frac{t^{3}}{3}] - 2 t + 2 \times \frac{1}{2 \sqrt{3}} \log |\frac{t - \sqrt{3}}{t + \sqrt{3}}| + C = \frac{2}{3} {(\sqrt{x + 2})}^{3} - 2 \sqrt{x + 2} + \frac{1}{\sqrt{3}} \log |\frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}}| + C = \frac{2}{3} {(x + 2)}^{\frac{3}{2}} - 2 \sqrt{x + 2} + \frac{1}{\sqrt{3}} \log |\frac{\sqrt{x + 2} - \sqrt{3}}{\sqrt{x + 2} + \sqrt{3}}| + C$

Page No 18.196:

Question 5:

$\int \frac{x}{(x - 3) \sqrt{x + 1}} d x$

Answer:

$We have, I = \int \frac{x d x}{(x - 3) \sqrt{x + 1}} Putting x + 1 = t^{2} \Rightarrow x = t^{2} - 1 Diff both sides d x = 2 t d t ∴ I = \int \frac{(t^{2} - 1) 2 t d t}{(t^{2} - 1 - 3) t} = 2 \int (\frac{t^{2} - 1}{t^{2} - 4}) d t = 2 \int (\frac{t^{2} - 4 + 3}{t^{2} - 4}) d t = 2 \int (\frac{t^{2} - 4}{t^{2} - 4}) d t + 6 \int \frac{d t}{t^{2} - 2^{2}} = 2 \int d t + 6 \int \frac{d t}{t^{2} - 2^{2}} = 2 t + 6 \times \frac{1}{2 \times 2} \log |\frac{t - 2}{t + 2}| + C = 2 \sqrt{x + 1} + \frac{3}{2} \log |\frac{t - 2}{t + 2}| + C = 2 \sqrt{x + 1} + \frac{3}{2} \log |\frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2}| + C$

Page No 18.196:

Question 6:

$\int \frac{1}{(x^{2} + 1) \sqrt{x}} d x$

Answer:

$We have, I = \int \frac{d x}{(x^{2} + 1) \sqrt{x}} Putting x = t^{2} d x = 2 t d t ∴ I = \int \frac{2 t d t}{[{(t^{2})}^{2} + 1] t} = 2 \int \frac{d t}{t^{4} + 1} = \int [\frac{(t^{2} + 1) - (t^{2} - 1)}{(t^{4} + 1)}] d t = \int (\frac{t^{2} + 1}{t^{4} + 1}) d t - \int (\frac{t^{2} - 1}{t^{4} + 1}) d t Dividing numerator & denominator by t^{2} I = \int (\frac{1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}}) d t - \int \frac{(1 - \frac{1}{t^{2}}) d t}{t^{2} + \frac{1}{t^{2}}} = \int \frac{(1 + \frac{1}{t^{2}}) d t}{t^{2} + \frac{1}{t^{2}} - 2 + 2} - \int \frac{(1 - \frac{1}{t^{2}}) d t}{t^{2} + \frac{1}{t^{2}} + 2 - 2} = \int \frac{(1 + \frac{1}{t^{2}}) d t}{{(t - \frac{1}{t})}^{2} + {(\sqrt{2})}^{2}} - \int \frac{(1 - \frac{1}{t^{2}}) d t}{{(t + \frac{1}{t})}^{2} - {(\sqrt{2})}^{2}} Putting t - \frac{1}{t} = p \Rightarrow (1 + \frac{1}{t^{2}}) d t = d p Putting t + \frac{1}{t} = q \Rightarrow (1 - \frac{1}{t^{2}}) d t = d q ∴ I = \int \frac{d p}{p^{2} + {(\sqrt{2})}^{2}} - \int \frac{d q}{q^{2} - {(\sqrt{2})}^{2}} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{p}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} \log |\frac{q - \sqrt{2}}{q + \sqrt{2}}| + C = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} \log |\frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}}| + C = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t^{2} - 1}{\sqrt{2} t}) - \frac{1}{2 \sqrt{2}} \log |\frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1}| + C = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{x - 1}{\sqrt{2 x}}) - \frac{1}{2 \sqrt{2}} \log |\frac{x - \sqrt{2 x} + 1}{x + \sqrt{2 x} + 1}| + C$

Page No 18.196:

Question 7:

$\int \frac{x}{(x^{2} + 2 x + 2) \sqrt{x + 1}} d x$

Answer:

$We have, I = \int \frac{x d x}{(x^{2} + 2 x + 2) \sqrt{x + 1}} = \int \frac{x d x}{[{(x + 1)}^{2} + 1] \sqrt{x + 1}} Putting x + 1 = t^{2} \Rightarrow x = t^{2} - 1 Diff both sides d x = 2 t d t ∴ I = \int \frac{(t^{2} - 1) 2 t d t}{[{(t^{2})}^{2} + 1] t} = 2 \int \frac{(t^{2} - 1) d t}{t^{4} + 1} Dividing numerator and denominator by t^{2} I = 2 (\frac{1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}}) d t$

$= 2 \int \frac{(1 - \frac{1}{t^{2}}) d t}{t^{2} + \frac{1}{t^{2}} + 2 - 2} = 2 \int \frac{(1 - \frac{1}{t^{2}}) d t}{{(t + \frac{1}{t})}^{2} - {(\sqrt{2})}^{2}} Putting t + \frac{1}{t} = p \Rightarrow (1 - \frac{1}{t^{2}}) d t = d p I = 2 \int \frac{d p}{p^{2} - {(\sqrt{2})}^{2}} = 2 \times \frac{1}{2 \sqrt{2}} \log |\frac{p - \sqrt{2}}{p + \sqrt{2}}| + C = \frac{1}{\sqrt{2}} \log |\frac{p - \sqrt{2}}{P + \sqrt{2}}| + C = \frac{1}{\sqrt{2}} \log |\frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}}| + C = \frac{1}{\sqrt{2}} \log |\frac{t^{2} - \sqrt{2} t + 1}{t^{2} + \sqrt{2} t + 1}| + C = \frac{1}{\sqrt{2}} \log |\frac{x + 1 - \sqrt{2 (x + 1)} + 1}{x + 1 + \sqrt{2 (x + 1)} + 1}| + C = \frac{1}{\sqrt{2}} \log |\frac{(x + 2) - \sqrt{2 (x + 1)}}{(x + 2) + \sqrt{2 (x + 1)}}| + C$

Page No 18.196:

Question 8:

$\int \frac{1}{(x - 1) \sqrt{x^{2} + 1}} d x$

Answer:

$We have, I = \int \frac{d x}{(x - 1) \sqrt{x^{2} + 1}} Putting x - 1 = \frac{1}{t} \Rightarrow d x = - \frac{1}{t^{2}} d t ∴ I = \int \frac{- \frac{1}{t^{2}} d t}{(\frac{1}{t}) \sqrt{{(1 + \frac{1}{t})}^{2} + 1}} = \int \frac{- \frac{1}{t} d t}{\sqrt{1 + \frac{1}{t^{2}} + \frac{2}{t} + 1}} = \int \frac{- \frac{1}{t} d t}{\frac{\sqrt{t^{2} + 1 + 2 t + t^{2}}}{t}} = \int \frac{- d t}{\sqrt{2 t^{2} + 2 t + 1}} = - \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{t^{2} + t + \frac{1}{2}}} = - \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{t^{2} + t + \frac{1}{4} - \frac{1}{4} + \frac{1}{2}}} = - \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{{(t + \frac{1}{2})}^{2} + {(\frac{1}{2})}^{2}}} = - \frac{1}{\sqrt{2}} \log |t + \frac{1}{2} + \sqrt{{(t + \frac{1}{2})}^{2} + \frac{1}{4}}| + C where t = \frac{1}{x - 1}$

Page No 18.196:

Question 9:

$\int \frac{1}{(x + 1) \sqrt{x^{2} + x + 1}} d x$

Answer:

$We have, I = \int \frac{d x}{(x + 1) \sqrt{x^{2} + x + 1}} Putting x + 1 = \frac{1}{t} \Rightarrow d x = - \frac{1}{t^{2}} d t ∴ I = \int \frac{- \frac{1}{t^{2}} d t}{\frac{1}{t} \sqrt{{(\frac{1}{t} - 1)}^{2} + - 1 + 1 \frac{1}{t}}} = \int \frac{- \frac{1}{t^{2}} d t}{\frac{1}{t} \sqrt{\frac{1}{t^{2}} - + 1 + \frac{2}{t} \frac{1}{t}}} = \int \frac{- \frac{1}{t} d t}{\frac{\sqrt{t^{2} + t - 2 t + 1}}{t}} = - \int \frac{d t}{\sqrt{t^{2} - t + 1}} = - \int \frac{d t}{\sqrt{t^{2} - t + \frac{1}{4} - \frac{1}{4} + 1}} = - \int \frac{d t}{\sqrt{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}} = - \log |t - \frac{1}{2} + \sqrt{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}| + C = - \log |t - \frac{1}{2} + \sqrt{t^{2} - t + 1}| + C = - \log |\frac{1}{x + 1} - \frac{1}{2} + \sqrt{\frac{1}{{(x + 1)}^{2}} - \frac{1}{x + 1} + 1}| + C = - \log |\frac{1}{x + 1} - \frac{1}{2} + \frac{\sqrt{{(x + 1)}^{2} - (x + 1) + 1}}{x + 1}| + C = - \log |\frac{1}{x + 1} - \frac{1}{2} + \frac{\sqrt{x^{2} + x + 1}}{x + 1}| + C$

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Question 10:

$\int \frac{1}{(x^{2} - 1) \sqrt{x^{2} + 1}} d x$

Answer:

$We have, I = \int \frac{d x}{(x^{2} - 1) \sqrt{x^{2} + 1}} Putting x = \frac{1}{t} \Rightarrow d x = - \frac{1}{t^{2}} d t ∴ I = \int \frac{- \frac{1}{t^{2}} d t}{(\frac{1}{t^{2}} - 1) \sqrt{\frac{1}{t^{2}} + 1}} = \int \frac{- \frac{1}{t^{2}} d t}{\frac{(1 - t^{2})}{t^{2}} \times \frac{\sqrt{1 + t^{2}}}{t}} = \int \frac{- t d t}{(1 - t^{2}) \sqrt{1 + t^{2}}} Putting 1 + t^{2} = u^{2} \Rightarrow t^{2} = u^{2} - 1 \Rightarrow 2 t d t = 2 u d u \Rightarrow t d t = u d u I = - \int \frac{u d u}{(1 - u^{2} + 1) u} = - \int \frac{d u}{2 - u^{2}} = - \int \frac{d u}{{(\sqrt{2})}^{2} - u^{2}} = - \frac{1}{2 \sqrt{2}} \log |\frac{u + \sqrt{2}}{u - \sqrt{2}}| + C = - \frac{1}{2 \sqrt{2}} \log |\frac{\sqrt{1 + t^{2}} + \sqrt{2}}{\sqrt{1 + t^{2}} - \sqrt{2}}| + C = - \frac{1}{2 \sqrt{2}} \log |\frac{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{2}}{\sqrt{1 + \frac{1}{x^{2}}} - \sqrt{2}}| + C = - \frac{1}{2 \sqrt{2}} \log |\frac{\sqrt{x^{2} + 1} + \sqrt{2} x}{\sqrt{x^{2} + 1} - \sqrt{2} x}| + C$

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Question 11:

$\int \frac{x}{(x^{2} + 4) \sqrt{x^{2} + 1}} d x$

Answer:

$We have, I = \int \frac{x d x}{(x^{2} + 4) \sqrt{x^{2} + 1}} Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{(t + 4) \sqrt{t + 1}} Again Putting t + 1 = p^{2} \Rightarrow t = p^{2} - 1 \Rightarrow d t = 2 p d p I = \frac{1}{2} \int \frac{2 p d p}{(p^{2} - 1 + 4) p} = \int \frac{d p}{p^{2} + 3} = \int \frac{d p}{p^{2} + {(\sqrt{3})}^{2}} = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{p}{\sqrt{3}}) + C = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{\sqrt{t + 1}}{\sqrt{3}}) + C = \frac{1}{\sqrt{3}} \tan^{- 1} (\sqrt{\frac{x^{2} + 1}{3}}) + C$

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Question 12:

$\int \frac{1}{(1 + x^{2}) \sqrt{1 - x^{2}}} d x$

Answer:

$We have, I = \int \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}} Putting x = \frac{1}{t} \Rightarrow d x = - \frac{1}{t^{2}} d t ∴ I = \int \frac{- \frac{1}{t^{2}} d t}{(1 + \frac{1}{t^{2}}) \sqrt{1 - \frac{1}{t^{2}}}} = \int \frac{- \frac{1}{t^{2}} d t}{\frac{(t^{2} + 1)}{t^{2}} \frac{\sqrt{t^{2} - 1}}{t}} = - \int \frac{t d t}{(t^{2} + 1) \sqrt{t^{2} - 1}} Again Putting t^{2} - 1 = u^{2} \Rightarrow 2 t d t = 2 u d u \Rightarrow t d t = u d u ∴ I = - \int \frac{u d u}{(u^{2} + 2) u} = - \int \frac{d u}{u^{2} + {(\sqrt{2})}^{2}} = - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{u}{\sqrt{2}}) + C = - \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\sqrt{t^{2} - 1}}{\sqrt{2}}) + C = - \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{\frac{\frac{1}{x^{2}} - 1}{2}}) + C = - \frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{\frac{1 - x^{2}}{2 x^{2}}}) + C$

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Question 13:

$\int \frac{1}{(2 x^{2} + 3) \sqrt{x^{2} - 4}} d x$

Answer:

$We have, I = \int \frac{d x}{(2 x^{2} + 3) \sqrt{x^{2} - 4}} Putting x = \frac{1}{t} \Rightarrow d x = - \frac{1}{t^{2}} d t ∴ I = \int \frac{- \frac{1}{t^{2}} d t}{(\frac{2}{t^{2}} + 3) \sqrt{\frac{1}{t^{2}} - 4}} = \int \frac{- \frac{1}{t^{2}} d t}{\frac{(2 + 3 t^{2})}{t^{2}} \times \frac{\sqrt{1 - 4 t^{2}}}{t}} = - \int \frac{t d t}{(2 + 3 t^{2}) \sqrt{1 - 4 t^{2}}} Again Putting 1 - 4 t^{2} = u^{2} \Rightarrow - 8 t d t = 2 u d u \Rightarrow t d t = - \frac{u}{4} d u ∴ I = \frac{1}{4} \int \frac{u d u}{[2 + 3 (\frac{1 - u^{2}}{4})] u} = \frac{1}{4} \int \frac{4 d u}{[8 + 3 - 3 u^{2}]} = \int \frac{d u}{11 - 3 u^{2}} = \frac{1}{3} \int \frac{d u}{\frac{11}{3} - u^{2}} = \frac{1}{3} \int \frac{d u}{{(\sqrt{\frac{11}{3}})}^{2} - u^{2}} = \frac{1}{3} \times \frac{1}{2 \times \frac{\sqrt{11}}{\sqrt{3}}} \log |\frac{\frac{\sqrt{11}}{\sqrt{3}} + u}{\frac{\sqrt{11}}{\sqrt{3}} - u}| + C = \frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} + \sqrt{3} u}{\sqrt{11} - \sqrt{3} u}| + C = \frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} + \sqrt{3} \sqrt{1 - 4 t^{2}}}{\sqrt{11} - \sqrt{3} \sqrt{1 - 4 t^{2}}}| + C = \frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} + \sqrt{3 - 12 t^{2}}}{\sqrt{11} - \sqrt{3 - 12 t^{2}}}| + C = \frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} + \sqrt{3 - \frac{12}{x^{2}}}}{\sqrt{11} - \sqrt{3 - \frac{12}{x^{2}}}}| + C = \frac{1}{2 \sqrt{33}} \log |\frac{\sqrt{11} x + \sqrt{3 x^{2} - 12}}{\sqrt{11} x - \sqrt{3 x^{2} - 12}}| + C$

Page No 18.196:

Question 14:

$\int \frac{x}{(x^{2} + 4) \sqrt{x^{2} + 9}} d x$

Answer:

$We have, I = \int \frac{x d x}{(x^{2} + 4) \sqrt{x^{2} + 9}} Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{(t + 4) \sqrt{t + 9}} Again Putting t + 9 = u^{2} \Rightarrow d t = 2 u d u ∴ I = \frac{1}{2} \int \frac{2 u d u}{(u^{2} - 9 + 4) u} = \int \frac{d u}{u^{2} - 5} = \int \frac{d u}{u^{2} - {(\sqrt{5})}^{2}} = \frac{1}{2 \sqrt{5}} \log |\frac{u - \sqrt{5}}{u + \sqrt{5}}| + C = \frac{1}{2 \sqrt{5}} \log |\frac{\sqrt{t + 9} - \sqrt{5}}{\sqrt{t + 9} + \sqrt{5}}| + C = \frac{1}{2 \sqrt{5}} \log |\frac{\sqrt{x^{2} + 9} - \sqrt{5}}{\sqrt{x^{2} + 9} + \sqrt{5}}| + C$

Page No 18.197:

Question 1:

$\int \frac{1}{\sqrt{x} + \sqrt{x + 1}} d x$

Answer:

$\int \frac{1}{\sqrt{x + 1} + \sqrt{x}} d x Rationalising the denominator, = \int \frac{(\sqrt{x + 1} - \sqrt{x})}{(\sqrt{x + 1} + \sqrt{x}) (\sqrt{x + 1} - \sqrt{x})} d x = \int \frac{(\sqrt{x + 1} - \sqrt{x})}{x + 1 - x} d x = \int (\sqrt{x + 1} - \sqrt{x}) d x = \int [{(x + 1)}^{\frac{1}{2}} - x^{\frac{1}{2}}] d x = \frac{{(x + 1)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C = \frac{2}{3} {(x + 1)}^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C = \frac{2}{3} [{(x + 1)}^{\frac{3}{2}} - x^{\frac{3}{2}}] + C$

Page No 18.197:

Question 2:

$\int \frac{1 - x^{4}}{1 - x} d x$

Answer:

$\int (\frac{1 - x^{4}}{1 - x}) d x = \int \frac{(1 - x^{2}) (1 + x^{2})}{(1 - x)} d x = \int \frac{(1 - x) (1 + x) (1 + x^{2})}{(1 - x)} d x = \int (1 + x) (1 + x^{2}) d x = \int (1 + x^{2} + x + x^{3}) d x = x + \frac{x^{3}}{3} + \frac{x^{2}}{2} + \frac{x^{4}}{4} + C$

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Question 3:

$\int \frac{x + 2}{{(x + 1)}^{3}} d x$

Answer:

$Let \int \frac{(x + 2)}{{(x + 1)}^{3}} d x Putting x + 1 = t \Rightarrow x = t - 1 \Rightarrow d x = d t ∴ I = \int (\frac{t - 1 + 2}{t^{3}}) d t = \int (\frac{1}{t^{2}} + \frac{1}{t^{3}}) d t = \int (t^{- 2} + t^{- 3}) d t = [\frac{t^{- 2 + 1}}{- 2 + 1} + \frac{t^{- 3 + 1}}{- 3 + 1}] + C = - \frac{1}{t} - \frac{2}{t^{2}} + C = - \frac{1}{x + 1} - \frac{1}{2 {(x + 1)}^{2}} + C (∵ t = x + 1)$

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Question 4:

$\int \frac{8 x + 13}{\sqrt{4 x + 7}} d x$

Answer:

$Let I = \int (\frac{8 x + 13}{\sqrt{4 x + 7}}) d x Putting 4 x + 7 = t \Rightarrow x = \frac{t - 7}{4} \Rightarrow 4 d x = d t \Rightarrow d x = \frac{d t}{4} ∴ I = \frac{1}{4} \int \{\frac{8 (\frac{t - 7}{4}) + 13}{\sqrt{t}}\} d t = \frac{1}{4} \int (\frac{2 t - 14 + 13}{\sqrt{t}}) d t = \frac{1}{4} \int (\frac{2 t - 1}{\sqrt{t}}) d t = \frac{1}{4} \int \frac{2 t}{\sqrt{t}} d t - \frac{1}{4} \int \frac{d t}{\sqrt{t}} = \frac{1}{2} \int t^{\frac{1}{2}} d t - \frac{1}{4} \int t^{- \frac{1}{2}} d t = \frac{1}{2} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] - \frac{1}{4} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = \frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}} - \frac{2}{4} t^{\frac{1}{2}} + C = \frac{1}{3} t^{\frac{3}{2}} - \frac{2}{4} t^{\frac{1}{2}} + C = \frac{1}{3} {(4 x + 7)}^{\frac{3}{2}} - \frac{1}{2} {(4 x + 7)}^{\frac{1}{2}} + C (∵ t = 4 x + 7) = \frac{1}{3} {(4 x + 7)}^{\frac{3}{2}} - \sqrt{4 x + 7} + C$

Page No 18.197:

Question 5:

$\int \frac{1 + x + x^{2}}{x^{2} (1 + x)} d x$

Answer:

$\int \frac{1 + x + x^{2}}{x^{2} (1 + x)} d x \Rightarrow \int \frac{(1 + x)}{x^{2} (1 + x)} d x + \int \frac{x^{2}}{x^{2} (1 + x)} d x \Rightarrow \int \frac{d x}{x^{2}} + \int \frac{d x}{1 + x} \Rightarrow \int x^{- 2} d x + \int \frac{1}{1 + x} d x \Rightarrow [\frac{x^{- 2 + 1}}{- 2 + 1}] + \ln (1 + x) + C \Rightarrow \frac{- 1}{x} + \ln |1 + x| + C$

Page No 18.197:

Question 6:

$\int \frac{{(2^{x} + 3^{x})}^{2}}{6^{x}} d x$

Answer:

$\int \frac{{(2^{x} + 3^{x})}^{2}}{6^{x}} d x = \int [\frac{{(2^{x})}^{2} + {(3^{x})}^{2} + 2 \cdot 2^{x} \cdot 3^{x}}{6^{x}}] d x = \int (\frac{{(2^{x})}^{2}}{2^{x} \cdot 3^{x}} + \frac{{(3^{x})}^{2}}{2^{x} \cdot 3^{x}} + \frac{2 \cdot 2^{x} \cdot 3^{x}}{2^{x} \cdot 3^{x}}) d x \Rightarrow \int [{(\frac{2}{3})}^{x} + {(\frac{3}{2})}^{x} + 2] d x \Rightarrow \frac{{(\frac{2}{3})}^{x}}{\ln (\frac{2}{3})} + \frac{{(\frac{3}{2})}^{x}}{\ln \frac{3}{2}} + 2 x + C (∵ \int a^{x} d x = \frac{a^{x}}{\ln a})$

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Question 7:

$\int \frac{\sin x}{1 + \sin x} d x$

Answer:

$\int \frac{\sin x}{1 + \sin x} d x Rationalising the denominator \Rightarrow \int \frac{\sin x}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} d x \Rightarrow \int (\frac{\sin x - \sin^{2} x}{1 - \sin^{2} x}) d x \Rightarrow \int (\frac{\sin x}{\cos^{2} x} - \tan^{2} x) d x \Rightarrow \int \{\frac{\sin x}{\cos x} \times \frac{1}{\cos x} - (\sec^{2} x - 1)\} d x \Rightarrow \int (\sec x \tan x - \sec^{2} x + 1) d x \Rightarrow \sec x - \tan x + x + C$

Page No 18.197:

Question 8:

$\int \frac{x^{4} + x^{2} - 1}{x^{2} + 1} d x$

Answer:

$\int (\frac{x^{4} + x^{2} - 1}{x^{2} + 1}) d x \Rightarrow \int (\frac{x^{4} + x^{2}}{x^{2} + 1}) d x - \int \frac{1}{x^{2} + 1} d x \Rightarrow \int \frac{x^{2} (x^{2} + 1)}{(x^{2} + 1)} d x - \int \frac{1}{x^{2} + 1} d x \Rightarrow \int x^{2} d x - \int \frac{1}{x^{2} + 1} d x \Rightarrow \frac{x^{3}}{3} - \tan^{- 1} (x) + C (∵ \int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C)$

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Question 9:

$\int \sec^{2} x \cos^{2} 2 x d x$

Answer:

$\int (\sec^{2} x \cdot \cos^{2} 2 x) d x = \int \sec^{2} x \times {(2 \cos^{2} x - 1)}^{2} d x = \int \sec^{2} x [4 \cos^{4} x - 4 \cos^{2} x + 1] d x \Rightarrow \int (4 \cos^{2} x - 4 + \sec^{2} x) d x = 4 \int \cos^{2} x d x + \int \sec^{2} x d x - 4 \int d x \Rightarrow 4 \int (\frac{1 + \cos 2 x}{2}) d x + \int \sec^{2} x - 4 \int d x \Rightarrow 2 [x + \frac{\sin 2 x}{2}] + \tan x - 4 x + C \Rightarrow \sin 2 x + \tan x - 2 x + C$

Page No 18.197:

Question 10:

$\int {cosec}^{2} x \cos^{2} 2 x d x$

Answer:

$\int {cosec}^{2} x \cdot \cos^{2} 2 x d x \Rightarrow \int {cosec}^{2} x {(1 - 2 \sin^{2} x)}^{2} d x \Rightarrow \int {cosec}^{2} x (1 + 4 \sin^{4} x - 4 \sin^{2} x) d x \Rightarrow \int ({cosec}^{2} x + 4 \sin^{2} x - 4) d x \Rightarrow \int {cosec}^{2} x d x + 4 \int (\frac{1 - \cos 2 x}{2}) d x - 4 \int d x \Rightarrow - \cot x + 2 [x - \frac{\sin 2 x}{2}] - 4 x + C \Rightarrow - \cot x + 2 x - \sin 2 x - 4 x + C \Rightarrow - \cot x - \sin 2 x - 2 x + C$

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Question 11:

$\int \sin^{4} 2 x d x$

Answer:

$\int \sin^{4} 2 x d x \Rightarrow \int {(\sin^{2} 2 x)}^{2} d x \Rightarrow \int {[\frac{1 - \cos 4 x}{2}]}^{2} d x \Rightarrow \frac{1}{4} \int {(1 - \cos 4 x)}^{2} \Rightarrow \frac{1}{4} \int (1 + \cos^{2} 4 x - 2 \cos 4 x) d x \Rightarrow \frac{1}{4} \int [1 + (\frac{1 + \cos 8 x}{2}) - 2 \cos 4 x] d x \Rightarrow \frac{1}{4} \int [\frac{3}{2} + \frac{\cos 8 x}{2} - 2 \cos 4 x] d x \Rightarrow \frac{1}{4} [\frac{3 x}{2} + \frac{\sin 8 x}{16} - \frac{2 \sin 4 x}{4}] + C \Rightarrow \frac{3 x}{8} + \frac{\sin 8 x}{64} - \frac{\sin 4 x}{8} + C$

Page No 18.197:

Question 12:

$\int \cos^{3} 3 x d x$

Answer:

$\int \cos^{3} (3 x) d x As we know that \cos 3 A = 4 \cos^{3} A - 3 \cos A \Rightarrow \frac{\cos 3 A + 3 \cos A}{4} = \cos^{3} A \Rightarrow \int [\frac{\cos 9 x + 3 \cos 3 x}{4}] d x \Rightarrow \frac{1}{4} \int \cos 9 x d x + \frac{3}{4} \int \cos 3 x d x \Rightarrow \frac{1}{4} [\frac{\sin 9 x}{9}] + \frac{3}{4} [\frac{\sin 3 x}{3}] + C \Rightarrow \frac{\sin 9 x}{36} + \frac{\sin 3 x}{4} + C \Rightarrow \frac{1}{36} [3 \sin 3 x - 4 \sin^{3} 3 x] + \frac{\sin 3 x}{4} + C (∵ \sin 3 x = 3 \sin x - 4 \sin^{3} x) \Rightarrow \frac{\sin 3 x}{12} + \frac{\sin 3 x}{4} - \frac{1}{9} \sin^{3} 3 x + C \Rightarrow \frac{\sin 3 x}{3} - \frac{\sin^{3} 3 x}{9} + C$

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Question 13:

$\int \frac{\sin 2 x}{a^{2} + b^{2} \sin^{2} x}$

Answer:

$Let I = \int \frac{\sin 2 x}{a^{2} + b^{2} \sin^{2} x} d x Putting a + b^{2} \sin^{2} x = t \Rightarrow b^{2} (2 \sin x \cos x) d x = d t \Rightarrow b^{2} \times \sin 2 x d x = d t ∴ I = \frac{1}{b^{2}} \int \frac{d t}{t} = \frac{1}{b^{2}} \ln |t| + C = \frac{1}{b^{2}} \ln |a^{2} + b^{2} \sin^{2} x| + C (∵ t = a + b^{2} \sin^{2} x)$

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Question 14:

$\int \frac{1}{(\sin^{- 1} x) \sqrt{1 - x^{2}}} d x$

Answer:

$Let I = \int \frac{1}{\sin^{- 1} x \cdot \sqrt{1 - x^{2}}} d x Putting \sin^{- 1} x = t \Rightarrow \frac{d x}{\sqrt{1 - x^{2}}} = d t ∴ I = \int \frac{d t}{t} = \ln |t| + C = \ln |\sin^{- 1} x| + C (∵ t = \sin^{- 1} x)$

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Question 15:

$\int \frac{{(\sin^{- 1} x)}^{3}}{\sqrt{1 - x^{2}}} d x$

Answer:

$Let I = \int \frac{{(\sin^{- 1} x)}^{3}}{\sqrt{1 - x^{2}}} d x Putting \sin^{- 1} x = t \Rightarrow \frac{d x}{\sqrt{1 - x^{2}}} = d t ∴ I = \int t^{3} \cdot d t = \frac{t^{4}}{4} + C = \frac{{(\sin^{- 1} x)}^{4}}{4} + C (∵ t = \sin^{- 1} x)$

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Question 16:

$\int \frac{1}{e^{x} + 1} d x$

Answer:

$\int \frac{1}{e^{x} + 1} d x . . . (1)$
Multiplying numerator and Denominator of eq (1) by e^x

$\Rightarrow \int \frac{e^{x} \cdot d x}{e^{x} (e^{x} + 1)} Putting e^{x} = t \Rightarrow e^{x} d x = d t \Rightarrow \int \frac{d t}{t (t + 1)} ∴ \frac{1}{t (t + 1)} = \frac{A}{t} + \frac{B}{t + 1} \frac{1}{t (t + 1)} = \frac{A (t + 1) + B t}{t (t + 1)} . . . (2) \Rightarrow 1 = A (t + 1) + B t Putting t + 1 = 0 or, t = - 1 in eq (2) we get, \Rightarrow 1 = A \times 0 + B (- 1) \Rightarrow B = - 1 Now, putting t = 0 in eq (2) we get, \Rightarrow 1 = A (0 + 1) + B \times 0 \Rightarrow A = 1 Putting the values of A and B in eq (2) we get, \frac{1}{t (t + 1)} = \frac{1}{t} - \frac{1}{t + 1} ∴ \int \frac{d t}{t (t + 1)} = \int \frac{d t}{t} - \int \frac{d t}{t + 1} = \ln |t| - \ln |t + 1| + C = \ln |\frac{t}{t + 1}| + C = \ln |\frac{e^{x}}{e^{x} + 1}| + C = \ln e^{x} - \ln |e^{x} + 1| + C = x - \ln |e^{x} + 1| + C$

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Question 17:

$\int \frac{e^{x} - 1}{e^{x} + 1} d x$

Answer:

$We have, I = \int \frac{e^{x} - 1}{e^{x} + 1} d x = \int \frac{2 e^{x} - (e^{x} + 1)}{e^{x} + 1} d x = \int \frac{2 e^{x}}{e^{x} + 1} d x - \int d x Putting e^{x} + 1 = t \Rightarrow e^{x} d x = d t ∴ I = \int \frac{2}{t} d t - \int d x = 2 \log |t| - x + C = 2 \log |e^{x} + 1| - x + C = 2 \log (e^{x} + 1) - x + C$

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Question 18:

$\int \frac{1}{e^{x} + e^{- x}} d x$

Answer:

$Let I = \int \frac{1}{e^{x} + e^{- x}} d x = \int \frac{d x}{e^{x} + \frac{1}{e^{x}}} = \int \frac{e^{x} d x}{e^{2 x} + 1} = \int \frac{e^{x} d x}{{(e^{x})}^{2} + 1} Putting e^{x} = t \Rightarrow e^{x} d x = d t ∴ I = \int \frac{d t}{t^{2} + 1} = \tan^{- 1} t + C (∵ \int \frac{d t}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C) = \tan^{- 1} e^{x} + C (∵ t = e^{x})$

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Question 19:

$\int \frac{\cos^{7} x}{\sin x} d x$

Answer:

$Let I = \int \frac{\cos^{7} x}{\sin x} d x = \int \frac{\cos^{6} x \cdot \cos x d x}{\sin x} = \int \frac{{(\cos^{2} x)}^{3} \cdot \cos x}{\sin x} d x = \int \frac{{(1 - \sin^{2} x)}^{3} \cdot \cos x}{\sin x} d x Let \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \frac{{(1 - t^{2})}^{3}}{t} d t = \int (\frac{1 - t^{6} - 3 t^{2} + 3 t^{4}}{t}) d t = \int (\frac{1}{t} - t^{5} - 3 t + 3 t^{3}) d t = \ln |t| - \frac{t^{6}}{6} - \frac{3 t^{2}}{2} + \frac{3 t^{4}}{4} + C = \ln |\sin x| - \frac{\sin^{6} x}{6} - \frac{3 \sin^{2} x}{2} + \frac{3}{4} \sin^{4} x + C (∵ t = \sin x)$

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Question 20:

$\int \sin x \sin 2 x \sin 3 x d x$

Answer:

$\int \sin x \cdot \sin 2 x \cdot \sin 3 x d x = \frac{1}{2} \int (2 \sin 2 x \cdot \sin x) \sin 3 x d x = \frac{1}{2} \int [\cos (2 x - x) - \cos (2 x + x)] \sin 3 x d x [∵ 2 \sin A \sin B = \cos (A - B) - \cos (A + B)] \Rightarrow = \frac{1}{2} \int [\cos x - \cos 3 x] \sin 3 x d x = \frac{1}{2} \int \sin 3 x \cdot \cos x d x - \frac{1}{2} \int \sin 3 x \cdot \cos 3 x d x = \frac{1}{4} \int 2 \sin 3 x \cdot \cos x d x - \frac{1}{4} \int 2 \sin 3 x \cdot \cos 3 x d x = \frac{1}{4} \int [\sin 4 x + \sin 2 x] d x - \frac{1}{4} \int \sin 6 x d x [∵ 2 \sin A \cos B = \sin (A + B) + \sin (A - B)] = \frac{1}{4} [\frac{- \cos 4 x}{4} - \frac{\cos 2 x}{2}] - \frac{1}{4} [- \frac{\cos 6 x}{6}] + C = - \frac{\cos 4 x}{16} - \frac{\cos 2 x}{8} + \frac{\cos 6 x}{24} + C$

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Question 21:

$\int \cos x \cos 2 x \cos 3 x d x$

Answer:

$\int \cos x \cdot \cos 2 x \cdot \cos 3 x d x \Rightarrow \frac{1}{2} \int [2 \cos 2 x \cdot \cos x] \cos 3 x d x \Rightarrow \frac{1}{2} \int [\cos (2 x + x) + \cos (2 x - x)] \cos 3 x d x [∵ 2 \cos A \cos B = \cos (A + B) + \cos (A - B)] \Rightarrow \frac{1}{2} \int (\cos 3 x + \cos x) \cos 3 x d x \Rightarrow \frac{1}{2} \int \cos^{2} 3 x d x + \frac{1}{2} \int \cos 3 x \cdot \cos x d x \Rightarrow \frac{1}{2} \int (\frac{1 + \cos 6 x}{2}) d x + \frac{1}{4} \int 2 \cos 3 x \cdot \cos x d x [∵ \cos 2 x = \cos^{2} x - 1] \Rightarrow \frac{1}{4} [x + \frac{\sin 6 x}{6}] + \frac{1}{4} \int (\cos 4 x + \cos 2 x) d x \Rightarrow \frac{1}{4} [x + \frac{\sin 6 x}{6}] + \frac{1}{4} [\frac{\sin 4 x}{4} + \frac{\sin 2 x}{2}] + C \Rightarrow \frac{x}{4} + \frac{\sin 6 x}{24} + \frac{\sin 4 x}{16} + \frac{\sin 2 x}{8} + C$

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Question 22:

$\int \frac{\sin x + \cos x}{\sqrt{\sin 2 x}} d x$

Answer:

$Let I = \int (\frac{\sin x + \cos x}{\sqrt{\sin 2 x}}) d x Putting \sin x - \cos x = t \Rightarrow (\cos x + \sin x) d x = d t Also {(\sin x - \cos x)}^{2} = t^{2} \Rightarrow \sin^{2} x + \cos^{2} x - 2 \sin x \cos x = t^{2} \Rightarrow 1 - t^{2} = \sin (2 x) ∴ I = \int \frac{d t}{\sqrt{1 - t^{2}}} = \sin^{- 1} t + C (\int \frac{d t}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a} + C) = \sin^{- 1} (\sin x - \cos x) + C (∵ t = \sin x - \cos x)$

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Question 23:

$\int \frac{\sin x - \cos x}{\sqrt{\sin 2 x}} d x$

Answer:

$Let I = \int (\frac{\sin x - \cos x}{\sqrt{\sin 2 x}}) d x Putting \sin x + \cos x = t \Rightarrow (\cos x - \sin x) d x = d t \Rightarrow (\sin x - \cos x) d x = - d t Also \sin x + \cos x = t Squaring both sides, {(\sin x + \cos x)}^{2} = t^{2} \Rightarrow \sin^{2} x + \cos^{2} x + 2 \sin x \cos x = t^{2} \Rightarrow 1 + \sin 2 x = t^{2} \Rightarrow \sin 2 x = t^{2} - 1 ∴ I = \int \frac{- d t}{\sqrt{t^{2} - 1}} = - \ln |t + \sqrt{t^{2} - 1}| + C (∵ \int \frac{d t}{\sqrt{x^{2} - a^{2}}} = \ln |x + \sqrt{x^{2} - a^{2}}| + C) = - \ln |(\sin x + \cos x) + \sqrt{{(\sin x + \cos x)}^{2} - 1}| + C (∵ t = \sin x + \cos x) = - \ln |(\sin x + \cos x) + \sqrt{\sin^{2} x + \cos^{2} x + 2 \sin \cos x - 1}| + C = - \ln |\sin x + \cos x + \sqrt{\sin 2 x}| + C$

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Question 24:

$\int \frac{1}{\sin (x - a) \sin (x - b)} d x$

Answer:

$\int \frac{1}{\sin (x - a) \cdot \sin (x - b)} d x = \frac{1}{\sin (b - a)} \int \frac{\sin (b - a)}{\sin (x - a) \cdot \sin (x - b)} d x = \frac{1}{\sin (b - a)} \int \frac{\sin [(x - a) - (x - b)]}{\sin (x - a) \cdot \sin (x - b)} d x = \frac{1}{\sin (b - a)} \int \frac{\sin (x - a) \cdot \cos (x - b) - \cos (x - a) \sin (x - b)}{\sin (x - a) \cdot \sin (x - b)} d x = \frac{1}{\sin (b - a)} \int [\frac{\sin (x - a) \cdot \cos (x - b)}{\sin (x - a) \cdot \sin (x - b)} - \frac{\cos (x - a) \sin (x - b)}{\sin (x - a) \sin (x - b)}] d x = \frac{1}{\sin (b - a)} \int [\cot (x - b) - \cot (x - a)] d x = \frac{1}{\sin (b - a)} \int \cot (x - b) d x - \int \cot (x - a) d x = \frac{1}{\sin (b - a)} [\ln |\sin (x - b)| - \ln |\sin (x - a)|] + C = \frac{1}{\sin (b - a)} [\ln |\frac{\sin (x - b)}{\sin (x - a)}|] + C = \frac{- 1}{\sin (a - b)} [\ln |\frac{\sin (x - b)}{\sin (x - a)}|] + C = \frac{1}{\sin (a - b)} \ln |\frac{\sin (x - a)}{\sin (x - b)}| + C$

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Question 25:

$\int \frac{1}{\cos (x - a) \cos (x - b)} d x$

Answer:

$\int \frac{1}{\cos (x - a) \cdot \cos (x - b)} d x = \frac{1}{\sin (a - b)} \int \frac{\sin (a - b)}{\cos (x - a) \cdot \cos (x - b)} d x = \frac{1}{\sin (a - b)} \int \frac{\sin [(x - b) - (x - a)]}{\cos (x - a) \cdot \cos (x - b)} d x = \frac{1}{\sin (a - b)} \int \frac{\sin (x - b) \cdot \cos (x - a) - \cos (x - b) \cdot \sin (x - a)}{\cos (x - a) \cdot \cos (x - b)} = \frac{1}{\sin (a - b)} \int [\frac{\sin (x - b) \cdot \cos (x - a)}{\cos (x - a) \cdot \cos (x - b)} - \frac{\cos (x - b) \cdot \sin (x - a)}{\cos (x - a) \cdot \cos (x - b)}] d x = \frac{1}{\sin (a - b)} \int [\tan (x - b) - \tan (x - a)] d x = \frac{1}{\sin (a - b)} \int \tan (x - b) d x - \int \tan (x - a) d x = \frac{1}{\sin (a - b)} [\ln |\sec (x - b)| - \ln |\sec (x - a)|] + C = \frac{1}{\sin (a - b)} [\ln |\cos (x - a)| - \ln |\cos (x - b)|] + C = \frac{1}{\sin (a - b)} [\ln |\frac{\cos (x - a)}{\cos (x - b)}|] + C$

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Question 26:

$\int \frac{\sin x}{\sqrt{1 + \sin x}} d x$

Answer:

$We have, I = \int \frac{\sin x}{\sqrt{1 + \sin x}} d x I = \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sqrt{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}} d x I = \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sqrt{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}}} d x I = \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} d x I = \int \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2} - 1}{\sin x + \cos x} d x I = \int \frac{\sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} - 1}{\sin x + \cos x} d x I = \int \frac{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2} - 1}{\sin \frac{x}{2} + \cos \frac{x}{2}} d x I = \int \frac{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} d x - \int \frac{1}{\sin \frac{x}{2} + \cos \frac{x}{2}} d x I = \int (\sin \frac{x}{2} + \cos \frac{x}{2}) d x - \int \frac{1}{\sin \frac{x}{2} + \cos \frac{x}{2}} d x I = 2 (- \cos \frac{x}{2} + \sin \frac{x}{2}) + C_{1} - \frac{1}{\sqrt{2}} \int \frac{1}{\frac{1}{\sqrt{2}} (\sin \frac{x}{2} + \cos \frac{x}{2})} d x I = 2 (- \cos \frac{x}{2} + \sin \frac{x}{2}) + C_{1} - \frac{1}{\sqrt{2}} \int \frac{1}{\sin \frac{x}{2} c o s \frac{π}{4} + \cos \frac{x}{2} s i n \frac{π}{4}} d x I = 2 (- \cos \frac{x}{2} + \sin \frac{x}{2}) + C_{1} - \frac{1}{\sqrt{2}} \int \frac{1}{\sin (\frac{x}{2} + \frac{π}{4})} d x I = 2 (- \cos \frac{x}{2} + \sin \frac{x}{2}) + C_{1} - \frac{1}{\sqrt{2}} \int cosec (\frac{x}{2} + \frac{π}{4}) d x I = 2 (- \cos \frac{x}{2} + \sin \frac{x}{2}) - \sqrt{2} \log |\tan (\frac{x}{4} + \frac{π}{8})| + C$

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Question 27:

$\int \frac{\sin x}{\cos 2 x} d x$

Answer:

$Let I = \int \frac{\sin x}{\cos 2 x} d x = \int (\frac{\sin x}{2 \cos^{2} x - 1}) d x [∵ \cos 2 x = 2 \cos^{2} x - 1] Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = \int \frac{- d t}{2 t^{2} - 1} = \frac{1}{2} \int \frac{- d t}{t^{2} - \frac{1}{2}} = \frac{- 1}{2} \int \frac{d t}{t^{2} - {(\frac{1}{\sqrt{2}})}^{2}} = \frac{- 1}{2} \times \frac{1}{2 \times \frac{1}{\sqrt{2}}} \ln |\frac{t - \frac{1}{\sqrt{2}}}{t + \frac{1}{\sqrt{2}}}| + C [∵ \int \frac{1}{x^{2} - a^{2}} = \frac{1}{2 a} \ln |\frac{x - a}{x + a}| + C] = - \frac{1}{2 \sqrt{2}} \ln |\frac{\sqrt{2} t - 1}{\sqrt{2} t + 1}| + C = - \frac{1}{2 \sqrt{2}} \ln |\frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1}| + C [∵ t = \cos x] = \frac{1}{2 \sqrt{2}} \ln |\frac{\sqrt{2} \cos x + 1}{\sqrt{2} \cos x - 1}| + C$

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Question 28:

$\int \tan^{3} x d x$

Answer:

$Let I = \int \tan^{3} x d x = \int \tan x \cdot \tan^{2} x d x = \int \tan x (\sec^{2} x - 1) d x = \int \tan x \cdot \sec^{2} x d x - \int \tan x d x Putting \tan x = t in the Ist integral \Rightarrow \sec^{2} x d x = d t ∴ I = \int t \cdot d t - \int \tan x d x = \frac{t^{2}}{2} - \ln |\sec x| + C = \frac{\tan^{2} x}{2} - \ln |\sec x| + C [∵ t = \tan x]$

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Question 29:

$\int \tan^{4} x d x$

Answer:

$L e t I = \int \tan^{4} x d x = \int \tan^{2} x \cdot \tan^{2} x d x = \int (\sec^{2} x - 1) \tan^{2} x d x = \int \sec^{2} x \cdot \tan^{2} x d x - \int \tan^{2} x d x = \int \tan^{2} x \cdot \sec^{2} x - \int (\sec^{2} x - 1) d x Putting \tan x = t in the Ist integral \Rightarrow \sec^{2} x d x = d t ∴ I = \int t^{2} \cdot d t - \int (\sec^{2} x - 1) d x = \frac{t^{3}}{3} - \tan x + x + C = \frac{\tan^{3} x}{3} - \tan x + x + C (∵ t = \tan x)$

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Question 30:

$\int \tan^{5} x d x$

Answer:

$Let I = \int \tan^{5} x d x = \int \tan^{3} x \cdot \tan^{2} x d x = \int \tan^{3} x (\sec^{2} x - 1) d x = \int \tan^{3} x \cdot \sec^{2} x d x - \int \tan^{3} x d x = \int \tan^{3} x \cdot \sec^{2} x d x - \int \tan x \cdot \tan^{2} x d x = \int \tan^{3} x \cdot \sec^{2} x d x - \int \tan x \cdot (\sec^{2} x - 1) d x = \int \tan^{3} x \cdot \sec^{2} x d x - \int \tan x \cdot \sec^{2} x d x + \int \tan x d x Putting \tan x = t in the Ist and IInd integral . \Rightarrow \sec^{2} x d x = d t ∴ I = \int t^{3} \cdot d t - \int t \cdot d t + \int \tan x d x = \frac{t^{4}}{4} - \frac{t^{2}}{2} + \ln |\sec x| + C = \frac{\tan^{4} x}{4} - \frac{\tan^{2} x}{2} + \ln |\sec x| + C [∵ t = \tan x]$

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Question 31:

$\int \cot^{4} x d x$

Answer:

$Let I = \int \cot^{4} x d x = \int \cot^{2} x \cdot \cot^{2} x d x = \int \cot^{2} x \cdot ({cosec}^{2} x - 1) d x = \int \cot^{2} x \cdot {cosec}^{2} x d x - \int \cot^{2} x d x = \int \cot^{2} x \cdot {cosec}^{2} x d x - \int ({cosec}^{2} x - 1) d x Putting \cot x = t in the Ist integral \Rightarrow - {cosec}^{2} x d x = d t ∴ I = - \int t^{2} d t - \int ({cosec}^{2} x - 1) d x = \frac{- t^{3}}{3} + \cot x + x + C = \frac{- \cot^{3} x}{3} + \cot x + x + C [∵ t = \cot x]$

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Question 32:

$\int \cot^{5} x d x$

Answer:

$Let I = \int \cot^{5} x d x = \int \cot^{2} x \cdot \cot^{3} x d x = \int ({cosec}^{2} x - 1) \cot^{3} x d x = \int \cot^{3} x \cdot {cosec}^{2} x d x - \int \cot^{3} x d x = \int \cot^{3} x \cdot {cosec}^{2} x d x - \int \cot x \cdot \cot^{2} x dx = \int \cot^{3} x \cdot {cosec}^{2} x d x - \int \cot x ({cosec}^{2} x - 1) d x = \int \cot^{3} x \cdot {cosec}^{2} x d x - \int \cot x \cdot {cosec}^{2} x d x + \int \cot x d x Putting \cot x = t in the Ist and IInd integral \Rightarrow - {cosec}^{2} x d x = d t \Rightarrow {cosec}^{2} x d x = - d t ∴ I = - \int t^{3} d t + \int t \cdot d t + \int \cot x d x = - \frac{t^{4}}{4} + \frac{t^{2}}{2} + \ln |\sin x| + C = - \frac{\cot^{4} x}{4} + \frac{\cot^{2} x}{2} + \ln |\sin x| + C (∵ t = \cot x)$

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Question 33:

$\int \frac{x^{2}}{{(x - 1)}^{3}} d x$

Answer:

$\int \frac{x^{2}}{{(x - 1)}^{3}} d x = \int [\frac{x^{2} - 1 + 1}{{(x - 1)}^{3}}] d x = \int [\frac{(x - 1) (x + 1)}{{(x - 1)}^{3}} + \frac{1}{{(x - 1)}^{3}}] d x = \int [\frac{x + 1}{{(x - 1)}^{2}} + \frac{1}{{(x - 1)}^{3}}] d x = \int [\frac{x - 1 + 2}{{(x - 1)}^{2}} + \frac{1}{{(x - 1)}^{3}}] d x = \int [\frac{1}{(x - 1)} + \frac{2}{{(x - 1)}^{2}} + \frac{1}{{(x - 1)}^{3}}] d x = \int \frac{1}{x - 1} d x + 2 \int {(x - 1)}^{- 2} d x + \int {(x - 1)}^{- 3} d x = \ln |x - 1| + 2 [\frac{{(x - 1)}^{- 2 + 1}}{- 2 + 1}] + [\frac{{(x - 1)}^{- 3 + 1}}{- 3 + 1}] + C = \ln |x - 1| - \frac{2}{(x - 1)} - \frac{{(x - 1)}^{- 2}}{2} + C = \ln |x - 1| - \frac{2}{x - 1} - \frac{1}{2 {(x - 1)}^{2}} + C$

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Question 34:

$\int x \sqrt{2 x + 3} d x$

Answer:

$Let I = \int x \sqrt{2 x + 3} d x Putting 2 x + 3 = t \Rightarrow x = \frac{t - 3}{2} \Rightarrow 2 d x = d t \Rightarrow d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int (\frac{t - 3}{2}) \sqrt{t} d t = \frac{1}{4} \int (t - 3) \sqrt{t} d t = \frac{1}{4} \int (t^{\frac{3}{2}} - 3 t^{\frac{1}{2}}) d t = \frac{1}{4} [\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} - 3 \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{1}{4} \times \frac{2}{5} t^{\frac{5}{2}} - \frac{3}{4} \times \frac{2}{3} t^{\frac{3}{2}} + C = \frac{1}{10} t^{\frac{5}{2}} - 2 t^{\frac{3}{2}} + C = \frac{1}{10} {(2 x + 3)}^{\frac{5}{2}} - \frac{1}{2} {(2 x + 3)}^{\frac{3}{2}} + C [∵ t = 2 x + 3] = \frac{1}{10} {(2 x + 3)}^{\frac{5}{2}} - \frac{1}{2} {(2 x + 3)}^{\frac{3}{2}} + C$

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Question 35:

$\int \frac{x^{3}}{{(1 + x^{2})}^{2}} d x$

Answer:

$Let I = \int \frac{x^{3}}{{(1 + x^{2})}^{2}} d x = \int \frac{x^{2} \times x}{{(1 + x^{2})}^{2}} d x Putting 1 + x^{2} = t \Rightarrow x^{2} = t - 1 \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{(t - 1)}{t^{2}} d t = \frac{1}{2} \int (\frac{1}{t} - \frac{1}{t^{2}}) d t = \frac{1}{2} \int \frac{d t}{t} - \frac{1}{2} \int t^{- 2} d t = \frac{1}{2} \ln |t| - \frac{1}{2} [\frac{t^{- 2 + 1}}{- 2 + 1}] + C = \frac{1}{2} \ln |t| + \frac{1}{2 t} + C = \frac{1}{2} \ln |1 + x^{2}| + \frac{1}{2 (1 + x^{2})} + C (∵ t = 1 + x^{2})$

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Question 36:

$\int x \sin^{5} x^{2} \cos x^{2} d x$

Answer:

$Let I = \int x \cdot \sin^{5} x^{2} \cdot \cos x^{2} d x Putting \sin x^{2} = t \Rightarrow \cos (x^{2}) \times 2 x d x = d t \Rightarrow \cos (x^{2}) \cdot x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int t^{5} \cdot d t = \frac{1}{2} [\frac{t^{6}}{6}] + C = \frac{t^{6}}{12} + C = \frac{\sin^{6} x^{2}}{12} + C [∵ t = \sin x^{2}]$

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Question 37:

$\int \sin^{3} x \cos^{4} x d x$

Answer:

$Let I = \int \sin^{3} x \cdot \cos^{4} x d x = \int \sin^{2} x \cdot \sin x \cdot \cos^{4} x d x = \int (1 - \cos^{2} x) \cdot \cos^{4} x \cdot \sin x d x = \int (\cos^{4} x - \cos^{6} x) \cdot \sin x d x Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = - \int (t^{4} - t^{6}) d t = \int (t^{6} - t^{4}) d t = \frac{t^{7}}{7} - \frac{t^{5}}{5} + C = \frac{\cos^{7} x}{7} - \frac{\cos^{5} x}{5} + C [∵ t = \cos x]$

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Question 38:

$\int \sin^{5} x d x$

Answer:

$Let I = \int \sin^{5} x d x = \int \sin^{4} x \cdot \sin x d x = \int {(\sin^{2} x)}^{2} \sin x d x = \int {(1 - \cos^{2} x)}^{2} \sin x d x = \int (\cos^{4} x - 2 \cos^{2} x + 1) \sin x d x Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = - \int (t^{4} - 2 t^{2} + 1) d t = - \int t^{4} d t + 2 \int t^{2} d t - \int d t = \frac{- t^{5}}{5} + \frac{2 t^{3}}{3} - t + C = \frac{- \cos^{5} x}{5} + \frac{2}{3} \cos^{3} x - \cos x + C [∵ t = \cos x]$

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Question 39:

$\int \cos^{5} x d x$

Answer:

$Let I = \int \cos^{5} x d x = \int \cos^{4} x \cdot \cos x d x = \int {(\cos^{2} x)}^{2} \cos x d x = \int {(1 - \sin^{2} x)}^{2} \cos x d x Putting \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int {(1 - t^{2})}^{2} \cdot d t = \int (t^{4} - 2 t^{2} + 1) d t = \int t^{4} \cdot d t - 2 \int t^{2} d t + \int d t = \frac{t^{5}}{5} - 2 \times \frac{t^{2 + 1}}{2 + 1} + t + C = \frac{t^{5}}{5} - \frac{2}{3} t^{3} + t + C = \frac{\sin^{5} x}{5} - \frac{2}{3} \sin^{3} x + \sin x + C [∵ t = \sin x]$

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Question 40:

$\int \sqrt{\sin x} \cos^{3} x d x$

Answer:

$Let I = \int \sqrt{\sin x} \cdot \cos^{3} x d x = \int \sqrt{\sin x} \cdot (\cos^{2} x) \cdot \cos x d x = \int \sqrt{\sin x} (1 - \sin^{2} x) \cdot \cos x d x Putting \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \sqrt{t} (1 - t^{2}) \cdot d t = \int t^{\frac{1}{2}} d t - \int t^{\frac{1}{2}} \cdot t^{2} d t = \int t^{\frac{1}{2}} d t - \int t^{\frac{5}{2}} d t = \frac{t^{\frac{3}{2}}}{\frac{3}{2}} - \frac{t^{\frac{7}{2}}}{\frac{7}{2}} + C = \frac{2}{3} t^{\frac{3}{2}} - \frac{2}{7} t^{\frac{7}{2}} + C = \frac{2}{3} \sin^{\frac{3}{2}} x - \frac{2}{7} \sin^{\frac{7}{2}} x + C [∵ t = \sin x]$

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Question 41:

$\int \frac{\sin 2 x}{\sin^{4} x + \cos^{4} x} d x$

Answer:

$Let I = \int \frac{\sin 2 x}{\sin^{4} x + \cos^{4} x} d x = \int \frac{2 \sin x \cdot \cos x d x}{\sin^{4} x + \cos^{4} x} Dividing numerator and denominator by \cos^{4} x \Rightarrow \int \frac{2 \frac{\sin x \cdot \cos x}{\cos^{4} x} d x}{1 + \tan^{4} x} \Rightarrow \int \frac{2 \tan x \cdot \sec^{2} x d x}{1 + {(\tan^{2} x)}^{2}} Putting \tan^{2} x = t \Rightarrow 2 \tan x \cdot \sec^{2} x d x ∴ I = \int \frac{d t}{1 + t^{2}} = \tan^{- 1} t + C = \tan^{- 1} (\tan^{2} x) + C [∵ t = \tan^{2} x]$

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Question 42:

$\int \frac{1}{\sqrt{x^{2} - a^{2}}} d x$

Answer:

$Let I = \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x Putting x = a \sec θ \Rightarrow d x = a \sec θ \tan θ d θ ∴ I = \int \frac{a \sec θ \tan θ d θ}{\sqrt{a^{2} \sec^{2} θ - a^{2}}} = \int \frac{a \sec θ \cdot \tan θ d θ}{a \cdot \tan θ} = \int \sec θ d θ = \ln |\sec θ + \tan θ| + C = \ln |\sec θ + \sqrt{\sec^{2} θ - 1}| + C = \ln |\frac{x}{a} + \sqrt{{(\frac{x}{a})}^{2} - 1}| + C = \ln |\frac{x + \sqrt{x^{2} - a^{2}}}{a}| + C = \ln |x + \sqrt{x^{2} - a^{2}}| - \ln a + C = \ln |x + \sqrt{x^{2} - a^{2}}| + C' where C' = C - \ln a$

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Question 43:

$\int \frac{1}{\sqrt{x^{2} + a^{2}}} d x$

Answer:

$Let I = \int \frac{d x}{\sqrt{x^{2} - a^{2}}} Putting x = a \tan θ \Rightarrow d x = a \sec^{2} θ d θ ∴ I = \int \frac{a \cdot s e c^{2} θ d θ}{\sqrt{a^{2} \tan^{2} θ + a^{2}}} = \int \frac{a \sec^{2} θ \cdot d θ}{a \sqrt{1 + \tan^{2} θ}} = \int \frac{\sec^{2} θ \cdot d θ}{\sec θ} = \int \sec θ \cdot d θ = \int \sec θ \cdot d θ = \ln |\sec θ + \tan θ| + C = \ln |\sec θ + \sqrt{\sec^{2} θ - 1}| + C = \ln |\frac{x}{a} + \sqrt{\frac{x^{2}}{a^{2}} - 1}| + C = \ln |x + \sqrt{x^{2} - a^{2}}| - \ln a + C = \ln |x + \sqrt{x^{2} - a^{2}}| + C' where C' = C - \ln a$

Page No 18.198:

Question 44:

$\int \frac{1}{4 x^{2} + 4 x + 5} d x$

Answer:

$Let I = \int \frac{d x}{4 x^{2} + 4 x + 1 + 4} = \int \frac{d x}{{(2 x)}^{2} + 2 \times 2 x + 1 + 22} = \int \frac{d x}{{(2 x + 1)}^{2} + 2^{2}} Putting (2 x + 1) = t \Rightarrow 2 d x = d t \Rightarrow d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{t^{2} + 2^{2}} = \frac{1}{2} \times \frac{1}{2} \tan^{- 1} (\frac{t}{2}) + C = \frac{1}{4} \tan^{- 1} (\frac{2 x + 1}{2}) + C [∵ t = (2 x + 1)]$

Page No 18.198:

Question 45:

$\int \frac{1}{x^{2} + 4 x - 5} d x$

Answer:

$\int \frac{1}{x^{2} + 4 x - 5} d x = \int \frac{1}{x^{2} + 4 x + 4 - 4 - 5} d x = \int \frac{1}{x^{2} + 4 x + 4 - 3^{2}} d x = \int \frac{1}{{(x + 2)}^{2} - 3^{2}} d x = \frac{1}{2 \times 3} \ln |\frac{x + 2 - 3}{x + 2 + 3}| + C [∵ \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \ln |\frac{x - a}{x + a}| + C] = \frac{1}{6} \ln |\frac{x - 1}{x + 5}| + C$

Page No 18.198:

Question 46:

$\int \frac{1}{1 - x - 4 x^{2}} d x$

Answer:

$We have, I = \int \frac{1}{1 - x - 4 x^{2}} d x = \frac{1}{4} \int \frac{1}{\frac{1}{4} - - x^{2} \frac{x}{4}} d x = \frac{1}{4} \int \frac{1}{\frac{1}{4} - (x^{2} + \frac{x}{4})} d x = \frac{1}{4} \int \frac{1}{\frac{1}{4} - \{x^{2} + + {(\frac{1}{8})}^{2} - {(\frac{1}{8})}^{2} \frac{x}{4}\}} d x = \frac{1}{4} \int \frac{1}{\frac{1}{4} - {(x + \frac{1}{8})}^{2} + \frac{1}{64}} d x = \frac{1}{4} \int \frac{1}{\frac{1}{4} + - {(x + \frac{1}{8})}^{2} \frac{1}{64}} d x = \frac{1}{4} \int \frac{1}{\frac{16 + 1}{64} - {(x + \frac{1}{8})}^{2}} d x = \frac{1}{4} \int \frac{1}{{(\frac{\sqrt{17}}{8})}^{2} - {(x + \frac{1}{8})}^{2}} d x = \frac{1}{4} \times \frac{1}{2 \times \frac{\sqrt{17}}{8}} \ln |\frac{\frac{\sqrt{17}}{8} + x + \frac{1}{8}}{\frac{\sqrt{17}}{8} - x - \frac{1}{8}}| + C [∵ \int \frac{1}{a^{2} - x^{2}} dx = \frac{1}{2 a} \ln |\frac{a + x}{a - x}| + C] = \frac{1}{\sqrt{17}} \ln |\frac{\frac{\sqrt{17} + 1}{8} + x}{\frac{\sqrt{17} - 1}{8} - x}| + C$

Page No 18.198:

Question 47:

$\int \frac{1}{3 x^{2} + 13 x - 10} d x$

Answer:

$\int \frac{1}{3 x^{2} + 13 x - 10} d x = \frac{1}{3} \int \frac{1}{x^{2} + \frac{13}{3} x - \frac{10}{3}} d x = \frac{1}{3} \int \frac{1}{x^{2} + \frac{13 x}{3} + {(\frac{13}{6})}^{2} - {(\frac{13}{6})}^{2} - \frac{10}{3}} d x = \frac{1}{3} \int \frac{1}{{(x + \frac{13}{6})}^{2} - \frac{169}{36} - \frac{10}{3}} d x = \frac{1}{3} \int \frac{1}{{(x + \frac{13}{6})}^{2} - \frac{169 - 120}{36}} d x = \frac{1}{3} \int \frac{1}{{(x + \frac{13}{6})}^{2} - {(\frac{17}{6})}^{2}} d x = \frac{1}{3} \times \frac{1}{2 \times \frac{17}{6}} \ln |\frac{x + \frac{13}{6} - \frac{17}{6}}{x + \frac{13}{6} + \frac{17}{6}}| [∵ \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \ln |\frac{x - a}{x + a}| + C] = \frac{1}{17} \ln |\frac{x - \frac{2}{3}}{x + 5}| + C = \frac{1}{17} \ln |\frac{3 x - 2}{3 x + 15}| + C$

Page No 18.198:

Question 48:

$\int \frac{\sin x}{\sqrt{\cos^{2} x - 2 \cos x - 3}} d x$

Answer:

$Let I = \int \frac{\sin x}{\sqrt{\cos^{2} x - 2 \cos x - 3}} d x Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = - \int \frac{d t}{\sqrt{t^{2} - 2 t - 3}} = - \int \frac{d t}{\sqrt{t^{2} - 2 t + 1 - 4}} = - \int \frac{d t}{\sqrt{{(t - 1)}^{2} - {(2)}^{2}}} = - \ln |t - 1 + \sqrt{{(t - 1)}^{2} - 4}| + C [∵ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \ln |x + \sqrt{x^{2} - a^{2}}| + C] = - \ln |(\cos x - 1) + \sqrt{\cos^{2} x - 2 \cos x - 3}| + C [∵ t = \cos x]$

Page No 18.198:

Question 49:

$\int \sqrt{cosec x - 1} d x$

Answer:

$Let I = \int \sqrt{cosec x - 1} d x = \int \sqrt{\frac{1}{\sin x} - 1} d x = \int \frac{\sqrt{1 - \sin x}}{\sqrt{\sin x}} d x = \int \frac{\sqrt{(1 - \sin x) (1 + \sin x)}}{\sqrt{\sin x (1 + \sin x)}} d x = \int \frac{\sqrt{1 - \sin^{2} x}}{\sqrt{\sin^{2} x + \sin x}} d x = \int \frac{\cos x}{\sqrt{\sin^{2} x + \sin x}} d x Putting \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \frac{d t}{\sqrt{t^{2} + t}} = \int \frac{d t}{\sqrt{t^{2} + t + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} = \int \frac{d t}{\sqrt{{(t + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} = \ln |t + \frac{1}{2} + \sqrt{{(t + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}| + C [∵ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \ln |x + \sqrt{x^{2} - a^{2}}| + C] = \ln |t + \frac{1}{2} + \sqrt{t^{2} + t}| + C = \ln |(\sin x + \frac{1}{2}) + \sqrt{\sin^{2} x + \sin x}| + C [∵ t = \sin x]$

Page No 18.198:

Question 50:

$\int \frac{1}{\sqrt{3 - 2 x - x^{2}}} d x$

Answer:

$Let I = \int \frac{1}{\sqrt{3 - 2 x - x^{2}}} d x = \int \frac{1}{\sqrt{3 - (x^{2} + 2 x + 1 - 1)}} d x = \int \frac{1}{\sqrt{4 - {(x + 1)}^{2}}} d x Putting (x + 1) = t \Rightarrow d x = d t ∴ I = \int \frac{d t}{\sqrt{2^{2} - t^{2}}} = \sin^{- 1} (\frac{t}{2}) + C [∵ \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \sin^{- 1} \frac{x}{a} + C] = \sin^{- 1} (\frac{x + 1}{2}) + C [∵ t = (x + 1)]$

Page No 18.198:

Question 51:

$\int \frac{x + 1}{x^{2} + 4 x + 5} d x$

Answer:

$Let I = \int \frac{(x + 1)}{x^{2} + 4 x + 5} d x & let (x + 1) = A \frac{d}{d x} (x^{2} + 4 x + 5) + B \Rightarrow x + 1 = A (2 x + 4) + B \Rightarrow x + 1 = (2 A) x + 4 A + B Equating the coefficients of like terms 2 A = 1 \Rightarrow A = \frac{1}{2} & 4 A + B = 1 \Rightarrow 4 \times \frac{1}{2} + B = 1 \Rightarrow B = - 1 ∴ (x + 1) = \frac{1}{2} (2 x + 4) - 1 ∴ I = \int [\frac{\frac{1}{2} (2 x + 4) - 1}{x^{2} + 4 x + 5}] d x = \frac{1}{2} \int \frac{(2 x + 4)}{x^{2} + 4 x + 5} d x - \int \frac{1}{x^{2} + 4 x + 5} d x Putting x^{2} + 4 x + 5 = t \Rightarrow (2 x + 4) d x = d t ∴ I = \frac{1}{2} \int \frac{1}{t} d t - \int \frac{1}{x^{2} + 4 x + 4 + 1} d x = \frac{1}{2} \int \frac{d t}{t} - \int \frac{1}{{(x + 2)}^{2} + 1^{2}} d x = \frac{1}{2} \ln |t| - \tan^{- 1} (\frac{x + 2}{1}) + C [∵ \int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C] = \frac{1}{2} \ln |x^{2} + 4 x + 5| - \tan^{- 1} (x + 2) + C [∵ t = x^{2} + 4 x + 5]$

Page No 18.198:

Question 52:

$\int \frac{5 x + 7}{\sqrt{(x - 5) (x - 4)}} d x$

Answer:

$We have, I = \int (\frac{5 x + 7}{\sqrt{(x - 5) (x - 4)}}) d x = \int (\frac{5 x + 7}{\sqrt{x^{2} - 9 x + 20}}) d x Let 5 x + 7 = A \frac{d}{d x} (x^{2} - 9 x + 20) + B \Rightarrow 5 x + 7 = A (2 x - 9) + B Equating Coefficients of like terms 2 A = 5 \Rightarrow A = \frac{5}{2} And - 9 A + B = 7 \Rightarrow - 9 \times \frac{5}{2} + B = 7 \Rightarrow B = 7 + \frac{45}{2} \Rightarrow B = \frac{59}{2} ∴ I = \int (\frac{\frac{5}{2} (2 x - 9) + \frac{59}{2}}{\sqrt{x^{2} - 9 x + 20}}) d x = \frac{5}{2} \int \frac{(2 x - 9) d x}{\sqrt{x^{2} - 9 x + 20}} + \frac{59}{2} \int \frac{d x}{\sqrt{x^{2} - 9 x + 20}} Putting x^{2} - 9 x + 20 = t \Rightarrow (2 x - 9) d x = d t I = \frac{5}{2} \int \frac{d t}{\sqrt{t}} + \frac{59}{2} \int \frac{d x}{\sqrt{x^{2} - 9 x + {(\frac{9}{2})}^{2} - {(\frac{9}{2})}^{2} + 20}} = \frac{5}{2} \int t^{- \frac{1}{2}} d t + \frac{59}{2} \int \frac{d x}{\sqrt{{(x - \frac{9}{2})}^{2} - \frac{81 + 80}{4}}} = \frac{5}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + \frac{59}{2} \int \frac{d x}{\sqrt{{(x - \frac{9}{2})}^{2} - {(\frac{1}{2})}^{2}}} = \frac{5}{2} \times 2 \sqrt{t} + \frac{59}{2} \log |(x - \frac{9}{2}) + \sqrt{{(x - \frac{9}{2})}^{2} - {(\frac{1}{2})}^{2}}| + C = 5 \sqrt{t} + \frac{59}{2} \log |(x - \frac{9}{2}) + \sqrt{x^{2} - 9 x + 20}| + C = 5 \sqrt{x^{2} - 9 x + 20} + \frac{59}{2} \log |(x - \frac{9}{2}) + \sqrt{x^{2} - 9 x + 20}| + C$

Page No 18.198:

Question 53:

$\int \sqrt{\frac{1 + x}{x}} d x$

Answer:

$Let I = \int \sqrt{\frac{1 + x}{x}} d x = \int \frac{\sqrt{1 + x}}{\sqrt{x}} \times \frac{\sqrt{1 + x}}{\sqrt{1 + x}} d x = \int (\frac{1 + x}{\sqrt{x^{2} + x}}) d x Let x + 1 = A \frac{d}{d x} (x^{2} + x) + B \Rightarrow x + 1 = A (2 x + 1) + B \Rightarrow x + 1 = (2 A) x + A + B Equating the coefficients of like terms 2 A = 1 \Rightarrow A = \frac{1}{2} & A + B = 1 \Rightarrow \frac{1}{2} + B = 1 ∴ B = \frac{1}{2} ∴ I = \int \frac{(x + 1)}{\sqrt{x^{2} + x}} d x = \int (\frac{\frac{1}{2} (2 x + 1) + \frac{1}{2}}{\sqrt{x^{2} + x}}) d x = \frac{1}{2} \int \frac{(2 x + 1)}{\sqrt{x^{2} + x}} d x + \frac{1}{2} \int \frac{1}{\sqrt{x^{2} + x}} d x$

$Putting x^{2} + x = t \Rightarrow (2 x + 1) d x = d t ∴ I = \frac{1}{2} \int \frac{1}{\sqrt{t}} d t + \frac{1}{2} \int \frac{1}{\sqrt{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} d x = \frac{1}{2} \int \frac{1}{\sqrt{t}} d t + \frac{1}{2} \int \frac{1}{\sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} d x = \frac{1}{2} \int t^{- \frac{1}{2}} d t + \frac{1}{2} \int \frac{1}{\sqrt{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} d x = \frac{1}{2} \times 2 \sqrt{t} + \frac{1}{2} \ln |x + \frac{1}{2} + \sqrt{{(x + \frac{1}{2})}^{2} - \frac{1}{4}}| + C [∵ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \ln |x + \sqrt{x^{2} - a^{2}}| + C] = \sqrt{t} + \frac{1}{2} \ln |x + \frac{1}{2} + \sqrt{x^{2} + x}| + C = \sqrt{x^{2} + x} + \frac{1}{2} \ln |(x + \frac{1}{2}) + \sqrt{x^{2} + x}| + C [∵ t = x^{2} + x]$

Page No 18.198:

Question 54:

$\int \sqrt{\frac{1 - x}{x}} d x$

Answer:

$Let I = \int \frac{\sqrt{1 - x}}{\sqrt{x}} d x = \int (\frac{\sqrt{1 - x} \cdot \sqrt{1 - x}}{\sqrt{x} \cdot \sqrt{1 - x}}) d x = \int \frac{(1 - x)}{\sqrt{x - x^{2}}} d x Let (1 - x) = A \frac{d}{d x} (x - x^{2}) + B \Rightarrow 1 - x = A (1 - 2 x) + B \Rightarrow 1 - x = - (2 A) x + A + B Equating coefficients of like terms - 2 A = - 1 \Rightarrow A = \frac{1}{2} & A + B = 1 \Rightarrow \frac{1}{2} + B = 1 ∴ B = \frac{1}{2} ∴ I = \int \frac{\frac{1}{2} (1 - 2 x) + \frac{1}{2}}{\sqrt{x - x^{2}}} d x = \frac{1}{2} \int \frac{(1 - 2 x)}{\sqrt{x - x^{2}}} d x + \frac{1}{2} \int \frac{1}{\sqrt{x - x^{2} + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} d x = \frac{1}{2} \int \frac{(1 - 2 x)}{\sqrt{x - x^{2}}} d x + \frac{1}{2} \int \frac{1}{\sqrt{{(\frac{1}{2})}^{2} - (x^{2} - x + \frac{1}{2^{2}})}} d x = \frac{1}{2} \int \frac{(1 - 2 x)}{\sqrt{x - x^{2}}} d x + \frac{1}{2} \int \frac{1}{\sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{1}{2})}^{2}}} d x$

$Putting x - x^{2} = t in the first integral \Rightarrow (1 - 2 x) d x = d t ∴ I = \frac{1}{2} \int \frac{1}{\sqrt{t}} d t + \frac{1}{2} \int \frac{1}{\sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{1}{2})}^{2}}} d x = \frac{1}{2} \int t^{- \frac{1}{2}} d t + \frac{1}{2} \int \frac{d x}{\sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{1}{2})}^{2}}} = \frac{1}{2} \times 2 t^{\frac{1}{2}} + \frac{1}{2} \times \sin^{- 1} (\frac{x - \frac{1}{2}}{\frac{1}{2}}) + C [∵ \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \sin^{- 1} \frac{x}{a} + C] = \sqrt{t} + \frac{1}{2} \sin^{- 1} (2 x - 1) + C = \sqrt{x - x^{2}} + \frac{1}{2} \sin^{- 1} (2 x - 1) + C [∵ t = x - x^{2}]$

Page No 18.198:

Question 55:

$\int \frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{a x}} d x$

Answer:

$We have, I = \int \frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{a x}} d x I = \frac{1}{\sqrt{a}} \int \frac{1 + a - 1 - \sqrt{a x}}{1 - \sqrt{a x}} d x I = \frac{1}{\sqrt{a}} \int \frac{1 - \sqrt{a x}}{1 - \sqrt{a x}} d x + \frac{1}{\sqrt{a}} \int \frac{a - 1}{1 - \sqrt{a x}} d x I = \frac{1}{\sqrt{a}} \int d x + \frac{a - 1}{\sqrt{a}} \int \frac{1}{1 - \sqrt{a x}} d x I = \frac{1}{\sqrt{a}} x + \frac{a - 1}{\sqrt{a}} \int \frac{1}{1 - \sqrt{a x}} d x Let, I_{1} = \int \frac{1}{1 - \sqrt{a x}} d x Put a x = z^{2} \Rightarrow a d x = 2 z d z I_{1} = \frac{1}{a} \int \frac{2 z}{1 - z} d z I_{1} = \frac{1}{a} \int \frac{2 z - 2 + 2}{1 - z} d z I_{1} = \frac{1}{a} \int \frac{2 z - 2}{1 - z} d z + \frac{1}{a} \int \frac{2}{1 - z} d z I_{1} = \frac{- 2}{a} \int \frac{1 - z}{1 - z} d z + \frac{1}{a} \int \frac{2}{1 - z} d z I_{1} = \frac{- 2}{a} \int d z + \frac{1}{a} \int \frac{2}{1 - z} d z I_{1} = \frac{- 2}{a} z - \frac{2}{a} \log |1 - z| + C_{1} I_{1} = \frac{- 2 \sqrt{a x}}{a} - \frac{2}{a} \log |1 - \sqrt{a x}| + C_{1} I = \frac{1}{\sqrt{a}} x + \frac{a - 1}{\sqrt{a}} (\frac{- 2 \sqrt{a x}}{a} - \frac{2}{a} \log |1 - \sqrt{a x}|) + C$

Note: The answer in indefinite integration may vary depending on the integral constant.

Page No 18.198:

Question 56:

$\int \frac{1}{(\sin x - 2 \cos x) (2 \sin x + \cos x)} d x$

Answer:

$Let I = \int \frac{1}{(\sin x - 2 \cos x) (2 \sin x + \cos x)} d x$
Dividing numerator and denominator by cos²x we get ,
$I = \int \frac{\frac{1}{\cos^{2} x}}{(\tan x - 2) (2 \tan x + 1)} d x = \int \frac{\sec^{2} x}{(\tan x - 2) (2 \tan x + 1)} d x Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{1}{(t - 2) (2 t + 1)} d t = \int \frac{1}{2 t^{2} + t - 4 t - 2} d t = \int \frac{1}{2 t^{2} - 3 t - 2} d t = \frac{1}{2} \int \frac{1}{t^{2} - \frac{3 t}{2} - 1} d t = \frac{1}{2} \int \frac{1}{t^{2} - \frac{3}{2} t + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{2} - 1} d t = \frac{1}{2} \int \frac{1}{{(t - \frac{3}{4})}^{2} - \frac{9}{16} - 1} d t = \frac{1}{2} \int \frac{1}{{(t - \frac{3}{4})}^{2} - {(\frac{5}{4})}^{2}} d t = \frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \ln |\frac{t - \frac{3}{4} - \frac{5}{4}}{t - \frac{3}{4} + \frac{5}{4}}| + C [∵ \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \ln |\frac{x - a}{x + a}| + C] = \frac{1}{5} \ln |\frac{t - 2}{t + \frac{1}{2}}| + C = \frac{1}{5} \ln |\frac{2 (t - 2)}{2 t + 1}| + C = \frac{1}{5} \ln |\frac{2 (\tan x - 2)}{2 \tan x + 1}| + C [∵ t = \tan x] = \frac{1}{5} \ln |\frac{\tan x - 2}{2 \tan x + 1}| + \frac{1}{5} \ln 2 + C = \frac{1}{5} \ln |\frac{\tan x - 2}{2 \tan x + 1}| + C' where C' = C + \frac{1}{5} \ln 2$

Page No 18.198:

Question 57:

$\int \frac{1}{4 \sin^{2} x + 4 \sin x \cos x + 5 \cos^{2} x} d x$

Answer:

$Let I = \int \frac{1}{4 \sin^{2} x + 4 \sin x \cdot \cos x + 5 \cos^{2} x} d x$

Dividing numerator and denominator by cos²x we get
$I = \int \frac{\sec^{2} x}{4 \tan^{2} x + 4 \tan x + 5} d x Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{4 t^{2} + 4 t + 5} = \frac{1}{4} \int \frac{d t}{t^{2} + t + \frac{5}{4}} = \frac{1}{4} \int \frac{d t}{t^{2} + t + \frac{1}{4} - \frac{1}{4} + \frac{5}{4}} = \frac{1}{4} \int \frac{d t}{{(t + \frac{1}{2})}^{2} + 1^{2}} = \frac{1}{4} \times \tan^{- 1} (t + \frac{1}{2}) + C [∵ \int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C] = \frac{1}{4} \tan^{- 1} (\frac{2 t + 1}{2}) + C = \frac{1}{4} \tan^{- 1} (\frac{2 \tan x + 1}{2}) + C [∵ t = \tan x] = \frac{1}{4} \tan^{- 1} (\tan x + \frac{1}{2}) + C$

Page No 18.198:

Question 58:

$\int \frac{1}{a + b \tan x} d x$

Answer:

$Let I = \int \frac{1}{a + b \tan x} d x = \int \frac{1}{a + b \frac{\sin x}{\cos x}} d x = \int \frac{\cos x \cdot}{a \cos x + b \sin x} d x Let \cos x = A \frac{d}{d x} (a \cos x + b \sin x) + B (a \cos x + b \sin x) \Rightarrow \cos x = A (- a \sin x + b \cos x) + B (a \cos x + b \sin x) 1 \cdot \cos x = (A b + B \cdot a) \cos x + \sin x (- A \cdot a + B \cdot b) Equating coefficients of like terms A \cdot b + B \cdot a = 1 . . . (1) - A \cdot a + B \cdot b = 0 . . . (2) Multiplying equation (1) by a and eq (2) by b and then adding them A \cdot a b + B \cdot a^{2} = a - A \cdot a \cdot b + B b^{2} = 0 \Rightarrow B = \frac{a}{a^{2} + b^{2}} Substituting the value of B in eq (1) \Rightarrow A \cdot b + \frac{a^{2}}{a^{2} + b^{2}} = 1 \Rightarrow A \cdot b = 1 - \frac{a^{2}}{a^{2} + b^{2}} \Rightarrow A = \frac{b}{a^{2} + b^{2}} ∴ I = \frac{b}{a^{2} + b^{2}} \int (\frac{- a \sin x + b \cos x}{a \cos x + b \sin x}) d x + \frac{a}{a^{2} + b^{2}} \int (\frac{a \cos x + b \sin x}{a \cos x + b \sin x}) d x = \frac{b}{a^{2} + b^{2}} \int (\frac{- a \sin x + b \cos x}{a \cos x + b \sin x}) d x + \frac{a}{a^{2} + b^{2}} \int d x Putting a \cos x + b \sin x = t in the Ist integral \Rightarrow (- a \sin x + b \cos x) d x = d t ∴ I = \frac{b}{a^{2} + b^{2}} \int \frac{d t}{t} + \frac{a}{a^{2} + b^{2}} \int d x = \frac{b}{a^{2} + b^{2}} \ln |t| + \frac{a x}{a^{2} + b^{2}} + C = \frac{b}{a^{2} + b^{2}} \ln |a \cos x + b \sin x| + \frac{a x}{a^{2} + b^{2}} + C [∵ t = a \cos x + b \sin x]$

Page No 18.198:

Question 59:

$\int \frac{1}{\sin^{2} x + \sin 2 x} d x$

Answer:

$L e t I = \int \frac{1}{\sin^{2} x + \sin 2 x} d x = \int \frac{1}{\sin^{2} x + 2 \sin x \cdot \cos x} d x$

Dividing numerator and denominator by cos²x, we get
$I = \int \frac{\frac{1}{\cos^{2} x}}{\tan^{2} x + 2 \tan x} d x = \int \frac{\sec^{2} x}{\tan^{2} x + 2 \tan x} d x Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{1}{t^{2} + 2 t} d t = \int \frac{1}{t^{2} + 2 t + 1 - 1} d t = \int \frac{1}{{(t + 1)}^{2} - 1^{2}} d t = \frac{1}{2} \ln |\frac{t + 1 - 1}{t + 1 + 1}| + C = \frac{1}{2} \ln |\frac{t}{t + 2}| + C = \frac{1}{2} \ln |\frac{\tan x}{\tan x + 2}| + C [∵ t = \tan x]$

Page No 18.198:

Question 60:

$\int \frac{\sin x + 2 \cos x}{2 \sin x + \cos x} d x$

Answer:

$\int (\frac{\sin x + 2 \cos x}{2 \sin x + \cos x}) d x Let \sin x + 2 \cos x = A \frac{d}{d x} (2 \sin x + \cos x) + B (2 \sin x + \cos x) \Rightarrow \sin x + 2 \cos x = A (2 \cos x - \sin x) + 2 B \sin x + B \cos x \Rightarrow \sin x + 2 \cos x = (2 A + B) \cos x + (2 B - A) \sin x Equating coefficients of like terms \Rightarrow 2 A + B = 2 . . . (1) \Rightarrow - A + 2 B = 1 . . . (2) Multiplying eq (2) by 2 and adding it to eq (1) we get, 5 B = 4 \Rightarrow B = \frac{4}{5} Putting B = \frac{4}{5} in eq (1) we get, 2 A + \frac{4}{5} = 2 \Rightarrow A = \frac{3}{5} ∴ \int (\frac{\sin x + 2 \cos x}{2 \sin x + \cos x}) d x = \int [\frac{\frac{3}{5} (2 \cos x - \sin x)}{2 \sin x + \cos x}] d x + \frac{4}{5} \int \frac{(2 \sin x + \cos x)}{(2 \sin x + \cos x)} d x = \frac{3}{5} \int (\frac{2 \cos x - \sin x}{2 \sin x + \cos x}) d x + \frac{4}{5} \int d x Putting 2 \sin x + \cos x = t \Rightarrow (2 \cos x - \sin x) d x = d t ∴ I = \frac{3}{5} \int \frac{d t}{t} + \frac{4}{5} \int d x = \frac{3}{5} \ln |t| + \frac{4 x}{5} + C = \frac{3}{5} \ln |2 \sin x + \cos x| + \frac{4 x}{5} + C [∵ t = 2 \sin x + \cos x]$

Page No 18.198:

Question 61:

$\int \frac{x^{3}}{\sqrt{x^{8} + 4}} d x$

Answer:

$Let I = \int \frac{x^{3}}{\sqrt{x^{8} + 2^{2}}} d x = \int \frac{x^{3}}{\sqrt{{(x^{4})}^{2} + 2^{2}}} d x Putting x^{4} = t \Rightarrow 4 x^{3} d x = d t \Rightarrow x^{3} \cdot d x = \frac{d t}{4} ∴ I = \frac{1}{4} \int \frac{1}{\sqrt{t^{2} + 2^{2}}} d t = \frac{1}{4} \ln |t + \sqrt{t^{2} + 4}| + C = \frac{1}{4} \ln |x^{4} + \sqrt{x^{8} + 4}| + C [∵ t = x^{4}]$

Page No 18.198:

Question 62:

$\int \frac{1}{2 - 3 \cos 2 x} d x$

Answer:

$Let I = \int \frac{1}{2 - 3 \cdot \cos 2 x} d x = \int \frac{1}{2 - 3 (2 \cos^{2} x - 1)} d x = \int \frac{1}{2 - 6 \cos^{2} x + 3} d x = \int \frac{1}{5 - 6 \cos^{2} x} d x$
Dividing numerator and denominator by cos²x, we get
$I = \int \frac{\sec^{2} x}{5 \sec^{2} x - 6} d x = \int \frac{\sec^{2} x}{5 (1 + \tan^{2} x) - 6} d x = \int \frac{\sec^{2} x}{5 \tan^{2} x - 1} d x Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{1}{5 t^{2} - 1} d t = \frac{1}{5} \int \frac{1}{t^{2} - {(\frac{1}{\sqrt{5}})}^{2}} d t = \frac{1}{5} \times \frac{1}{2 \times \frac{1}{\sqrt{5}}} \ln |\frac{t - \frac{1}{\sqrt{5}}}{t + \frac{1}{\sqrt{5}}}| + C [∵ \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \ln |\frac{x - a}{x + a}| + C] = \frac{1}{2 \sqrt{5}} \ln |\frac{\sqrt{5} t - 1}{\sqrt{5} t + 1}| + C = \frac{1}{2 \sqrt{5}} \ln |\frac{\sqrt{5} \tan x - 1}{\sqrt{5} \tan x + 1}| + C [∵ t = \tan x]$

Page No 18.198:

Question 63:

$\int \frac{\cos x}{\frac{1}{4} - \cos^{2} x} d x$

Answer:

$Let I = \int \frac{\cos x}{\frac{1}{4} - \cos^{2} x} d x = \int \frac{\cos x}{\frac{1}{4} - (1 - \sin^{2} x)} d x = \int \frac{\cos x}{\sin^{2} x - \frac{3}{4}} d x = \int \frac{\cos x}{\sin^{2} x - {(\frac{\sqrt{3}}{2})}^{2}} d x Putting \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \frac{1}{t^{2} - {(\frac{\sqrt{3}}{2})}^{2}} d t = \frac{1}{2 \times \frac{\sqrt{3}}{2}} \ln |\frac{t - \frac{\sqrt{3}}{2}}{t + \frac{\sqrt{3}}{2}}| + C = \frac{1}{\sqrt{3}} \ln |\frac{2 t - \sqrt{3}}{2 t + \sqrt{3}}| + C = \frac{1}{\sqrt{3}} \ln |\frac{2 \sin x - \sqrt{3}}{2 \sin x + \sqrt{3}}| + C [∵ t = \sin x]$

Page No 18.198:

Question 64:

$\int \frac{1}{1 + 2 \cos x} d x$

Answer:

$Let I = \int \frac{1}{1 + 2 \cos x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} ∴ I = \int \frac{1}{1 + 2 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{1 + \tan^{2} \frac{x}{2} + 2 - 2 \tan^{2} \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{3 - \tan^{2} \frac{x}{2}} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) \cdot d x = 2 d t ∴ I = \int \frac{2}{3 - t^{2}} d t = 2 \int \frac{1}{{(\sqrt{3})}^{2} - t^{2}} d t = 2 \times \frac{1}{2 \sqrt{3}} \ln |\frac{\sqrt{3} + t}{\sqrt{3} + t}| + C [∵ \int \frac{1}{a^{2} - x^{2}} d x = \frac{1}{2 a} \ln |\frac{a + x}{a - x}| + C] = \frac{1}{\sqrt{3}} \ln |\frac{\sqrt{3} + \tan \frac{x}{2}}{\sqrt{3} - \tan \frac{x}{2}}| + C [∵ t = \tan \frac{x}{2}]$

Page No 18.198:

Question 65:

$\int \frac{1}{1 - 2 \sin x} d x$

Answer:

$Let I = \int \frac{1}{1 - 2 \sin x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} ∴ I = \int \frac{1}{1 - 2 (\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{1 + \tan^{2} \frac{x}{2} - 4 \tan \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} - 4 \tan \frac{x}{2}} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) \cdot d x = 2 d t ∴ I = \int \frac{2}{t^{2} - 4 t + 1} d t = 2 \int \frac{1}{t^{2} - 4 t + 4 - 4 + 1} d t = 2 \int \frac{1}{{(t - 2)}^{2} - {(\sqrt{3})}^{2}} d t = 2 \times \frac{1}{2 \sqrt{3}} \ln |\frac{t - 2 - \sqrt{3}}{t - 2 + \sqrt{3}}| + C [∵ \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \ln |\frac{x - a}{x + a}| + C] = \frac{1}{\sqrt{3}} \ln |\frac{\tan \frac{x}{2} - 2 - \sqrt{3}}{\tan \frac{x}{2} - 2 + \sqrt{3}}| + C [∵ t = \tan \frac{x}{2}]$

Page No 18.198:

Question 66:

$\int \frac{1}{\sin x (2 + 3 \cos x)} d x$

Answer:

$Let I = \int \frac{1}{\sin x (2 + 3 \cos x)} d x = \int \frac{\sin x}{\sin^{2} x (2 + 3 \cos x)} d x = \int \frac{\sin x}{(1 - \cos^{2} x) (2 + 3 \cos x)} d x = \int \frac{\sin x}{(1 - \cos x) (1 + \cos x) (2 + 3 \cos x)} d x Putting \cos x = t \Rightarrow - \sin x d x = d t ∴ I = \int \frac{- 1}{(1 - t) (1 + t) (2 + 3 t)} d t = \int \frac{1}{(t - 1) (t + 1) (3 t + 2)} d t Let \frac{1}{(t - 1) (t + 1) (3 t + 2)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{3 t + 2} \Rightarrow \frac{1}{(t - 1) (t + 1) (3 t + 2)} = \frac{A (t + 1) (3 t + 2) + B (t - 1) (3 t + 2) + C (t + 1) (t - 1)}{(t - 1) (t + 1) (3 t + 2)} \Rightarrow 1 = A (t + 1) (3 t + 2) + B (t - 1) (3 t + 2) + C (t + 1) (t - 1) Putting t + 1 = 0 or t = - 1 \Rightarrow 1 = A \times 0 + B (- 1 - 1) (3 \times - 1 + 2) + C \times 0 ∴ B = \frac{1}{2} Now, putting t - 1 = 0 or t = 1 \Rightarrow 1 = A (1 + 1) (3 + 2) + B \times 0 + C \times 0 ∴ A = \frac{1}{10} Now, putting 3 t + 2 = 0 or t = \frac{- 2}{3} \Rightarrow 1 = A \times 0 + B \times 0 + C (- \frac{2}{3} + 1) (- \frac{2}{3} - 1) \Rightarrow 1 = C (\frac{1}{3}) (\frac{- 5}{3}) ∴ C = \frac{- 9}{5} ∴ I = \int \frac{1}{10 (t - 1)} d t + \frac{1}{2} \int \frac{1}{t + 1} d t - \frac{9}{5} \int \frac{1}{3 t + 2} d t = \frac{1}{10} \ln |t - 1| + \frac{1}{2} \ln |t + 1| - \frac{9}{5} \ln \frac{|3 t + 2|}{3} + C = \frac{1}{10} \ln |t - 1| + \frac{1}{2} \log |t + 1| - \frac{3}{5} \ln |3 t + 2| + C = \frac{1}{10} + \ln |\cos x - 1| + \frac{1}{2} \ln |\cos x + 1| - \frac{3}{5} \ln |3 \cos x + 2| + C [∵ t = \cos x]$

Page No 18.198:

Question 67:

$\int \frac{1}{\sin x + \sin 2 x} d x$

Answer:

$Let I = \int \frac{1}{\sin x + \sin 2 x} d x = \int \frac{1}{\sin x + 2 \sin x \cos x} d x = \int \frac{1}{\sin x (1 + 2 \cos x)} d x = \int \frac{\sin x}{\sin^{2} x (1 + 2 \cos x)} d x = \int \frac{\sin x}{(1 - \cos^{2} x) (1 + 2 \cos x)} d x = \int \frac{\sin x d x}{(1 - \cos x) (1 + \cos x) (1 + 2 \cos x)} Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t$

$∴ I = - \int \frac{1}{(1 - t) (1 + t) (1 + 2 t)} d t = \int \frac{1}{(t - 1) (t + 1) (2 t + 1)} d t ∴ \frac{1}{(t - 1) (t + 1) (2 t + 1)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{C}{2 t + 1} \Rightarrow 1 = A (t + 1) (2 t + 1) + B (t - 1) (2 t + 1) + C (t - 1) (t + 1) Putting t + 1 = 0 or t = - 1 \Rightarrow 1 = A \times 0 + B (- 1 - 1) (- 2 + 1) + C \times 0 \Rightarrow 1 = B (2) ∴ B = \frac{1}{2} Now, putting t - 1 = 0 or t = 1 \Rightarrow 1 = A (2) (3) + B \times 0 + C \times 0 ∴ A = \frac{1}{6} Now, putting 2 t + 1 = 0 or t = - \frac{1}{2} \Rightarrow 1 = A \times 0 + B \times 0 + C (- \frac{1}{2} - 1) (- \frac{1}{2} + 1) \Rightarrow 1 = C (- \frac{3}{2}) (\frac{1}{2}) ∴ C = - \frac{4}{3} ∴ I = \frac{1}{6} \int \frac{1}{t - 1} d t + \frac{1}{2} \int \frac{1}{t + 1} d t - \frac{4}{3} \int \frac{1}{2 t + 1} d t = \frac{1}{6} \ln |t - 1| + \frac{1}{2} \log |t + 1| - \frac{4}{3} \ln \frac{|2 t + 1|}{2} + C = \frac{1}{6} \ln |t - 1| + \frac{1}{2} \ln |t + 1| - \frac{2}{3} \ln |2 t + 1| + C = \frac{1}{6} \ln |\cos x - 1| + \frac{1}{2} \ln |\cos x + 1| - \frac{2}{3} \ln |2 \cos x + 1| + C [∵ t = \cos x]$

Page No 18.198:

Question 68:

$\int \frac{1}{\sin^{4} x + \cos^{4} x} d x$

Answer:

$We have, I = \int \frac{d x}{\sin^{4} x + \cos^{4} x}$
Dividing numerator and denominator by cos⁴x
$I = \int \frac{\sec^{4} x d x}{\tan^{4} x + 1} = \int \frac{\sec^{2} x \sec^{2} x d x}{\tan^{4} x + 1} = \int \frac{(1 + \tan^{2} x) \sec^{2} x d x}{\tan^{4} x + 1} Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{(1 + t^{2}) d t}{t^{4} + 1} = \int \frac{(\frac{1}{t^{2}} + 1) d t}{t^{2} + \frac{1}{t^{2}}} = \int \frac{(1 + \frac{1}{t^{2}})}{{(t - \frac{1}{t})}^{2} + 2} d t Putting t - \frac{1}{t} = p \Rightarrow (1 + \frac{1}{t^{2}}) d t = d p ∴ I = \int \frac{1}{p^{2} + {(\sqrt{2})}^{2}} d p = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{p}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) + C = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t^{2} - 1}{\sqrt{2} t}) + C = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{\tan^{2} x - 1}{\sqrt{2} \tan x}) + C = \frac{1}{\sqrt{2}} \tan^{- 1} (- \sqrt{2} \times \frac{1 - \tan^{2} x}{2 \tan x}) + C = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{- \sqrt{2}}{\tan 2 x}) + C = \frac{1}{\sqrt{2}} \tan^{- 1} (- \sqrt{2} \cot 2 x) + C$

Page No 18.198:

Question 69:

$\int \frac{1}{5 - 4 \sin x} d x$

Answer:

$Let I = \int \frac{1}{5 - 4 \sin x} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} ∴ I = \int \frac{1}{5 - 4 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) - 8 \tan \frac{x}{2}} d x = \int \frac{\sec^{2} \frac{x}{2}}{5 \tan^{2} \frac{x}{2} - 8 \tan \frac{x}{2} + 5} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t$

$\Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = 2 \int \frac{1}{5 t^{2} - 8 t + 5} d t = \frac{2}{5} \int \frac{1}{t^{2} - \frac{8}{5} t + 1} d t = \frac{2}{5} \int \frac{1}{t^{2} - \frac{8 t}{5} + {(\frac{4}{5})}^{2} - {(\frac{4}{5})}^{2} + 1} d t = \frac{2}{5} \int \frac{1}{{(t - \frac{4}{5})}^{2} - \frac{16}{25} + 1} d t = \frac{2}{5} \int \frac{1}{{(t - \frac{4}{5})}^{2} + {(\frac{3}{5})}^{2}} d t = \frac{2}{5} \times \frac{5}{3} \tan^{- 1} (\frac{t - \frac{4}{5}}{\frac{3}{5}}) + C [∵ \int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C] = \frac{2}{3} \tan^{- 1} (\frac{5 t - 4}{3}) + C = \frac{2}{3} \tan^{- 1} (\frac{5 \tan \frac{x}{2} - 4}{3}) + C [∵ t = \tan \frac{x}{2}]$

Page No 18.198:

Question 70:

$\int \sec^{4} x d x$

Answer:

$Let I = \int \sec^{4} x d x = \int \sec^{2} x \cdot \sec^{2} x d x = \int (1 + \tan^{2} x) \cdot \sec^{2} x d x Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int (1 + t^{2}) d t = \int d t + \int t^{2} d t = t + \frac{t^{3}}{3} + C = \tan x + \frac{1}{3} \tan^{3} x + C [∵ t = \tan x]$

Page No 18.198:

Question 71:

$\int {cosec}^{4} 2 x d x$

Answer:

$Let I = \int {cosec}^{4} 2 x d x = \int {cosec}^{2} 2 x \cdot {cosec}^{2} 2 x d x = \int (1 + \cot^{2} 2 x) \cdot {cosec}^{2} 2 x d x Putting \cot 2 x = t \Rightarrow - {cosec}^{2} (2 x) \cdot 2 d x = d t \Rightarrow {cosec}^{2} (2 x) \cdot d x = \frac{- d t}{2} ∴ I = - \frac{1}{2} \int (1 + t^{2}) \cdot d t = - \frac{1}{2} [t + \frac{t^{3}}{3}] + C = - \frac{1}{2} \cot 2 x + \frac{1}{6} \cot^{3} 2 x + C [∵ t = \cot 2 x]$

Page No 18.198:

Question 72:

$\int \frac{1 + \sin x}{\sin x (1 + \cos x)} d x$

Answer:

$Let I = \int \frac{(1 + \sin x)}{\sin x (1 + \cos x)} d x Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} and \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} ∴ I = \int \frac{(1 + \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})}{\frac{(2 \tan \frac{x}{2})}{(1 + \tan^{2} \frac{x}{2})} (1 + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \int \frac{(1 + \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2}) (1 + \tan^{2} \frac{x}{2})}{(2 \tan \frac{x}{2}) (1 + \tan^{2} \frac{x}{2} + 1 - \tan^{2} \frac{x}{2})} d x = \frac{1}{4} \int \frac{(1 + \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2}) \sec^{2} \frac{x}{2}}{\tan \frac{x}{2}} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \frac{1}{4} \int \frac{(1 + t^{2} + 2 t) \cdot (2 d t)}{t} = \frac{1}{2} \int (\frac{1}{t} + t + 2) d t = \frac{1}{2} [\ln |t| + \frac{t^{2}}{2} + 2 t] + C = \frac{1}{2} [\ln |\tan \frac{x}{2}| + \frac{\tan^{2} (\frac{x}{2})}{2} + 2 \tan (\frac{x}{2})] + C [∵ t = \tan \frac{x}{2}] = \frac{1}{2} \ln |\tan \frac{x}{2}| + \frac{1}{4} \tan^{2} \frac{x}{2} + \tan \frac{x}{2} + C$

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Question 73:

$\int \frac{1}{2 + \cos x} d x$

Answer:

$Let I = \int \frac{1}{2 + \cos x} d x Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} ∴ I = \int \frac{1}{2 + (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \int \frac{(1 + \tan^{2} \frac{x}{2})}{2 (1 + \tan^{2} \frac{x}{2}) + 1 - \tan^{2} (\frac{x}{2})} d x = \int \frac{\sec^{2} (\frac{x}{2})}{2 + 2 \tan^{2} (\frac{x}{2}) + 1 - \tan^{2} (\frac{x}{2})} d x = \frac{\sec^{2} (\frac{x}{2})}{3 + \tan^{2} (\frac{x}{2})} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \int \frac{2}{3 + t^{2}} d t = 2 \int \frac{1}{t^{2} + {(\sqrt{3})}^{2}} d t = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{t}{\sqrt{3}}) + C = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{\tan \frac{x}{2}}{\sqrt{3}}) + C [∵ t = \tan \frac{x}{2}]$

Page No 18.198:

Question 74:

$\int \sqrt{\frac{a + x}{x}} d x$

Answer:

$Let I = \int \sqrt{\frac{a + x}{x}} d x = \int \frac{\sqrt{(a + x) (a + x)}}{\sqrt{x (a + x)}} = \int (\frac{a + x}{\sqrt{x^{2} + a x}}) d x = a \int \frac{1}{\sqrt{x^{2} + a x}} d x + \int \frac{x}{\sqrt{x^{2} + a x}} d x = a \int \frac{1}{\sqrt{x^{2} + a x + {(\frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \int \frac{x}{\sqrt{x^{2} + a x}} d x = a \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} + a x}} d x = a \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int (\frac{2 x + a - a}{\sqrt{x^{2} + a x}}) d x = a \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int \frac{(2 x + a)}{\sqrt{x^{2} + a x}} d x - \frac{a}{2} \int \frac{1}{\sqrt{x^{2} + a x}} d x = \frac{a}{2} \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int \frac{(2 x + a)}{\sqrt{x^{2} + a x}} d x Putting x^{2} + a x = t in the Ist integral \Rightarrow (2 x + a) d x = d t ∴ I = \frac{a}{2} \int \frac{1}{\sqrt{{(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}}} d x + \frac{1}{2} \int \frac{1}{\sqrt{t}} d t = \frac{a}{2} \ln |x + \frac{a}{2} + \sqrt{x^{2} + a x}| + \frac{1}{2} \times 2 \sqrt{t} + C [∵ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \ln |x + \sqrt{x^{2} - a^{2}}| + C] = \frac{a}{2} \ln |x + \frac{a}{2} + \sqrt{x^{2} + a x}| + \sqrt{x^{2} + a x} + C [∵ t = x^{2} + a x]$

Page No 18.198:

Question 75:

$\int \frac{6 x + 5}{\sqrt{6 + x - 2 x^{2}}} d x$

Answer:

$\int \frac{(6 x + 5) d x}{\sqrt{6 + x - 2 x^{2}}} Let 6 x + 5 = A \frac{d}{d x} (6 + x - 2 x^{2}) + B \Rightarrow 6 x + 5 = A (- 4 x + 1) + B \Rightarrow 6 x + 5 = - 4 A x + (A + B) Equating coefficients of like terms - 4 A = 6 \Rightarrow A = - \frac{3}{2} & A + B = 5 \Rightarrow - \frac{3}{2} + B = 5 \Rightarrow B = 5 + \frac{3}{2} \Rightarrow B = \frac{13}{2} Then, 6 x + 5 = - \frac{3}{2} (- 4 x + 1) + \frac{13}{2} ∴ \int \frac{(6 x + 5)}{\sqrt{6 + x - 2 x^{2}}} d x = \int (\frac{\frac{- 3}{2} (- 4 x + 1) + \frac{13}{2}}{\sqrt{6 + x - 2 x^{2}}}) d x = - \frac{3}{2} \int \frac{(- 4 x + 1)}{\sqrt{6 + x - 2 x^{2}}} d x + \frac{13}{2} \int \frac{1}{\sqrt{6 + x - 2 x^{2}}} d x Putting 6 + x - 2 x^{2} = t in the Ist integral \Rightarrow (- 4 x + 1) d x = d t ∴ \int \frac{(6 x + 5)}{\sqrt{6 + x - 2 x^{2}}} d x = - \frac{3}{2} \int \frac{1}{\sqrt{t}} d t + \frac{13}{2 \times \sqrt{2}} \int \frac{1}{\sqrt{3 + \frac{x}{2} - x^{2}}} d x = - \frac{3}{2} \int t^{- \frac{1}{2}} \cdot d t + \frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3 + \frac{x}{2} - x^{2} - {(\frac{1}{4})}^{2} + {(\frac{1}{4})}^{2}}} d x = - \frac{3}{2} \int t^{- \frac{1}{2}} \cdot d t + \frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3 + \frac{1}{16} - {(x - \frac{1}{4})}^{2}}} d x = - \frac{3}{2} \int t^{- \frac{1}{2}} \cdot d t + \frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{{(\frac{7}{4})}^{2} - {(x - \frac{1}{4})}^{2}}} d x = - 3 [t^{\frac{1}{2}}] + \frac{13}{2 \sqrt{2}} \times \sin^{- 1} (\frac{x - \frac{1}{4}}{\frac{7}{4}}) + C [∵ \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \sin^{- 1} \frac{x}{a} + C] = - 3 \sqrt{6 + x - 2 x^{2}} + \frac{13}{2 \sqrt{2}} \times \sin^{- 1} (\frac{x - \frac{1}{4}}{\frac{7}{4}}) + C = - 3 \sqrt{6 + x - 2 x^{2}} + \frac{13}{2 \sqrt{2}} \sin^{- 1} (\frac{4 x - 1}{7}) + C$

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Question 76:

$\int \frac{\sin^{5} x}{\cos^{4} x} d x$

Answer:

$Let I = \int \frac{\sin^{5} x}{\cos^{4} x} d x = \int (\frac{\sin^{4} x . \sin x}{\cos^{4} x}) d x = \int \frac{{(\sin^{2} x)}^{2} \cdot \sin x}{\cos^{4} x} d x = \int (\frac{{(1 - \cos^{2} x)}^{2} . \sin x}{\cos^{4} x}) d x = \int (\frac{1 + \cos^{4} x - 2 \cos^{2} x}{\cos^{4} x}) \sin x d x Putting \cos x = t \Rightarrow - \sin x d x = d t ∴ I = - \int \frac{(1 + t^{4} - 2 t^{2}) d t}{t^{4}} = - \int t^{- 4} d t - \int d t + 2 \int t^{- 2} d t = - [\frac{t^{- 4 + 1}}{- 4 + 1}] - t + 2 [\frac{t^{- 2 + 1}}{- 2 + 1}] + C = \frac{1}{3 t^{3}} - t - \frac{2}{t} + C = \frac{1}{3 \cos^{3} x} - \cos x - \frac{2}{\cos x} + C [∵ t = \cos x]$

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Question 77:

$\int \frac{\cos^{5} x}{\sin x} d x$

Answer:

$Let I = \int \frac{\cos^{5} x d x}{\sin x} = \int \frac{\cos^{4} x \cdot \cos x d x}{\sin x} = \int \frac{{(\cos^{2} x)}^{2} \cdot \cos x d x}{\sin x} = \int \frac{{(1 - \sin^{2} x)}^{2} \cos x d x}{\sin x} = \int (\frac{1 + \sin^{4} x - 2 \sin^{2} x}{\sin x}) \cos x d x Putting \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int (\frac{1 + t^{4} - 2 t^{2}}{t}) d t = \int \frac{d t}{t} + \int t^{3} d t - 2 \int t d t = \ln |t| + \frac{t^{4}}{4} - \frac{2 t^{2}}{2} + C = \ln |t| + \frac{t^{4}}{4} - t^{2} + C = \ln |\sin x| + \frac{1}{4} \sin^{4} x - \sin^{2} x + C [∵ t = \sin x]$

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Question 78:

$\int \frac{\sin^{6} x}{\cos x} d x$

Answer:

$Let I = \int \frac{\sin^{6} x \cdot}{\cos x} d x = \int \frac{\sin^{6} x \cdot \cos x}{\cos^{2} x} d x = \int \frac{\sin^{6} x}{(1 - \sin^{2} x)} \cos x d x Putting \sin x = t \Rightarrow \cos x d x = d t ∴ I = \int \frac{t^{6}}{(1 - t^{2})} d t = \int (\frac{t^{6} - 1 + 1}{1 - t^{2}}) d t = \int \frac{[{(t^{2})}^{3} - 1^{3}]}{1 - t^{2}} d t + \int \frac{1}{1 - t^{2}} d t = \int \frac{(t^{2} - 1) (1 + t^{2} + t^{4})}{(1 - t^{2})} + \int \frac{1}{1 - t^{2}} d t = - \int (t^{4} + t^{2} + 1) d t + \int \frac{1}{1 - t^{2}} d t = - [\frac{t^{5}}{5} + \frac{t^{3}}{3} + t] + \frac{1}{2} \ln |\frac{1 + t}{1 - t}| + C = - \frac{1}{5} \sin^{5} x - \frac{1}{3} \sin^{3} x - \sin x + \frac{1}{2} \ln |\frac{1 + \sin x}{1 - \sin x}| + C [∵ t = \sin x]$

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Question 79:

$\int \frac{\sin^{2} x}{\cos^{6} x} d x$

Answer:

$Let I = \int \frac{\sin^{2} x}{\cos^{6} x} d x = \int \frac{\sin^{2} x}{\cos^{2} x \cdot \cos^{4} x} d x = \int \tan^{2} x \cdot \sec^{4} x d x = \int \tan^{2} x \sec^{2} x \cdot \sec^{2} x d x = \int \tan^{2} x (1 + \tan^{2} x) \sec^{2} x d x Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int t^{2} (1 + t^{2}) d t = \int (t^{2} + t^{4}) d t = \frac{t^{3}}{3} + \frac{t^{5}}{5} + C = \frac{1}{3} \tan^{3} x + \frac{1}{5} \tan^{5} x + C [∵ t = \tan x]$

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Question 80:

$\int \sec^{6} x d x$

Answer:

$Let I = \int \sec^{6} x d x = \int \sec^{4} x \cdot \sec^{2} x d x = \int {(\sec^{2} x)}^{2} \cdot \sec^{2} x d x = \int {(1 + \tan^{2} x)}^{2} \sec^{2} x d x Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int {(1 + t^{2})}^{2} \cdot d t = \int (1 + t^{4} + 2 t^{2}) d t = \int d t + \int t^{4} d t + 2 \int t^{2} d t = t + \frac{t^{5}}{5} + \frac{2 t^{3}}{3} + C = \tan x + \frac{1}{5} \tan^{5} x + \frac{2}{3} \tan^{3} x + C [∵ t = \tan x]$

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Question 81:

$\int \tan^{5} x \sec^{3} x d x$

Answer:

$Let I = \int \tan^{5} x \cdot \sec^{3} x d x = \int \tan^{4} x \cdot \sec^{2} x \cdot \sec x \tan x d x = \int {(\sec^{2} x - 1)}^{2} \cdot \sec^{2} x \cdot \sec x \tan x d x Putting \sec x = t \Rightarrow \sec x \tan x d x = d t ∴ I = \int {(t^{2} - 1)}^{2} \cdot t^{2} \cdot d t = \int (t^{4} - 2 t^{2} + 1) t^{2} d t = \int (t^{6} - 2 t^{4} + t^{2}) d t = \frac{t^{7}}{7} - \frac{2 t^{5}}{5} + \frac{t^{3}}{3} + C = \frac{1}{7} \sec^{7} x - \frac{2}{5} \sec^{5} x + \frac{1}{3} \sec^{3} x + C [∵ t = \sec x]$

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Question 82:

$\int \tan^{3} x \sec^{4} x d x$

Answer:

$Let I = \int \tan^{3} x \cdot \sec^{4} x d x = \int \tan^{3} x \cdot \sec^{2} x \cdot \sec^{2} x d x = \int \tan^{3} x (1 + \tan^{2} x) \cdot \sec^{2} x d x = \int (\tan^{3} x + \tan^{5} x) \sec^{2} x d x Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int (t^{3} + t^{5}) d t = \frac{t^{4}}{4} + \frac{t^{6}}{6} + C = \frac{\tan^{4} x}{4} + \frac{\tan^{6} x}{6} + C [∵ t = \tan x]$

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Question 83:

$\int \frac{1}{\sec x + cosec x} d x$

Answer:

$We have, I = \int \frac{1}{\sec x + cosec x} d x I = \int \frac{1}{\frac{1}{\cos x} + \frac{1}{\sin x}} d x I = \frac{1}{2} \int \frac{2 \sin x \cos x}{\sin x + \cos x} d x I = \frac{1}{2} \int \frac{1 + 2 \sin x \cos x - 1}{\sin x + \cos x} d x I = \frac{1}{2} \int \frac{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x - 1}{\sin x + \cos x} d x I = \frac{1}{2} \int \frac{{(\sin x + \cos x)}^{2} - 1}{\sin x + \cos x} d x I = \frac{1}{2} \int \frac{{(\sin x + \cos x)}^{2}}{\sin x + \cos x} d x - \frac{1}{2} \int \frac{1}{\sin x + \cos x} d x I = \frac{1}{2} \int (\sin x + \cos x) d x - \frac{1}{2} \int \frac{1}{\sin x + \cos x} d x I = \frac{1}{2} (- \cos x + \sin x) + C_{1} - \frac{1}{2 \sqrt{2}} \int \frac{1}{\frac{1}{\sqrt{2}} (\sin x + \cos x)} d x I = \frac{1}{2} (- \cos x + \sin x) + C_{1} - \frac{1}{2 \sqrt{2}} \int \frac{1}{\sin x \cos \frac{π}{4} + \cos x \sin \frac{π}{4}} d x I = \frac{1}{2} (- \cos x + \sin x) + C_{1} - \frac{1}{2 \sqrt{2}} \int \frac{1}{\sin (x + \frac{π}{4})} d x I = \frac{1}{2} (- \cos x + \sin x) + C_{1} - \frac{1}{2 \sqrt{2}} \int cosec (x + \frac{π}{4}) d x I = \frac{1}{2} (- \cos x + \sin x) - \frac{1}{2 \sqrt{2}} \log |\tan (\frac{x}{2} + \frac{π}{8})| + C$

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Question 84:

$\int \sqrt{a^{2} + x^{2}} d x$

Answer:

$Let I = \int \underset{II}{1 \cdot} \underset{I}{\sqrt{a^{2} + x^{2}}} d x = \sqrt{a^{2} + x^{2}} \int 1 d x - \int (\frac{d}{d x} (\sqrt{a^{2} + x^{2}}) \int 1 d x) d x = \sqrt{a^{2} + x^{2}} \cdot x - \int \frac{1 \times 2 x}{2 \sqrt{a^{2} + x^{2}}} \cdot x d x = \sqrt{a^{2} + x^{2}} \cdot x - \int (\frac{x^{2} + a^{2} - a^{2}}{\sqrt{a^{2} + x^{2}}}) d x = x \sqrt{a^{2} + x^{2}} - \int \sqrt{a^{2} + x^{2}} d x + a^{2} \int \frac{1}{\sqrt{a^{2} + x^{2}}} d x = x \sqrt{a^{2} + x^{2}} - I + a^{2} \int \frac{1}{\sqrt{a^{2} + x^{2}}} d x ∴ 2 I = x \sqrt{a^{2} + x^{2}} + a^{2} \ln |x + \sqrt{x^{2} + a^{2}}| \Rightarrow I = \frac{x}{2} \sqrt{a^{2} + x^{2}} + \frac{a^{2}}{2} \ln |x + \sqrt{x^{2} + a^{2}}| + C$

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Question 85:

$\int \sqrt{x^{2} - a^{2}} d x$

Answer:

$Let I = \int \underset{II}{1 \cdot} \underset{I}{\sqrt{x^{2} - a^{2}}} d x = \sqrt{x^{2} - a^{2}} \int 1 d x - \int (\frac{d}{d x} (\sqrt{x^{2} - a^{2}}) \int 1 d x) d x = \sqrt{x^{2} - a^{2}} \cdot x - \int \frac{1 \times 2 x}{2 \sqrt{x^{2} - a^{2}}} \cdot x d x = \sqrt{x^{2} - a^{2}} \cdot x - \int (\frac{x^{2} - a^{2} + a^{2}}{\sqrt{x^{2} - a^{2}}}) d x = \sqrt{x^{2} - a^{2}} \cdot x - \int \sqrt{x^{2} - a^{2}} d x - a^{2} \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = x \sqrt{x^{2} - a^{2}} - I - a^{2} \int \frac{d x}{\sqrt{x^{2} - a^{2}}} ∴ 2 I = x \sqrt{x^{2} - a^{2}} - a^{2} \ln |x + \sqrt{x^{2} - a^{2}}| \Rightarrow I = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \ln |x + \sqrt{x^{2} - a^{2}}| + C$

Page No 18.198:

Question 86:

$\int \sqrt{a^{2} - x^{2}} d x$

Answer:

$Let I = \int \sqrt{a^{2} - x^{2}} d x = \int \underset{II}{1 \cdot} \underset{I}{\sqrt{a^{2} - x^{2}}} d x = \underset{}{\sqrt{a^{2} - x^{2}}} \int 1 d x - \int (\frac{d}{d x} (\sqrt{a^{2} - x^{2}}) \int 1 d x) d x = \sqrt{a^{2} - x^{2}} \cdot x + \int \frac{1 \times 2 x}{2 \sqrt{a^{2} - x^{2}}} \cdot x d x = \sqrt{a^{2} - x^{2}} \cdot x + \int (\frac{x^{2} - a^{2} + a^{2}}{\sqrt{a^{2} - x^{2}}}) d x = x \sqrt{a^{2} - x^{2}} - \int \sqrt{a^{2} - x^{2}} d x + a^{2} \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = x \sqrt{a^{2} - x^{2}} - I + a^{2} \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x ∴ 2 I = x \sqrt{a^{2} - x^{2}} + a^{2} \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x \Rightarrow I = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a}) + C$

Page No 18.198:

Question 87:

$\int \sqrt{3 x^{2} + 4 x + 1} d x$

Answer:

$\int \sqrt{3 x^{2} + 4 x + 1} d x = \sqrt{3} \int \sqrt{x^{2} + \frac{4}{3} x + \frac{1}{3}} d x = \sqrt{3} \int \sqrt{x^{2} + \frac{4}{3} x + {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{2} + \frac{1}{3}} d x = \sqrt{3} \int \sqrt{{(x + \frac{2}{3})}^{2} - \frac{4}{9} + \frac{1}{3}} d x = \sqrt{3} \int \sqrt{{(x + \frac{2}{3})}^{2} - {(\frac{1}{3})}^{2}} d x = \sqrt{3} [\frac{1}{2} (x + \frac{2}{3}) \sqrt{{(x + \frac{2}{3})}^{2} - {(\frac{1}{3})}^{2}} - \frac{1}{2} \times {(\frac{1}{3})}^{2} \ln |(x + \frac{2}{3}) + \sqrt{{(x + \frac{2}{3})}^{2} - {(\frac{1}{3})}^{2}}| + C] [∵ \int \sqrt{x^{2} - a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} - a^{2}} - \frac{1}{2} a^{2} \ln |x + \sqrt{x^{2} - a^{2}}| + C] = \frac{1}{6} (3 x + 2) \sqrt{3 x^{2} + 4 x + 1} - \frac{\sqrt{3}}{18} \ln |(x + \frac{2}{3}) + \sqrt{x^{2} + \frac{4}{3} x + \frac{1}{3}}| + C$

Page No 18.198:

Question 88:

$\int \sqrt{1 + 2 x - 3 x^{2}} d x$

Answer:

\int \sqrt{1 + 2 x - 3 x^{2}} d x = \sqrt{3} \int \sqrt{\frac{1}{3} + \frac{2}{3} x - x^{2}} d x = \sqrt{3} \int \sqrt{\frac{1}{3} - (x^{2} - \frac{2}{3} x)} d x = \sqrt{3} \int \sqrt{\frac{1}{3} - \{x^{2} - \frac{2}{3} x + {(\frac{1}{3})}^{2} - {(\frac{1}{3})}^{2}\}} d x = \sqrt{3} \int \sqrt{\frac{1}{3} + \frac{1}{9} - {(x - \frac{1}{3})}^{2}} d x = \sqrt{3} \int \sqrt{\frac{4}{9} - {(x - \frac{1}{3})}^{2}} d x = \sqrt{3} \int \sqrt{{(\frac{2}{3})}^{2} - {(x - \frac{1}{3})}^{2}} d x = \sqrt{3} [\frac{(x - \frac{1}{3})}{2} \sqrt{{(\frac{2}{3})}^{2} - {(x - \frac{1}{3})}^{2}} + \frac{{(\frac{2}{3})}^{2}}{2} \sin^{- 1} (\frac{x - \frac{1}{3}}{\frac{2}{3}})] + C [∵ \int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{1}{2} a^{2} \sin^{- 1} \frac{x}{a} + C] = (\frac{3 x - 1}{6}) \sqrt{1 + 2 x - 3 x^{2}} + \frac{2 \sqrt{3}}{9} \sin^{- 1} (\frac{3 x - 1}{2}) + C

Page No 18.198:

Question 89:

$\int x \sqrt{1 + x - x^{2}} d x$

Answer:

$Let I = \int x \sqrt{1 + x - x^{2}} d x & let x = A \frac{d}{d x} (1 + x - x^{2}) + B \Rightarrow x = A (- 2 x + 1) + B By equating the coefficients of like terms we get, x = (- 2 A) x \Rightarrow A = - \frac{1}{2} & A + B = 0 \Rightarrow B = \frac{1}{2} By substituting the values of A and B in eq (1) we get, I = \int [- \frac{1}{2} (- 2 x + 1) + \frac{1}{2}] \sqrt{1 + x - x^{2}} d x = - \frac{1}{2} \int (- 2 x + 1) \sqrt{1 + x - x^{2}} d x + \frac{1}{2} \sqrt{1 + x - x^{2}} d x Putting 1 + x - x^{2} = t \Rightarrow (- 2 x + 1) d x = d t ∴ I = - \frac{1}{2} \int \sqrt{t} \cdot d t + \frac{1}{2} \int \sqrt{1 + x - x^{2}} d x = - \frac{1}{2} \int \sqrt{t} d t + \frac{1}{2} \int \sqrt{1 - (x^{2} - x)} d x = - \frac{1}{2} \int t^{\frac{1}{2}} \cdot d t + \frac{1}{2} \int \sqrt{1 - (x^{2} - x + \frac{1}{4} - \frac{1}{4})} d x = - \frac{1}{2} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + \frac{1}{2} \int \sqrt{1 - {(x - \frac{1}{2})}^{2} + \frac{1}{4}} d x = - \frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}} + \frac{1}{2} \int \sqrt{{(\frac{\sqrt{5}}{2})}^{2} - {(x - \frac{1}{2})}^{2}} d x = - \frac{1}{3} t^{\frac{3}{2}} + \frac{1}{2} [(\frac{x - \frac{1}{2}}{2}) \sqrt{{(\frac{\sqrt{5}}{2})}^{2} - {(x - \frac{1}{2})}^{2}} + \frac{{(\frac{\sqrt{5}}{2})}^{2}}{2} \sin^{- 1} (\frac{x - \frac{1}{2}}{\frac{\sqrt{5}}{2}})] + C [∵ \int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{1}{2} a^{2} \sin^{- 1} \frac{x}{a} + C] = - \frac{1}{3} {(1 + x - x^{2})}^{\frac{3}{2}} + \frac{1}{2} [(\frac{2 x - 1}{4}) \sqrt{1 + x - x^{2}} + \frac{5}{8} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}})] + C = \frac{- (1 + x - x^{2}) \sqrt{1 + x - x^{2}}}{3} + \frac{(2 x - 1)}{8} \sqrt{1 + x - x^{2}} + \frac{5}{16} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}}) + C = \sqrt{1 + x - x^{2}} [\frac{- (1 + x - x^{2})}{3} + \frac{2 x - 1}{8}] + \frac{5}{16} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}}) + C = \sqrt{1 + x - x^{2}} [\frac{- 8 - 8 x + 8 x^{2} + 6 x - 3}{24}] + \frac{5}{16} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}}) + C = \sqrt{1 + x - x^{2}} [\frac{8 x^{2} - 2 x - 11}{24}] + \frac{5}{16} \sin^{- 1} (\frac{2 x - 1}{\sqrt{5}}) + C$

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Question 90:

$\int (2 x + 3) \sqrt{4 x^{2} + 5 x + 6} d x$

Answer:

$Let I = \int (2 x + 3) \sqrt{4 x^{2} + 5 x + 6} d x & let 2 x + 3 = A \frac{d}{d x} (4 x^{2} + 5 x + 6) + B \Rightarrow 2 x + 3 = A (8 x + 5) + B . . . (1) By equating coefficients of like terms we get, 2 x = 8 A x \Rightarrow A = \frac{1}{4} & 5 A + B = 3 \Rightarrow \frac{5}{4} + B = 3 \Rightarrow B = 3 - \frac{5}{4} = \frac{7}{4} Thus, by substituting the values of A and B in eq (1) we get I = \int (2 x + 3) \sqrt{4 x^{2} + 5 x + 6} d x = \int [\frac{1}{4} (8 x + 5) + \frac{7}{4}] \sqrt{4 x^{2} + 5 x + 6} d x = \frac{1}{4} \int (8 x + 5) \sqrt{4 x^{2} + 5 x + 6} d x + \frac{7}{4} \int \sqrt{4 x^{2} + 5 x + 6} d x Putting 4 x^{2} + 5 x + 6 = t in the first integral \Rightarrow (8 x + 5) d x = d t ∴ I = \frac{1}{4} \int \sqrt{t} \cdot d t + \frac{7 \times 2}{4} \int \sqrt{x^{2} + \frac{5 x}{4} + \frac{3}{2}} d x = \frac{1}{4} \int t^{\frac{1}{2}} \cdot d t + \frac{7}{2} \int \sqrt{x^{2} - \frac{5 x}{4} + {(\frac{5}{8})}^{2} - {(\frac{5}{8})}^{2} + \frac{3}{2}} d x = \frac{1}{4} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + \frac{7}{2} \int \sqrt{{(x + \frac{5}{8})}^{2} - \frac{25}{64} + \frac{3}{2}} d x = \frac{1}{4} \times \frac{2}{3} t^{\frac{3}{2}} + \frac{7}{2} \int \sqrt{{(x + \frac{5}{8})}^{2} + \frac{- 25 + 96}{64}} = \frac{1}{6} t^{\frac{3}{2}} + \frac{7}{2} \int \sqrt{{(x + \frac{5}{8})}^{2} + {(\frac{\sqrt{71}}{8})}^{2}} = \frac{1}{6} {(4 x^{2} + 5 x + 6)}^{\frac{3}{2}} + \frac{7}{2} [\frac{x + \frac{5}{8}}{2} \sqrt{{(x + \frac{5}{8})}^{2} + {(\frac{\sqrt{71}}{8})}^{2}} + \frac{71}{64 \times 2} \ln |x + \frac{5}{8} + \sqrt{{(x + \frac{5}{8})}^{2} + {(\frac{\sqrt{71}}{8})}^{2}}|] + C [∵ \int \sqrt{a^{2} + x^{2}} d x = \frac{1}{2} x \sqrt{a^{2} + x^{2}} + \frac{1}{2} a^{2} \ln |x + \sqrt{x^{2} + a^{2}}| + C] = \frac{1}{6} {(4 x^{2} + 5 x + 6)}^{\frac{3}{2}} + \frac{7}{2} \frac{(8 x + 5)}{16} \sqrt{x^{2} + \frac{5}{4} x + \frac{3}{2}} + \frac{71 \times 7}{2 \times 128} \ln |x + \frac{5}{8} + \sqrt{x^{2} + \frac{5}{4} x + \frac{3}{2}}| + C = \frac{1}{6} {(4 x^{2} + 5 x + 6)}^{\frac{3}{2}} + \frac{7 \times 2 (8 x + 5)}{4 \times 16} \sqrt{x^{2} + \frac{5}{4} x + \frac{3}{2}} + \frac{497}{256} \ln |x + \frac{5}{8} + \sqrt{x^{2} + \frac{5}{4} x + \frac{3}{2}}| + C = \frac{1}{6} (4 x^{2} + 5 x + 6) \sqrt{4 x^{2} + 5 x + 6} + \frac{7}{64} (8 x + 5) \sqrt{4 x^{2} + 5 x + 6} + \frac{497}{256} \ln |x + \frac{5}{6} + \sqrt{x^{2} + \frac{5}{4} x + \frac{3}{2}}| + C = \sqrt{4 x^{2} + 5 x + 6} [\frac{4 x^{2} + 5 x + 6}{6} + \frac{7}{64} (8 x + 5)] + \frac{497}{256} \ln |x + \frac{5}{8} + \sqrt{x^{2} + \frac{5}{4} x + \frac{3}{2}}| + C = \sqrt{4 x^{2} + 5 x + 6} [\frac{128 x^{2} + 328 x + 297}{192}] + \ln |x + \frac{5}{8} + \sqrt{x^{2} + \frac{5}{4} x + \frac{3}{2}}| + C$

Page No 18.198:

Question 91:

$\int (1 + x^{2}) \cos 2 x d x$

Answer:

$Let I = \int (1 + x^{2}) \cdot \cos 2 x \cdot d x = \int \cos 2 x d x + \int x^{2} \cdot \cos 2 x d x = \frac{\sin 2 x}{2} + I_{1} where I_{1} = \int x^{2} \cos 2 x d x . . . (1) I_{1} = \int \underset{I}{x^{2}} \underset{II}{\cos 2 x} d x = x^{2} \int \cos 2 x d x - \int \{\frac{d}{d x} (x^{2}) \int \cos 2 x d x\} d x = x^{2} \cdot \frac{\sin 2 x}{2} - \int \frac{2 x \times \sin 2 x}{2} d x = \frac{x^{2} \cdot \sin 2 x}{2} - \int \underset{I}{x} \cdot \underset{II}{\sin 2 x} d x = \frac{x^{2} \cdot \sin 2 x}{2} - [x \int \sin 2 x d x - \int \{\frac{d}{d x} (x) \int \sin 2 x d x\} d x] = \frac{x^{2} \cdot \sin 2 x}{2} - [x (\frac{- \cos 2 x}{2}) - \int 1 \cdot (\frac{- \cos 2 x}{2}) d x] = \frac{x^{2} \cdot \sin 2 x}{2} + \frac{x \cdot \cos 2 x}{2} - \frac{\sin 2 x}{4} . . . (2) From (1) and (2) ∴ I = \frac{\sin 2 x}{2} + \frac{x^{2} \sin 2 x}{2} + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4} + C = (x^{2} + 1) \frac{\sin 2 x}{2} + \frac{x \cos 2 x}{2} - \frac{\sin 2 x}{4} + C$

Page No 18.198:

Question 92:

$\int \log_{10} x d x$

Answer:

$\int \log_{10} x d x = \int \frac{\log_{e} x}{\log_{e} 10} d x = \frac{1}{\log_{e} 10} \int \underset{II}{1} \cdot \underset{I}{\log x} d x = \frac{1}{\log_{e} 10} [\log_{e} x \int 1 d x - \int \{\frac{d}{d x} (\log_{e} x) \int 1 d x\} d x] = \frac{1}{\log_{e} 10} [\log_{e} x \cdot x - \int \frac{1}{x} \times x d x] = \frac{1}{\log_{e} 10} [x \log_{e} x - x] + C = \frac{1}{\log_{e} 10} \times x (\log_{e} x - 1) + C = x (\log_{e} x - 1) \cdot \log_{10} e + C$

Page No 18.198:

Question 93:

$\int \frac{\log (\log x)}{x} d x$

Answer:

$Let I = \int \frac{\log (\log x) d x}{x} Putting \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = \int \underset{II}{1} \cdot \underset{I}{\log t \cdot d} t = \log t \int 1 d t - \int \{\frac{d}{d t} (\log t) \int 1 d t\} d t = \log t \cdot t - \int \frac{1}{t} \times t d t = \log t \cdot t - \int d t = \log t \cdot t - t + C = t (\log t - 1) + C = \log x (\log (\log x) - 1) + C$

Page No 18.198:

Question 94:

$\int x \sec^{2} 2 x d x$

Answer:

$\int \underset{I}{x} \cdot \underset{II}{\sec^{2} 2 x d x} = x \int \sec^{2} 2 x d x - \int \{\frac{d}{d x} (x) \int \sec^{2} 2 x d x\} d x = \frac{x \tan 2 x}{2} - \int 1 \cdot \frac{\tan 2 x}{2} d x = \frac{x \tan 2 x}{2} - \frac{1}{2} \frac{\ln |\sec 2 x|}{2} + C = \frac{x \tan 2 x}{2} - \frac{1}{4} \ln |\sec 2 x| + C$

Page No 18.198:

Question 95:

$\int x \sin^{3} x d x$

Answer:

$\int x \cdot \sin^{3} x d x = \int x \cdot [\frac{1}{4} (3 \sin x - \sin 3 x)] d x [\sin^{3} A - \frac{1}{4} \{3 \sin A - \sin (3 A)\}] = \frac{3}{4} \int \underset{I}{x} \cdot \underset{II}{\sin x} d x - \frac{1}{4} \int x \cdot \sin 3 x d x = \frac{3}{4} [x \int \sin x d x - \int \{\frac{d}{d x} (x) \int \sin x d x\} d x] - \frac{1}{4} [x \int \sin 3 x d x - \int \{\frac{d}{d x} (x) \int \sin 3 x d x\} d x] = \frac{3}{4} [x (- \cos x) - \int 1 \cdot (- \cos x) d x] - \frac{1}{4} [x (\frac{- \cos 3 x}{3}) - \int 1 \cdot (\frac{- \cos 3 x}{3}) d x] = \frac{- 3 x}{4} \cos x + \frac{3}{4} \sin x + \frac{x \cos 3 x}{12} - \frac{\sin 3 x}{36} + C = \frac{1}{4} [- 3 x \cos x + 3 \sin x + \frac{x \cos 3 x}{3} - \frac{\sin 3 x}{9} + C]$

Page No 18.198:

Question 96:

$\int {(x + 1)}^{2} e^{x} d x$

Answer:

$\int \underset{I}{{(x + 1)}^{2}} \underset{II}{e^{x}} d x = {(x + 1)}^{2} \int e^{x} d x - \int \{\frac{d}{d x} {(x + 1)}^{2} \int e^{x} d x\} d x = {(x + 1)}^{2} \cdot e^{x} - \int 2 (x + 1) \cdot e^{x} d x = {(x + 1)}^{2} e^{x} - 2 \int \underset{I}{x} \underset{II}{e^{x}} d x - 2 \int e^{x} d x = {(x + 1)}^{2} e^{x} - 2 [x \cdot e^{x} - \int 1 \cdot e^{x} d x] - 2 e^{x} = {(x + 1)}^{2} e^{x} - 2 x e^{x} + 2 e^{x} - 2 e^{x} + C = [{(x + 1)}^{2} - 2 x] e^{x} + C = (x^{2} + 1) e^{x} + C$

Page No 18.198:

Question 97:

$\int \log (x + \sqrt{x^{2} + a^{2}}) d x$

Answer:

$Let I = \int \underset{}{\underset{II}{1} \cdot \log} \underset{I}{(x + \sqrt{x^{2} + a^{2}})} d x = \log (x + \sqrt{x^{2} + a^{2}}) \int 1 d x - \int [\frac{d}{d x} \{\log (x + \sqrt{x^{2} + a^{2}})\} \int 1 d x] = \log (x + \sqrt{x^{2} + a^{2}}) \cdot x - \int (\frac{1}{x + \sqrt{x^{2} + a^{2}}}) \times (1 + \frac{1 \times 2 x}{2 \sqrt{x^{2} + a^{2}}}) \cdot x \cdot d x = \log (x + \sqrt{x^{2} + a^{2}}) \cdot x - \int \frac{x}{\sqrt{x^{2} + a^{2}}} d x Putting x^{2} + a^{2} = t in the second integral \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = x \cdot \log (x + \sqrt{x^{2} + a^{2}}) - \frac{1}{2} \int \frac{1}{\sqrt{t}} d t = x \cdot \log (x + \sqrt{x^{2} + a^{2}}) - \frac{1}{2} \int t^{- \frac{1}{2}} d t = x \cdot \log (x + \sqrt{x^{2} + a^{2}}) - \frac{1}{2} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = x \cdot \log (x + \sqrt{x^{2} + a^{2}}) - \sqrt{t} + C = x \cdot \log (x + \sqrt{x^{2} + a^{2}}) - \sqrt{x^{2} - a^{2}} + C [∵ t = x^{2} + a^{2}]$

Page No 18.198:

Question 98:

$\int \frac{\log x}{x^{3}} d x$

Answer:

$\int \frac{\log x}{x^{3}} d x = \int \underset{II}{\frac{1}{x^{3}}} \underset{I}{\log x} d x = \log x \int \frac{1}{x^{3}} d x - \int \{\frac{d}{d x} (\log x) \int \frac{1}{x^{3}} d x\} d x = \log x \int x^{- 3} d x - \int \frac{1}{x} \times (\frac{x^{- 3 + 1}}{- 3 + 1}) d x = \log x [\frac{x^{- 3 + 1}}{- 3 + 1}] + \frac{1}{2} \int \frac{1}{x^{3}} d x = \log x (- \frac{1}{2 x^{2}}) + \frac{1}{2} \int x^{- 3} d x = \log x (- \frac{1}{2 x^{2}}) + \frac{1}{2} [\frac{x^{- 3 + 1}}{- 3 + 1}] + C = \log x (- \frac{1}{2 x^{2}}) - \frac{1}{4 x^{2}} + C = - \frac{1}{4 x^{2}} (2 \log x + 1) + C$

Page No 18.198:

Question 99:

$\int \frac{\log (1 - x)}{x^{2}} d x$

Answer:

$Let I = \int \frac{\log (1 - x)}{x^{2}} d x = \int \underset{II}{\frac{1}{x^{2}}} \underset{I}{\log (1 - x)} d x = \log (1 - x) \int x^{- 2} d x - \int \frac{- 1}{1 - x} \times (\frac{x^{- 2 + 1}}{- 2 + 1}) d x = \log (1 - x) [\frac{x^{- 2 + 1}}{- 2 + 1}] + \int \frac{- 1}{(1 - x) x} d x = \log (1 - x) \times (- \frac{1}{x}) + \int \frac{1}{x^{2} - x} d x = - \frac{\log (1 - x)}{x} + \int \frac{1}{x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d x = - \frac{\log (1 - x)}{x} + \int \frac{1}{{(x - \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d x = - \frac{\log (1 - x)}{x} + \frac{1}{2 \times \frac{1}{2}} \log |\frac{x - \frac{1}{2} - \frac{1}{2}}{x - \frac{1}{2} + \frac{1}{2}}| + C = - \frac{\log (1 - x)}{x} + \log |\frac{x - 1}{x}| + C = - \frac{\log (1 - x)}{x} + \log |(x - 1)| - \log x + C = - \frac{\log |1 - x|}{x} + \log |1 - x| - \log |x| + C = (1 - \frac{1}{x}) \log |1 - x| - \log |x| + C$

Page No 18.198:

Question 100:

$\int x^{3} {(\log x)}^{2} d x$

Answer:

$\int \underset{II}{x^{3}} \cdot \underset{I}{{(\log x)}^{2}} \cdot d x = (\log x^{2}) \int x^{3} d x - \int \frac{2 \log x}{x} \times \frac{x^{4}}{4} d x = {(\log x)}^{2} \times \frac{x^{4}}{4} - \frac{1}{2} \int \underset{I}{\log x} \cdot \underset{II}{x^{3}} d x = {(\log x)}^{2} \times \frac{x^{4}}{4} - \frac{1}{2} [\log x \int x^{3} d x - \int \{\frac{d}{d x} (\log x) \int x^{3} d x\} d x] = {(\log x)}^{2} \times \frac{x^{4}}{4} - \frac{1}{2} [\log x \cdot \frac{x^{4}}{4} - \int \frac{1}{x} \times \frac{x^{4}}{4} d x] = {(\log x)}^{2} \times \frac{x^{4}}{4} - \frac{1}{2} [\log x \cdot \frac{x^{4}}{4} - \frac{1}{4} \int x^{3} d x] = {(\log x)}^{2} \times \frac{x^{4}}{4} - \frac{1}{2} [\log x \cdot \frac{x^{4}}{4} - \frac{x^{4}}{16}] + C = {(\log x)}^{2} \times \frac{x^{4}}{4} - \frac{\log x \cdot x^{4}}{8} + \frac{x^{4}}{32} + C$

Page No 18.198:

Question 101:

$\int \frac{1}{x \sqrt{1 + x^{n}}} d x$

Answer:

$We have, I = \int \frac{d x}{x \sqrt{1 + x^{n}}} = \int \frac{x^{n - 1} d x}{x^{n - 1} x^{1} \sqrt{1 + x^{n}}} = \int \frac{x^{n - 1} d x}{x^{n} \sqrt{1 + x^{n}}} Putting x^{n} = t \Rightarrow n x^{n - 1} d x = d t \Rightarrow x^{n - 1} d x = \frac{d t}{n} ∴ I = \frac{1}{n} \int \frac{d t}{t \sqrt{1 + t}} let 1 + t = p^{2} \Rightarrow d t = 2 p d p ∴ I = \frac{1}{n} \int \frac{2 p d p}{(p^{2} - 1) p} = \frac{2}{n} \int \frac{d p}{p^{2} - 1^{2}} = \frac{2}{n} \times \frac{1}{2} \log |\frac{p - 1}{p + 1}| + C = \frac{1}{n} \log |\frac{\sqrt{1 + t} - 1}{\sqrt{1 + t} + 1}| + C = \frac{1}{n} \log |\frac{\sqrt{1 + x^{n}} - 1}{\sqrt{1 + x^{n}} + 1}| + C$

Page No 18.198:

Question 102:

$\int \frac{x^{2}}{\sqrt{1 - x}} d x$

Answer:

$We have, I = \int \frac{x^{2}}{\sqrt{1 - x}} d x Let, 1 - x = t^{2} Differentiating both sides we get - d x = 2 t d t Now, integration becomes, I = - \int \frac{{(1 - t^{2})}^{2}}{t} 2 t d t = - 2 \int {(1 - t^{2})}^{2} d t = - 2 \int (1 - 2 t^{2} + t^{4}) d t = - 2 [t - \frac{2 t^{3}}{3} + \frac{t^{5}}{5}] + C = \frac{- 2}{15} t [3 t^{4} - 10 t^{2} + 15] + C = \frac{- 2}{15} \sqrt{1 - x} [3 {(1 - x)}^{2} - 10 (1 - x) + 15] + C = \frac{- 2}{15} \sqrt{1 - x} [3 (1 - 2 x + x^{2}) - 10 (1 - x) + 15] + C = \frac{- 2}{15} \sqrt{1 - x} [3 x^{2} - 6 x + 3 - 10 + 10 x + 15] + C = \frac{- 2}{15} \sqrt{1 - x} [3 x^{2} + 4 x + 8] + C$

Page No 18.199:

Question 103:

$\int \frac{x^{5}}{\sqrt{1 + x^{3}}} d x$

Answer:

$We have, I = \int \frac{x^{5} d x}{\sqrt{1 + x^{3}}} = \int \frac{x^{3} x^{2} d x}{\sqrt{1 + x^{3}}} Putting x^{3} = t \Rightarrow 3 x^{2} d x = d t \Rightarrow x^{2} d x = \frac{d t}{3} ∴ I = \frac{1}{3} \int \frac{t d t}{\sqrt{1 + t}} = \frac{1}{3} \int (\frac{1 + t - 1}{\sqrt{1 + t}}) d t = \frac{1}{3} \int (\sqrt{1 + t} - \frac{1}{\sqrt{1 + t}}) d t = \frac{1}{3} [\frac{{(1 + t)}^{\frac{3}{2}}}{\frac{3}{2}} - \frac{{(1 + t)}^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = \frac{2}{9} {(1 + t)}^{\frac{3}{2}} - \frac{2}{3} {(1 + t)}^{\frac{1}{2}} + C = \frac{2}{9} {(1 + x^{3})}^{\frac{3}{2}} - \frac{2}{3} {(1 + x^{3})}^{\frac{1}{2}} + C = \frac{2}{9} {(1 + x^{3})}^{\frac{1}{2}} (1 + x^{3} - 3) + C = \frac{2}{9} \sqrt{1 + x^{3}} (x^{3} - 2) + C$

Page No 18.199:

Question 104:

$\int \frac{1 + x^{2}}{\sqrt{1 - x^{2}}} d x$

Answer:

$We have, I = \int \frac{1 + x^{2}}{\sqrt{1 - x^{2}}} d x = \int \frac{2 - (1 - x^{2})}{\sqrt{1 - x^{2}}} d x = 2 \int \frac{1}{\sqrt{1 - x^{2}}} d x - \int \frac{1 - x^{2}}{\sqrt{1 - x^{2}}} d x = 2 \int \frac{1}{\sqrt{1 - x^{2}}} d x - \int \sqrt{1 - x^{2}} d x = 2 \sin^{- 1} x - [\frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{- 1} x] + C = 2 \sin^{- 1} x - \frac{x}{2} \sqrt{1 - x^{2}} - \frac{1}{2} \sin^{- 1} x + C = \frac{3}{2} \sin^{- 1} x - \frac{x}{2} \sqrt{1 - x^{2}} + C$

Page No 18.199:

Question 105:

$\int x \sqrt{\frac{1 - x}{1 + x}} d x$

Answer:

$We have, I = \int x \sqrt{\frac{1 - x}{1 + x}} d x \Rightarrow I = \int x \sqrt{\frac{(1 - x) (1 - x)}{(1 + x) (1 - x)}} d x \Rightarrow I = \int x \frac{(1 - x)}{\sqrt{1 - x^{2}}} d x \Rightarrow I = \int \frac{x - x^{2}}{\sqrt{1 - x^{2}}} d x \Rightarrow I = \int \frac{x - x^{2} - 1 + 1}{\sqrt{1 - x^{2}}} d x \Rightarrow I = \int \frac{- x^{2} + 1}{\sqrt{1 - x^{2}}} d x + \int \frac{x - 1}{\sqrt{1 - x^{2}}} d x \Rightarrow I = \int \sqrt{1 - x^{2}} d x + \int \frac{x}{\sqrt{1 - x^{2}}} d x - \int \frac{1}{\sqrt{1 - x^{2}}} d x \Rightarrow I = \frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{- 1} (x) + C_{1} - \sqrt{1 - x^{2}} + C_{2} - \sin^{- 1} (x) + C_{3} [∵ \int \frac{x}{\sqrt{1 - x^{2}}} d x = - \sqrt{1 - x^{2}} + C_{2}] \Rightarrow I = \sqrt{1 - x^{2}} (\frac{x}{2} - 1) - \frac{1}{2} \sin^{- 1} (x) + C$

Page No 18.199:

Question 106:

$\int \frac{1}{x \sqrt{1 + x^{3}}} d x$

Answer:

$We have, I = \int \frac{d x}{x \sqrt{1 + x^{3}}} = \int \frac{x^{2} d x}{x^{3} \sqrt{1 + x^{3}}} putting x^{3} = t \Rightarrow 3 x^{2} d x = d t \Rightarrow x^{2} d x = \frac{d t}{3} ∴ I = \frac{1}{3} \int \frac{d t}{t \sqrt{1 + t}} let 1 + t = p^{2} \Rightarrow d t = 2 p d p I = \frac{1}{3} \int \frac{2 p d p}{(p^{2} - 1) \times p} = \frac{2}{3} \int \frac{d p}{p^{2} - 1} = \frac{2}{3} \times \frac{1}{2} \log |\frac{p - 1}{p + 1}| + C = \frac{1}{3} \log |\frac{p - 1}{p + 1}| + C = \frac{1}{3} \log |\frac{\sqrt{1 + t} - 1}{\sqrt{1 + t} + 1}| + C = \frac{1}{3} \log |\frac{\sqrt{1 + x^{3}} - 1}{\sqrt{1 + x^{3}} + 1}| + C$

Page No 18.199:

Question 107:

$\int \frac{\sin x + \cos x}{\sin^{4} x + \cos^{4} x} d x$

Answer:

$We have, I = \int \frac{\sin x + \cos x}{\sin^{4} x + \cos^{4} x} d x = \int \frac{\sin x + \cos x}{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x} d x = \int \frac{\sin x + \cos x}{1 - 2 \sin^{2} x \cos^{2} x} d x = \int \frac{\sin x + \cos x}{1 - \frac{1}{2} {(2 \sin x \cos x)}^{2}} d x = \int \frac{\sin x + \cos x}{1 - \frac{1}{2} \sin^{2} 2 x} d x$

$Putting \sin x - \cos x = t . . . . . (1) \Rightarrow {(\sin x - \cos x)}^{2} = t^{2} \Rightarrow \sin^{2} x + \cos^{2} x - 2 \sin x \cos x = t^{2} \Rightarrow 1 - 2 \sin x \cos x = t^{2} \Rightarrow \sin 2 x = 1 - t^{2} Differentiating (1), we get (\cos x + \sin x) d x = d t ∴ I = \int \frac{1}{1 - \frac{1}{2} {(1 - t^{2})}^{2}} d t = \int \frac{2}{2 - {(1 - t^{2})}^{2}} d t = \int \frac{2}{{(\sqrt{2})}^{2} - {(1 - t^{2})}^{2}} d t = 2 \int \frac{1}{(\sqrt{2} + 1 - t^{2}) (\sqrt{2} - 1 + t^{2})} d t$

$= \frac{2}{2 \sqrt{2}} \int [\frac{1}{\sqrt{2} + 1 - t^{2}} + \frac{1}{\sqrt{2} - 1 + t^{2}}] d t = \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{2} + 1 - t^{2}} d t + \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{2} - 1 + t^{2}} d t = \frac{1}{\sqrt{2}} \int \frac{1}{{(\sqrt{\sqrt{2} + 1})}^{2} - t^{2}} d t + \frac{1}{\sqrt{2}} \int \frac{1}{{(\sqrt{\sqrt{2} - 1})}^{2} + t^{2}} d t = \frac{1}{\sqrt{2}} \times \frac{1}{2 \sqrt{\sqrt{2} + 1}} \log |\frac{\sqrt{\sqrt{2} + 1} + t}{\sqrt{\sqrt{2} + 1} - t}| + \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{\sqrt{2} + 1}} \tan^{- 1} \frac{t}{\sqrt{\sqrt{2} + 1}} + C = \frac{1}{\sqrt{2}} [\frac{1}{2 \sqrt{\sqrt{2} + 1}} \log |\frac{\sqrt{\sqrt{2} + 1} + t}{\sqrt{\sqrt{2} + 1} - t}| + \frac{1}{\sqrt{\sqrt{2} + 1}} \tan^{- 1} \frac{t}{\sqrt{\sqrt{2} + 1}}] + C, where t = \sin x - \cos x$

Page No 18.199:

Question 108:

$\int x^{2} \tan^{- 1} x d x$

Answer:

$We have, I = \int x^{2} \tan^{- 1} x d x Considering \tan^{- 1} x as first function and x^{2} as second function I = \tan^{- 1} x \frac{x^{3}}{3} - \int (\frac{1}{1 + x^{2}} \times \frac{x^{3}}{3}) d x = \tan^{- 1} x \frac{x^{3}}{3} - \frac{1}{3} \int \frac{x^{3} d x}{1 + x^{2}} = \tan^{- 1} x \frac{x^{3}}{3} - \frac{1}{3} \int (\frac{x^{2} x}{1 + x^{2}}) d x Putting 1 + x^{2} = t \Rightarrow x^{2} = t - 1 \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \tan^{- 1} x \frac{x^{3}}{3} - \frac{1}{6} \int (\frac{t - 1}{t}) d t = \frac{x^{3}}{3} \tan^{- 1} x - \frac{1}{6} \int d t + \frac{1}{6} \int \frac{d t}{t} = \frac{x^{3}}{3} \tan^{- 1} x - \frac{1}{6} t + \frac{1}{6} \log |t| + C = \frac{x^{3}}{3} \tan^{- 1} x - \frac{1}{6} (1 + x^{2}) + \frac{1}{6} \log |1 + x^{2}| + C = \frac{x^{3}}{3} \tan^{- 1} x - \frac{x^{2}}{6} + \frac{1}{6} \log |x^{2} + 1| + C' Where C' = C - \frac{1}{6}$

Page No 18.199:

Question 109:

$\int \tan^{- 1} \sqrt{x} d x$

Answer:

$We have, I = \int \tan^{- 1} \sqrt{x} d x Putting \sqrt{x} = \tan θ \Rightarrow x = \tan^{2} θ \Rightarrow d x = 2 \tan θ \sec^{2} θ d θ ∴ I = \int [\tan^{- 1} (\tan θ) 2 \tan θ \sec^{2} θ] d θ = 2 \int \underset{i}{θ} \underset{ii}{\tan θ s e c^{2} θ} d θ = 2 [θ \times \frac{\tan^{2} θ}{2} - \int 1 \frac{\tan^{2} θ d θ}{2}] (∵ \int \tan θ \sec^{2} θ d θ = \frac{\tan^{2} θ}{2}) = 2 [θ \frac{\tan^{2} θ}{2} - \frac{1}{2} \int (\sec^{2} θ - 1) d θ] = θ \tan^{2} θ - \frac{2 \times \tan θ}{2} + \frac{2 \times θ}{2} + C = \tan^{- 1} \sqrt{x} \times x - \sqrt{x} + \tan^{- 1} \sqrt{x} + C = (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C$

Page No 18.199:

Question 110:

$\int \sin^{- 1} \sqrt{x} d x$

Answer:

$We have, I = \int \sin^{- 1} \sqrt{x} d x Putting \sqrt{x} = \sin θ \Rightarrow x = \sin^{2} θ \Rightarrow d x = 2 \sin θ \cos θ d θ \Rightarrow d x = \sin (2 θ) d θ ∴ I = \int θ \sin (2 θ) d θ = θ [\frac{- \cos 2 θ}{2}] - \int 1 (\frac{- \cos 2 θ}{2}) d θ = - \frac{θ \cos (2 θ)}{2} + \frac{1}{2} \int \cos (2 θ) d θ = - \frac{θ \cos (2 θ)}{2} + \frac{1}{2} [\frac{\sin (2 θ)}{2}] + C = \frac{- \sin^{- 1} \sqrt{x} (1 - 2 \sin^{2} θ)}{2} + \frac{1}{2} [\frac{2 \sin θ \cos θ}{2}] + C = \frac{- \sin \sqrt{x} (1 - 2 x)}{2} + \frac{\sin θ \sqrt{1 - \sin^{2} θ}}{2} + C = \frac{- \sin^{- 1} \sqrt{x} (1 - 2 x)}{2} + \frac{\sqrt{x} \sqrt{1 - x}}{2} + C = - \frac{1}{2} \sin^{- 1} (\sqrt{x}) (1 - 2 x) + \frac{1}{2} \sqrt{x - x^{2}} + C$

Page No 18.199:

Question 111:

$\int \sec^{- 1} \sqrt{x} d x$

Answer:

$We have, I = \int \sec^{- 1} \sqrt{x} d x Putting \sqrt{x} = \sec θ \Rightarrow x = \sec^{2} θ \Rightarrow d x = 2 \sec θ \sec θ \tan θ d θ = 2 \sec^{2} θ \tan θ d θ ∴ I = 2 \int θ \sec^{2} θ \tan θ d θ = 2 \int θ \tan θ \sec^{2} θ d θ Considering θ as first fucction and \tan θ \sec^{2} θ as second function I = 2 [θ \frac{\tan^{2} θ}{2} - \int 1 \frac{\tan^{2} θ}{2} d θ] (∵ \int \tan θ \sec^{2} θ d θ = \frac{\tan^{2} θ}{2}) = θ \tan^{2} θ - \int (\sec^{2} θ - 1) d θ = θ \tan^{2} θ - \tan θ + θ + C = θ (1 + \tan^{2} θ) - \tan θ + C = θ \sec^{2} θ - \sqrt{se c^{2} θ - 1} + C = \sec^{- 1} \sqrt{x} x - \sqrt{x - 1} + C = x \sec^{- 1} \sqrt{x} - \sqrt{x - 1} + C$

Page No 18.199:

Question 112:

$\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} d x$

Answer:

$We have, I = \int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} d x Putting x = \cos θ \Rightarrow d x = - \sin θ d θ ∴ I = \int \tan^{- 1} \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} (- \sin θ d θ) = \int \tan^{- 1} \sqrt{\frac{2 \sin^{2} \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}}} (- \sin θ) d θ = \int \tan^{- 1} (\tan \frac{θ}{2}) (- \sin θ) d θ = - \frac{1}{2} \int θ \sin θ d θ Considering θ as first function and \sin θ as second function I = - \frac{1}{2} [θ (- \cos θ) - \int 1 (- \cos θ) d θ] = - \frac{1}{2} (θ (- \cos θ) + \int \cos θ d θ) = - \frac{1}{2} (- θ \cos θ + \sin θ) + C = - \frac{1}{2} [- θ \cos θ + \sqrt{1 - \cos^{2} θ}] + C = - \frac{1}{2} [- \cos^{- 1} x \times x + \sqrt{1 - x^{2}}] + C = \frac{1}{2} [x \cos^{- 1} x - \sqrt{1 - x^{2}}] + C$

Page No 18.199:

Question 113:

$\int \sin^{- 1} \sqrt{\frac{x}{a + x}} d x$

Answer:

$We have, I = \int \sin^{- 1} \sqrt{\frac{x}{a + x}} d x Putting x = a \tan^{2} θ \Rightarrow \tan θ = \sqrt{\frac{x}{a}} \Rightarrow d x = a (2 \tan θ) \sec^{2} θ d θ ∴ I = \int \sin^{- 1} \sqrt{\frac{a \tan^{2} θ}{a + a \tan^{2} θ}} (2 a \tan θ \sec^{2} θ) d θ = \int \sin^{- 1} \sqrt{\frac{\tan^{2} θ}{\sec^{2} θ}} (2 a \tan θ \sec^{2} θ) d θ = 2 a \int \sin^{- 1} (\sin θ) \tan θ \sec^{2} θ d θ = 2 a \int θ \tan θ \sec^{2} θ d θ Considering θ as first function and \tan θ \sec^{2} θ as second function I = 2 a [θ \frac{\tan^{2} θ}{2} - \int 1 \frac{\tan^{2} θ}{2} d θ] = a [θ \tan^{2} θ - \int (\sec^{2} θ - 1) d θ] = a [θ \tan^{2} θ - \tan θ + θ] + C = a [θ \times (1 + \tan^{2} θ) - \tan θ] + C = a [\tan^{- 1} (\frac{\sqrt{x}}{\sqrt{a}}) (1 + \frac{x}{a}) - \frac{\sqrt{x}}{\sqrt{a}}] + C = (x + a) \tan^{- 1} (\frac{\sqrt{x}}{\sqrt{a}}) - \sqrt{a x} + C$

Page No 18.199:

Question 114:

$\int \sin^{- 1} (3 x - 4 x^{3}) d x$

Answer:

$We have, I = \int \sin^{- 1} (3 x - 4 x^{3}) d x Putting x = \sin θ \Rightarrow θ = \sin^{- 1} x \Rightarrow d x = \cos θ d θ ∴ I = \int \sin^{- 1} (3 \sin θ - 4 \sin^{3} θ) \cos θ d θ = \int \sin^{- 1} (\sin 3 θ) \cos θ d θ = 3 \int \underset{I}{θ} \underset{II}{\cos θ} d θ = 3 [θ (\sin θ) - \int 1 \sin θ d θ] = 3 [θ \sin θ + \cos θ] + C = 3 [θ \sin θ + \sqrt{1 - \sin^{2} θ}] + C = 3 [\sin^{- 1} x \times x + \sqrt{1 - x^{2}}] + C = 3 [x \sin^{- 1} x + \sqrt{1 - x^{2}}] + C$

Page No 18.199:

Question 115:

$\int {(\sin^{- 1} x)}^{3} d x$

Answer:

We have, I = \int {(\sin^{- 1} x)}^{3} d x Let, \sin^{- 1} x = t \Rightarrow \sin t = x \Rightarrow \cos t = \sqrt{1 - x^{2}} Differentiating both sides we get \cos t d t = d x Now, integral becomes I = \int {(\sin^{- 1} x)}^{3} d x = \int t^{3} \cos t d t = t^{3} \sin t - \int 3 t^{2} \sin t d t (Using by parts) = t^{3} \sin t - 3 \int t^{2} \sin t d t = t^{3} \sin t - 3 [- t^{2} \cos t - \int - 2 t \cos t d t] = t^{3} \sin t + 3 t^{2} \cos t - 6 \int t \cos t d t = t^{3} \sin t + 3 t^{2} \cos t - 6 [t \sin t - \int \sin t d t] = t^{3} \sin t + 3 t^{2} \cos t - 6 [t \sin t + \cos t] + C = {(\sin^{- 1} x)}^{3} x + 3 {(\sin^{- 1} x)}^{2} \sqrt{1 - x^{2}} - 6 (\sin^{- 1} x) x - 6 \sqrt{1 - x^{2}} + C = x (\sin^{- 1} x) [{(\sin^{- 1} x)}^{2} - 6] + 3 [{(\sin^{- 1} x)}^{2} - 2] \sqrt{1 - x^{2}} + C

Page No 18.199:

Question 116:

$\int \cos^{- 1} (1 - 2 x^{2}) d x$

Answer:

$We have, I = \int \cos^{- 1} (1 - 2 x^{2}) d x Putting x = \sin θ \Rightarrow d x = \cos θ d θ ∴ I = \int \cos^{- 1} (1 - 2 \sin^{2} θ) \cos θ d θ = \int \cos^{- 1} (\cos 2 θ) \cos θ d θ = 2 \int \underset{I}{θ} \underset{II}{\cos θ} d θ = 2 [θ \sin θ - \int 1 \sin θ d θ] = 2 [θ \sin θ + \cos θ] + C = 2 [\sin^{- 1} x \times x + \sqrt{1 - x^{2}}] + C = 2 [x \sin^{- 1} x + \sqrt{1 - x^{2}}] + C$

Page No 18.199:

Question 117:

$\int \frac{x \sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} d x$

Answer:

$We have, I = \int \frac{x \sin^{- 1} x}{{(1 - x^{2})}^{\frac{3}{2}}} d x Putting \sin^{- 1} x = θ \Rightarrow x = \sin θ \Rightarrow d x = \cos θ d θ ∴ I = \int \frac{\sin θ θ \cos θ d θ}{{(1 - \sin^{2} θ)}^{\frac{3}{2}}} = \int \frac{θ \sin θ \cos θ d θ}{{(\cos^{2} θ)}^{\frac{3}{2}}} = \int θ \frac{\sin θ}{\cos^{2} θ} d θ = \int \underset{I}{θ} \underset{II}{\sec θ \tan θ} d θ = θ \times \sec θ - \int 1 \times \sec θ d θ = θ \times \sec θ - \int \sec θ d θ = θ \times \sec θ - \log |\sec θ + \tan θ| + C = \frac{θ}{\cos θ} - \log |\frac{1}{\cos θ} + \frac{\sin θ}{\cos θ}| + C = \frac{θ}{\sqrt{1 - \sin^{2} θ}} - \log |\frac{1 + \sin θ}{\cos θ}| + C = \frac{θ}{\sqrt{1 - \sin^{2} θ}} - \log |\frac{1 + \sin θ}{\sqrt{1 - \sin^{2} θ}}| + C = \frac{θ}{\sqrt{1 - \sin^{2} θ}} - \log |\frac{\sqrt{1 + \sin θ}}{\sqrt{1 - \sin θ}}| + C = \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}} - \log |\frac{\sqrt{1 + x}}{\sqrt{1 - x}}| + C = \frac{\sin^{- 1} x}{\sqrt{1 - x^{2}}} - \frac{1}{2} \log |\frac{1 + x}{1 - x}| + C$

Page No 18.199:

Question 118:

$\int e^{2 x} (\frac{1 + \sin 2 x}{1 + \cos 2 x}) d x$

Answer:

$We have, I = \int e^{2 x} (\frac{1 + \sin 2 x}{1 + \cos 2 x}) d x = \int e^{2 x} (\frac{1}{1 + \cos 2 x} + \frac{\sin 2 x}{1 + \cos 2 x}) d x = \int e^{2 x} (\frac{1}{2 \cos^{2} x} + \frac{2 \sin x \cos x}{2 \cos^{2} x}) d x = \int e^{2 x} (\frac{\sec^{2} x}{2} + \tan x) d x Let e^{2 x} \tan x = t \Rightarrow (e^{2 x} \sec^{2} x + 2 e^{2 x} \tan x) d x = d t = [\frac{e^{2 x} \sec^{2} x}{2} + e^{2 x} \tan x] d x = \frac{d t}{2} ∴ I = \int \frac{d t}{2} = \frac{t}{2} + C = \frac{e^{2 x} \tan x}{2} + C$

Page No 18.199:

Question 119:

$\int \frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x / 2} d x$

Answer:

$We have, I = \int \frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- \frac{x}{2}} d x = \int (\frac{\sqrt{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}}}{1 + \cos x}) e^{- \frac{x}{2}} d x = \int \frac{\sqrt{{(\cos \frac{x}{2} - \sin \frac{x}{2})}^{2}} e^{- \frac{x}{2}}}{2 \cos^{2} \frac{x}{2}} d x = \int \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} e^{- \frac{x}{2}} d x = \frac{1}{2} \int (\sec \frac{x}{2} - \tan \frac{x}{2} \sec \frac{x}{2}) e^{- \frac{x}{2}} d x Let e^{- \frac{x}{2}} \sec (\frac{x}{2}) = t \Rightarrow [e^{- \frac{x}{2}} (\sec \frac{x}{2} \tan \frac{x}{2} \times \frac{1}{2}) - e^{- \frac{x}{2}} \frac{\sec (\frac{x}{2})}{2}] d x = d t \Rightarrow \frac{1}{2} (\sec \frac{x}{2} \tan \frac{x}{2} - \sec \frac{x}{2}) e^{- \frac{x}{2}} d x = d t ∴ I = - \int d t = - t + C = - e^{- \frac{x}{2}} \sec (\frac{x}{2}) + C$

Page No 18.199:

Question 120:

$\int e^{x} \frac{{(1 - x)}^{2}}{{(1 + x^{2})}^{2}} d x$

Answer:

$We have, I = \int \frac{e^{x} {(1 - x)}^{2}}{{(1 + x^{2})}^{2}} d x = \int \frac{e^{x} (1 + x^{2} - 2 x)}{{(1 + x^{2})}^{2}} d x = \int e^{x} [\frac{(1 + x^{2})}{{(1 + x^{2})}^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x = \int e^{x} (\frac{1}{1 + x^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}) d x = \frac{e^{x}}{1 + x^{2}} + C [∵ \int e^{x} \{f (x) + f' (x)\} d x = e^{x} f (x) + C where, f (x) = \frac{1}{1 + x^{2}} \Rightarrow f' (x) = - \frac{2 x}{{(1 + x^{2})}^{2}}]$

Page No 18.199:

Question 121:

$\int \frac{e^{m \tan^{- 1} x}}{{(1 + x^{2})}^{3 / 2}} d x$

Answer:

$We have, I = \int \frac{e^{m \tan^{- 1} x}}{{(1 + x^{2})}^{\frac{3}{2}}} d x Putting \tan^{- 1} x = t \Rightarrow x = \tan t \Rightarrow \frac{1}{1 + x^{2}} d x = d t \Rightarrow d x = (1 + x^{2}) d t \Rightarrow d x = (1 + \tan^{2} t) d t ∴ I = \int \frac{e^{m t}}{{(1 + \tan^{2} t)}^{\frac{3}{2}}} (1 + \tan^{2} t) d t = \int \frac{e^{m t} d t}{\sqrt{1 + \tan^{2} t}} = \int {\underset{II}{e}}^{m t} \underset{I}{\cos} t d t = \cos t \frac{e^{m t}}{m} - \int (- \sin t) \frac{e^{m t}}{m} d t = \cos t \frac{e^{m t}}{m} + \frac{1}{m} \int e^{m t} \sin t d t = \cos t \frac{e^{m t}}{m} + \frac{1}{m} I_{1} . . . . . (1) Where, I_{1} = \int {\underset{II}{e}}^{m t} \underset{I}{\sin t} d t = \sin t \frac{e^{m t}}{m} - \int \cos t \frac{e^{m t}}{m} d t I_{1} = \sin t \frac{e^{m t}}{m} - \frac{1}{m} I . . . . . (2) from (1) and (2) I = \cos t \frac{e^{m t}}{m} + \frac{1}{m} [\sin t \frac{e^{m t}}{m} - \frac{1}{m} I] \Rightarrow I = \cos t \frac{e^{m t}}{m} + \frac{\sin t e^{m t}}{m^{2}} - \frac{1}{m^{2}} I \Rightarrow I + \frac{I}{m^{2}} = \frac{e^{m t} (m \cos t + \sin t)}{m^{2}} \Rightarrow I = \frac{e^{m t} (m \cos t + \sin t)}{1 + m^{2}} + C \Rightarrow I = \frac{e^{m t}}{\sqrt{1 + m^{2}}} [\cos t \frac{m}{\sqrt{1 + m^{2}}} + \sin t \frac{1}{\sqrt{1 + m^{2}}}] + C Let \frac{m}{\sqrt{1 + m^{2}}} = \cos θ Then, \sin θ = \frac{1}{\sqrt{1 + m^{2}}} \Rightarrow \cot θ = m \Rightarrow θ = \cot^{- 1} m ∴ I = \frac{e^{m t}}{\sqrt{1 + m^{2}}} \{\cos t \cos θ + \sin t \sin θ\} + C = \frac{e^{m t}}{\sqrt{1 + m^{2}}} \{\cos (t - θ)\} + C = \frac{e^{m t}}{\sqrt{1 + m^{2}}} \{\cos (\tan^{- 1} x - \cot^{- 1} m)\} + C$

Page No 18.199:

Question 122:

$\int \frac{x^{2}}{{(x - 1)}^{3} (x + 1)} d x$

Answer:

$We have, I = \int \frac{x^{2}}{{(x - 1)}^{3} (x + 1)} d x Let \frac{x^{2}}{{(x - 1)}^{3} (x + 1)} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}} + \frac{D}{x + 1} . . . . . (1) \Rightarrow x^{2} = A {(x - 1)}^{2} (x + 1) + B (x - 1) (x + 1) + C (x + 1) + D {(x - 1)}^{3} . . . . . (2) Putting x = 1 in (2), we get 1 = 2 C \Rightarrow C = \frac{1}{2} Putting x = - 1 in (2), we get 1 = - 8 D \Rightarrow D = \frac{- 1}{8}$

$Putting x = 2 in (2), we get 4 = 3 A + 3 B + 3 C + D \Rightarrow 4 = 3 A + 3 B + \frac{3}{2} - \frac{1}{8} \Rightarrow 3 A + 3 B = 4 - \frac{3}{2} + \frac{1}{8} \Rightarrow 3 A + 3 B = \frac{32 - 12 + 1}{8} \Rightarrow 3 A + 3 B = \frac{21}{8} \Rightarrow A + B = \frac{7}{8} And putting x = 0 in (2), we get 0 = A - B + C - D \Rightarrow 0 = A - B + \frac{1}{2} + \frac{1}{8} [∵ C = \frac{1}{2}, D = \frac{1}{8}] \Rightarrow A - B = - \frac{5}{8}$
$Here, A + B = \frac{7}{8} and A - B = - \frac{5}{8} \Rightarrow A = \frac{1}{8} and B = \frac{3}{4} Therefore, (1) becomes, \frac{x^{2}}{{(x - 1)}^{3} (x + 1)} = \frac{1}{8 (x - 1)} + \frac{3}{4 {(x - 1)}^{2}} + \frac{1}{2 {(x - 1)}^{3}} - \frac{1}{8 (x + 1)} Now, integral becomes I = \int [\frac{1}{8 (x - 1)} + \frac{3}{4 {(x - 1)}^{2}} + \frac{1}{2 {(x - 1)}^{3}} - \frac{1}{8 (x + 1)}] d x = \frac{1}{8} \log |x - 1| - \frac{3}{4 (x - 1)} - \frac{1}{4 {(x - 1)}^{2}} - \frac{1}{8} \log |x + 1| + C = \frac{1}{8} \log |\frac{x - 1}{x + 1}| - \frac{3}{4 (x - 1)} - \frac{1}{4 {(x - 1)}^{2}} + C$

Page No 18.199:

Question 123:

$\int \frac{x}{x^{3} - 1} d x$

Answer:

$We have, I = \int \frac{x d x}{x^{3} - 1} = \int \frac{x d x}{(x - 1) (x^{2} + x + 1)} Let \frac{x}{(x - 1) (x^{2} + x + 1)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + x + 1} \Rightarrow \frac{x}{(x - 1) (x^{2} + x + 1)} = \frac{A (x^{2} + x + 1) + (B x + C) (x - 1)}{(x - 1) (x^{2} + x + 1)} \Rightarrow x = A (x^{2} + x + 1) + B x^{2} - B x + C x - C \Rightarrow x = (A + B) x^{2} + (A - B + C) x + A - C Equating Coefficient of like terms A + B = 0 . . . . . (1) A - B + C = 1 . . . . . (2) A - C = 0 . . . . . (3) Solving (1), (2) and (3), we get A = \frac{1}{3} B = - \frac{1}{3} C = \frac{1}{3} ∴ \frac{x}{(x - 1) (x^{2} + x + 1)} = \frac{1}{3 (x - 1)} + \frac{- \frac{1}{3} x + \frac{1}{3}}{x^{2} + x + 1} = \frac{1}{3 (x - 1)} + \frac{1}{3} [\frac{- x + 1}{x^{2} + x + 1}] = \frac{1}{3 (x - 1)} - \frac{1}{3} [\frac{x - 1}{x^{2} + x + 1}] = \frac{1}{3 (x - 1)} - \frac{1}{6} [\frac{2 x - 2}{x^{2} + x + 1}] = \frac{1}{3 (x - 1)} - \frac{1}{6} [\frac{2 x + 1}{x^{2} + x + 1}] - \frac{1}{6} \times \frac{- 3}{x^{2} + x + 1} = \frac{1}{3 (x - 1)} - \frac{1}{6} [\frac{2 x + 1}{x^{2} + x + 1}] + \frac{1}{2} \times \frac{1}{x^{2} + x + 1} ∴ I = \frac{1}{3} \int \frac{d x}{x - 1} - \frac{1}{6} \int \frac{(2 x + 1) d x}{x^{2} + x + 1} + \frac{1}{2} \int \frac{d x}{x^{2} + x + \frac{1}{4} - \frac{1}{4} + 1} Putting x^{2} + x + 1 = t \Rightarrow (2 x + 1) d x = d t ∴ I = \frac{1}{3} \log |x - 1| - \frac{1}{6} \log |t| + \frac{1}{2} \int \frac{d x}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{1}{3} \log |x - 1| - \frac{1}{6} \log |x^{2} + x + 1| + \frac{1}{2} [\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}})] + C = \frac{1}{3} \log |x - 1| - \frac{1}{6} \log |x^{2} + x + 1| + \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) + C$

Page No 18.199:

Question 124:

$\int \frac{1}{1 + x + x^{2} + x^{3}} d x$

Answer:

$We have, I = \int \frac{d x}{1 + x + x^{2} + x^{3}} = \int \frac{d x}{(1 + x) + x^{2} (1 + x)} = \int \frac{d x}{(1 + x) (1 + x^{2})} Let \frac{1}{(x + 1) (1 + x^{2})} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} + 1} \Rightarrow \frac{1}{(x + 1) (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (x + 1)}{(x + 1) (x^{2} + 1)} \Rightarrow 1 = A (x^{2} + 1) + B x^{2} + B x + C x + C \Rightarrow 1 = (A + B) x^{2} + (B + C) x + (A + C) Equating Coefficient of like terms A + B = 0 . . . . . (1) B + C = 0 . . . . . (2) A + C = 1 . . . . . (3) Solving (1), (2) and (3), we get, A = \frac{1}{2} B = - \frac{1}{2} C = \frac{1}{2} ∴ \frac{1}{(x + 1) (x^{2} + 1)} = \frac{1}{2 (x + 1)} + \frac{- \frac{x}{2} + \frac{1}{2}}{x^{2} + 1} \Rightarrow \frac{1}{(x + 1) (x^{2} + 1)} = \frac{1}{2 (x + 1)} - \frac{1}{2} (\frac{x}{x^{2} + 1}) + \frac{1}{2 (x^{2} + 1)} ∴ I = \frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{2} \int \frac{x d x}{(x^{2} + 1)} + \frac{1}{2} \int \frac{d x}{x^{2} + 1} Putting x^{2} + 1 = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d x}{x + 1} - \frac{1}{4} \int \frac{d t}{t} + \frac{1}{2} \int \frac{d x}{x^{2} + 1} = \frac{1}{2} \log |x + 1| - \frac{1}{4} \log |t| + \frac{1}{2} \tan^{- 1} x + C = \frac{1}{2} \log |x + 1| - \frac{1}{4} \log |x^{2} + 1| + \frac{1}{2} \tan^{- 1} (x) + C = \frac{1}{2} \log |x + 1| - \frac{1}{2} \log (\sqrt{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (x) + C = \frac{1}{2} \log (\frac{|x + 1|}{\sqrt{x^{2} + 1}}) + \frac{1}{2} \tan^{- 1} (x) + C$

Page No 18.199:

Question 125:

$\int \frac{1}{(x^{2} + 2) (x^{2} + 5)} d x$

Answer:

$We have, I = \int \frac{d x}{(x^{2} + 2) (x^{2} + 5)} Putting x^{2} = t ∴ \frac{1}{(x^{2} + 2) (x^{2} + 5)} = \frac{1}{(t + 2) (t + 5)} Let \frac{1}{(t + 2) (t + 5)} = \frac{A}{t + 2} + \frac{B}{t + 5} \Rightarrow \frac{1}{(t + 2) (t + 5)} = \frac{A (t + 5) + B (t + 2)}{(t + 2) (t + 5)} \Rightarrow 1 = A (t + 5) + B (t + 2) Putting t = - 5 ∴ 1 = B (- 5 + 2) \Rightarrow B = - \frac{1}{3} Putting t = - 2 ∴ 1 = A (- 2 + 5) + B \times 0 \Rightarrow A = \frac{1}{3} ∴ I = \frac{1}{3} \int \frac{d x}{x^{2} + 2} - \frac{1}{3} \int \frac{d x}{x^{2} + 5} = \frac{1}{3} \int \frac{d x}{x^{2} + {(\sqrt{2})}^{2}} - \frac{1}{3} \int \frac{d x}{x^{2} + {(\sqrt{5})}^{2}} = \frac{1}{3 \sqrt{2}} \tan^{- 1} (\frac{x}{\sqrt{2}}) - \frac{1}{3 \sqrt{5}} \tan^{- 1} (\frac{x}{\sqrt{5}}) + C$

Page No 18.199:

Question 126:

$\int \frac{x^{2} - 2}{x^{5} - x} d x$

Answer:

$We have, I = \int (\frac{x^{2} - 2}{x^{5} - x}) d x = \int \frac{(x^{2} - 2)}{x (x^{4} - 1)} d x = \int \frac{x (x^{2} - 2)}{x^{2} (x^{2} - 1) (x^{2} + 1)} d x Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{(t - 2)}{t (t - 1) (t + 1)} d t Let \frac{t - 2}{t (t - 1) (t + 1)} = \frac{A}{t} + \frac{B}{t - 1} + \frac{C}{t + 1} \Rightarrow \frac{t - 2}{t (t - 1) (t + 1)} = \frac{A (t - 1) (t + 1) + B t (t + 1) + C t \cdot (t - 1)}{t (t - 1) (t + 1)} \Rightarrow t - 2 = A (t - 1) (t + 1) + B t (t + 1) + C t (t - 1) Putting t = 1 ∴ 1 - 2 = B \times 2 \Rightarrow B = - \frac{1}{2} Putting t = 0 ∴ - 2 = A (- 1) \Rightarrow A = 2 Putting t = - 1 ∴ - 3 = C (- 1) (- 2) \Rightarrow C = - \frac{3}{2} ∴ I = \frac{2}{2} \int \frac{d t}{t} - \frac{1}{2 \times 2} \int \frac{d t}{t - 1} - \frac{3}{2 \times 2} \int \frac{d}{t + 1} = \log |t| - \frac{1}{4} \log |t - 1| - \frac{3}{4} \log |t + 1| + C = \log |x^{2}| - \frac{1}{4} \log |x^{2} - 1| - \frac{3}{4} \log |x^{2} + 1| + C = 2 \log |x| - \frac{1}{4} \log |x^{2} - 1| - \frac{3}{4} \log |x^{2} + 1| + C$

Page No 18.199:

Question 127:

$\int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} d x$

Answer:

$We have, I = \int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} d x Putting \sqrt{x} = \cos θ \Rightarrow x = \cos^{2} θ \Rightarrow d x = - 2 \cos θ \sin θ d θ \Rightarrow d x = - \sin (2 θ) d θ ∴ I = \int \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} (- \sin 2 θ) d θ = \int \sqrt{\frac{2 \sin^{2} \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}}} (- 2 \sin θ \cos θ) d θ = \int (\frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}}) (- 2 \times 2 \sin \frac{θ}{2} \cos \frac{θ}{2} \cos θ) d θ = - 4 \int \sin^{2} \frac{θ}{2} \times \cos θ d θ = - 4 \int (\frac{1 - \cos θ}{2}) \cos θ d θ = - 2 \int (\cos θ - \cos^{2} θ) d θ = - 2 \int \{\cos θ - (\frac{1 + \cos 2 θ}{2})\} d θ = - 2 \int \cos θ d θ + \int (1 + \cos 2 θ) d θ = - 2 \sin θ + θ + \frac{\sin 2 θ}{2} + C = - 2 \sqrt{1 - \cos^{2} θ} + θ + \frac{2 \sin θ \cos θ}{2} + C = - 2 \sqrt{1 - \cos^{2} θ} + θ + \sin θ \cos θ + C = - 2 \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{1 - x} \sqrt{x} + C = - 2 \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + \sqrt{x} \sqrt{1 - x} + C$

Page No 18.199:

Question 128:

$\int \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} d x$

Answer:

$We have, I = \int \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} d x Let \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{C}{x + 2} . . . . . (1) \Rightarrow x^{2} + x + 1 = A (x + 1) (x + 2) + B (x + 2) + C {(x + 1)}^{2} . . . . . (2) Putting x = - 1 in (2), we get B = 1 Putting x = - 2 in (2), we get C = 3 Putting x = 0 in (2), we get 1 = 2 A + 2 B + C \Rightarrow 1 = 2 A + 2 + 3 \Rightarrow - 4 = 2 A \Rightarrow A = - 2 Now, (1) becomes \frac{x^{2} + x + 1}{{(x + 1)}^{2} (x + 2)} = \frac{- 2}{x + 1} + \frac{1}{{(x + 1)}^{2}} + \frac{3}{x + 2} Therefore, integral becomes I = \int [\frac{- 2}{x + 1} + \frac{1}{{(x + 1)}^{2}} + \frac{3}{x + 2}] d x = - 2 \log |x + 1| - \frac{1}{(x + 1)} + 3 \log |x + 2| + C$

Page No 18.199:

Question 129:

$\int \frac{\sin 4 x - 2}{1 - \cos 4 x} e^{2 x} d x$

Answer:

$We have, I = \int (\frac{\sin 4 x - 2}{1 - \cos 4 x}) e^{2 x} d x = \int (\frac{2 \sin 2 x \cos 2 x - 2}{2 \sin^{2} 2 x}) e^{2 x} d x = \int [\cot (2 x) - {cosec}^{2} (2 x)] e^{2 x} d x Let e^{2 x} \cot (2 x) = t \Rightarrow [2 e^{2 x} \cot (2 x) + e^{2 x} \{- {cosec}^{2} (2 x)\} \times 2] d x = d t \Rightarrow e^{2 x} [\cot 2 x - {cosec}^{2} (2 x)] d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int d t = \frac{t}{2} + C = \frac{1}{2} e^{2 x} \cot (2 x) + C$

Page No 18.199:

Question 130:

$\int \frac{\cot x + \cot^{3} x}{1 + \cot^{3} x} d x$

Answer:

$We have, I = \int (\frac{\cot x + \cot^{3} x}{1 + \cot^{3} x}) d x = \int [\frac{\cot x (1 + \cot^{2} x)}{1 + \cot^{3} x}] d x = \int (\frac{\cot x {cosec}^{2} x}{1 + \cot^{3} x}) d x Putting \cot x = t \Rightarrow - {cosec}^{2} x d x = d t \Rightarrow {cosec}^{2} x d x = - d t ∴ I = - \int \frac{t d t}{1 + t^{3}} = - \int \frac{t d t}{(1 + t) (t^{2} - t + 1)} Let \frac{t}{(1 + t) (t^{2} - t + 1)} = \frac{A}{t + 1} + \frac{B t + C}{t^{2} - t + 1} \Rightarrow \frac{t}{(1 + t) (t^{2} - t + 1)} = \frac{A (t^{2} - t + 1) + (B t + C) (t + 1)}{(t + 1) (t^{2} - t + 1)} \Rightarrow t = A (t^{2} - t + 1) + B t^{2} + B t + C t + C \Rightarrow t = (A + B) t^{2} + (B + C - A) t + A + C Equating Coefficients of like terms A + B = 0 . . . . . (1) B + C - A = 1 . . . . . (2) A + C = 0 . . . . . (3) Solving (1), (2) and (3), we get A = - \frac{1}{3} B = \frac{1}{3} C = \frac{1}{3} ∴ \frac{t}{(1 + t) (t^{2} - t + 1)} = - \frac{1}{3 (t + 1)} + \frac{1}{3} (\frac{t + 1}{t^{2} - t + 1}) \Rightarrow \frac{t}{(1 + t) (t^{2} - t + 1)} = - \frac{1}{3 (t + 1)} + \frac{1}{6} [\frac{2 t + 2}{t^{2} - t + 1}] \Rightarrow \frac{t}{(1 + t) (t^{2} - t + 1)} = - \frac{1}{3 (t + 1)} + \frac{1}{6} [\frac{2 t - 1 + 3}{t^{2} - t + 1}] ∴ I = - [- \frac{1}{3} \int \frac{d t}{t + 1} + \frac{1}{6} \int (\frac{2 t - 1}{t^{2} - t + 1}) d t + \frac{1}{2} \int \frac{d t}{t^{2} - t + 1}] = + \frac{1}{3} \int \frac{d t}{t + 1} - \frac{1}{6} \int (\frac{2 t - 1}{t^{2} - t + 1}) d t - \frac{1}{2} \int \frac{d t}{t^{2} - t + \frac{1}{4} - \frac{1}{4} + 1} = \frac{1}{3} \int \frac{d t}{t + 1} - \frac{1}{6} \int \frac{(2 t - 1) d t}{(t^{2} - t + 1)} - \frac{1}{2} \int \frac{d t}{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} let t^{2} - t + 1 = p \Rightarrow (2 t - 1) d t = d p ∴ I = \frac{1}{3} \int \frac{d t}{t + 1} - \frac{1}{6} \int \frac{d p}{p} - \frac{1}{2} \int \frac{d t}{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{1}{3} \log |t + 1| - \frac{1}{6} \log |p| - \frac{1}{2} \times \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{t - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C = \frac{1}{3} \log |t + 1| - \frac{1}{6} \log |p| - \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 t - 1}{\sqrt{3}}) + C = \frac{1}{3} \log |\cot x + 1| - \frac{1}{6} \log |\cot^{2} x - \cot x + 1| - \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 c o t x - 1}{\sqrt{3}}) + C$

Page No 18.203:

Question 1:

$\int \frac{x}{4 + x^{4}} d x$ is equal to

(a) $\frac{1}{4} \tan^{- 1} x^{2} + C$

(b) $\frac{1}{4} \tan^{- 1} (\frac{x^{2}}{2})$

(c) $\frac{1}{2} \tan^{- 1} (\frac{x^{2}}{2})$

(d) none of these

Answer:

(b) $\frac{1}{4} \tan^{- 1} (\frac{x^{2}}{2})$

$Let I = \int \frac{x}{4 + x^{4}} d x = \int \frac{x d x}{2^{2} + {(x^{2})}^{2}} Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{2^{2} + t^{2}} = \frac{1}{2} \times \frac{1}{2} \tan^{- 1} (\frac{t}{2}) + C (∵ \int \frac{1}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a}) = \frac{1}{4} \tan^{- 1} (\frac{x^{2}}{2}) + C (∵ t = x^{2})$

Page No 18.203:

Question 2:

$\int \frac{1}{\cos x + \sqrt{3} \sin x} d x$ is equal to

(a) $\log \tan (\frac{π}{3} + \frac{x}{2}) + C$

(b) $\log \tan (\frac{x}{2} - \frac{π}{3}) + C$

(c) $\frac{1}{2} \log \tan (\frac{x}{2} + \frac{π}{3}) + C$

(d) none of these

Answer:

(d) none of these

$\int \frac{1}{\cos x + \sqrt{3} \sin x} d x = \frac{1}{2} \int \frac{d x}{\cos x \times \frac{1}{2} + \sin x \times \frac{\sqrt{3}}{2}} = \frac{1}{2} \int \frac{d x}{\cos x \cdot \cos \frac{π}{3} + \sin x \cdot \sin \frac{π}{3}} = \frac{1}{2} \int \frac{d x}{\cos (x - \frac{π}{3})} = \frac{1}{2} \int \sec (x - \frac{π}{3}) d x = \frac{1}{2} \ln |\tan \{\frac{π}{4} + \frac{1}{2} (x - \frac{π}{3})\}| + C = \frac{1}{2} \ln |\tan (\frac{π}{4} + \frac{x}{2} - \frac{π}{6})| + C = \frac{1}{2} \ln |\tan (\frac{x}{2} + \frac{π}{12})| + C$

Page No 18.203:

Question 3:

$\int x \sec x^{2} d x is equal to$

(a) $\frac{1}{2}$ log (sec x² + tan x²) + C

(b) $\frac{x^{2}}{2}$ log (sec x² + tan x²) + C

(c) 2 log (sec x² + tan x²) + C

(d) none of these

Answer:

(a) $\frac{1}{2}$ log (sec x² + tan x²) + C

$Let I = \int x \sec x^{2} d x Putting x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \sec t \cdot d t = \frac{1}{2} \log |\sec t + \tan t| + C = \frac{1}{2} \log |\sec x^{2} + \tan x^{2}| + C (∵ t = x^{2})$

Page No 18.203:

Question 4:

If $\int \frac{1}{5 + 4 \sin x} d x = A \tan^{- 1} (B \tan \frac{x}{2} + \frac{4}{3}) + C,$ then

(a) A = $\frac{2}{3}$ , B = $\frac{5}{3}$

(b) A = $\frac{1}{3}$ , B = $\frac{2}{3}$

(c) A = $- \frac{2}{3}$ , B = $\frac{5}{3}$

(d) A = $\frac{1}{3}$ , B = $- \frac{5}{3}$

Answer:

(a) A = $\frac{2}{3}$ , B = $\frac{5}{3}$

$\int \frac{1}{5 + 4 \sin x} d x = A \tan^{- 1} (B \tan \frac{x}{2} + \frac{4}{3}) + C . . . . (1) Considering the LHS of eq (1) Putting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} \Rightarrow \int \frac{1}{5 + \frac{8 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x \Rightarrow \int \frac{(1 + \tan^{2} \frac{x}{2})}{5 (1 + \tan^{2} \frac{x}{2}) + 8 \tan \frac{x}{2}} d x \Rightarrow \int \frac{\sec^{2} \frac{x}{2}}{5 \tan^{2} \frac{x}{2} + 8 \tan \frac{x}{2} + 5} d x . . . (2) Let \tan \frac{x}{2} = t \Rightarrow \sec^{2} \frac{x}{2} \times \frac{1}{2} d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ Putting \tan \frac{x}{2} = t and \sec^{2} (\frac{x}{2}) d x = 2 d t we get, \int \frac{2 d t}{5 t^{2} + 8 t + 5} \Rightarrow \frac{2}{5} \int \frac{d t}{t^{2} + \frac{8}{5} t + 1} \Rightarrow \frac{2}{5} \int \frac{d t}{t^{2} + \frac{8}{5} t + {(\frac{4}{5})}^{2} - {(\frac{4}{5})}^{2} + 1} \Rightarrow \frac{2}{5} \int \frac{d t}{{(t + \frac{4}{5})}^{2} + 1 - \frac{16}{25}} \Rightarrow \frac{2}{5} \int \frac{d t}{{(t + \frac{4}{5})}^{2} + {(\frac{3}{5})}^{2}} \Rightarrow \frac{2}{5} \times \frac{5}{3} \tan^{- 1} (\frac{t + \frac{4}{5}}{\frac{3}{5}}) + C \Rightarrow \frac{2}{3} \tan^{- 1} (\frac{5 t + 4}{3}) + C \Rightarrow \frac{2}{3} \tan^{- 1} (\frac{5}{3} \tan \frac{x}{2} + \frac{4}{3}) + C (∵ t = \tan \frac{x}{2}) . . . (3) Comparing eq (3) with the RHS of eq (1) we get, ∴ A = \frac{2}{3}, B = \frac{5}{3}$

Page No 18.203:

Question 5:

$\int x^{\sin x} (\frac{\sin x}{x} + \cos x . \log x) d x is equal to$

(a) x^sin^x + C

(b) x^sin^x cos x + C

(c) $\frac{{(x^{\sin x})}^{2}}{2} + C$

(d) none of these

Answer:

(a) x^sin^x + C

$Let I = \int x^{\sin x} (\frac{\sin x}{x} + \cos x \cdot \log x) d x Putting x^{\sin x} = t \Rightarrow \ln {(x)}^{\sin x} = \ln t \Rightarrow \sin x \cdot \ln x = \ln t \Rightarrow (\sin x \times \frac{1}{x} + \cos x \ln x) d x = \frac{1}{t} d t ∴ I = \int t \cdot \frac{d t}{t} = t + C = x^{\sin x} + C (∵ t = x^{\sin x})$

Page No 18.204:

Question 6:

Integration of $\frac{1}{1 + {(\log_{e} x)}^{2}}$ with respect to log_e x is
(a) $\frac{\tan^{- 1} (\log_{e} x)}{x} + C$

(b) $\tan^{- 1} (\log_{e} x) + C$

(c) $\frac{\tan^{- 1} x}{x} + C$

(d) none of these

Answer:

(b) $\tan^{- 1} (\log_{e} x) + C$

$We have to integrate \frac{1}{1 + {(\log_{e} x)}^{2}} with respect to \log_{e} x Let I = \int \frac{d (\log_{e} x)}{1 + {(\log_{e} x)}^{2}} Putting \log_{e} x = t d (\log_{e} x) = d t ∴ I = \int \frac{d t}{1 + t^{2}} = \tan^{- 1} (t) + C = \tan^{- 1} (\log_{e} x) + C (∵ t = \log_{e} x)$

Page No 18.204:

Question 7:

If $\int \frac{\cos 8 x + 1}{\tan 2 x - \cot 2 x} d x$ = a cos 8x + C, then a =

(a) $- \frac{1}{16}$

(b) $\frac{1}{8}$

(c) $\frac{1}{16}$

(d) $- \frac{1}{8}$

Answer:

(c) $\frac{1}{16}$

$If \int (\frac{\cos 8 x + 1}{\tan 2 x - \cot 2 x}) d x = a \cos 8 x + C . . . (1) Considering the LHS of eq (1) \int (\frac{\cos 8 x + 1}{\tan 2 x - \cot 2 x}) d x \Rightarrow \int (\frac{2 \cos^{2} 4 x}{\frac{\sin 2 x}{\cos 2 x} - \frac{\cos 2 x}{\sin 2 x}}) d x \Rightarrow \int \frac{2 \cos^{2} 4 x}{(\sin^{2} 2 x - \cos^{2} 2 x)} \times \sin 2 x \cos 2 x \Rightarrow \int [\frac{- \cos^{2} 4 x \times 2 \sin 2 x \cdot \cos 2 x}{\cos^{2} 2 x - \sin^{2} 2 x}] d x \Rightarrow \int \frac{- \cos^{2} 4 x \times \sin 4 x}{\cos 4 x} d x (∵ \cos 2 x = \cos^{2} x - \sin^{2} x) \Rightarrow \frac{1}{2} \int - 2 \sin 4 x \cos 4 x d x \Rightarrow \frac{- 1}{2} \int \sin 8 x d x \Rightarrow - \frac{1}{2} [\frac{- \cos 8 x}{8}] + C = \frac{1}{16} [\cos 8 x] + C . . . . (2) Comparing RHS of eq (1) with the eq (2) ∴ a = \frac{1}{16}$

Page No 18.204:

Question 8:

If $\int \frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x} d x$ = a sin 2x + C, then a =

(a) −1/2
(b) 1/2
(c) −1
(d) 1

Answer:

(a) −1/2

$If \int (\frac{\sin^{8} x - \cos^{8} x}{1 - 2 \sin^{2} x \cos^{2} x}) d x = a \sin 2 x + C . . . . (1) Considering LHS of eq (1) \Rightarrow \int \frac{(\sin^{4} x - \cos^{4} x) (\sin^{4} x + \cos^{4} x)}{(1 - 2 \sin^{2} x \cos^{2} x)} \Rightarrow \int \frac{(\sin^{2} x - \cos^{2} x) (\sin^{2} x + \cos^{2} x) \cdot (\sin^{4} x + \cos^{4} x) d x}{\{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x\}} \Rightarrow \int \frac{(\sin^{2} x - \cos^{2} x) \cdot (\sin^{4} x + \cos^{4} x) d x}{(\sin^{4} x + \cos^{4} x + 2 \sin^{2} x \cos^{2} x - 2 \sin^{2} x \cos^{2} x)} \Rightarrow - \int \frac{(\cos^{2} x - \sin^{2} x) \times (\sin^{4} x + \cos^{4} x) d x}{(\sin^{4} x + \cos^{4} x)} \Rightarrow - \int \cos (2 x) d x (∵ \cos^{2} x - \sin^{2} x = \cos 2 x) . . . (2) Comparing the RHS of eq (1) with eq (2) we get, a = - \frac{1}{2}$

Page No 18.204:

Question 9:

$\int (x - 1) e^{- x} d x$ is equal to

(a) − xe^x + C

(b) xe^x + C

(c) − xe^−x + C

(d) xe^−x + C

Answer:

$\int \underset{I}{(x - 1)} \underset{II}{e^{- x}} d x = (x - 1) \int e^{- x} d x - \int \{\frac{d}{d x} (x - 1) \int e^{- x} d x\} d x = (x - 1) \cdot e^{- x} (- 1) - \int 1 \cdot e^{- x} \times - 1 d x = - (x - 1) e^{- x} + \int e^{- x} d x = (1 - x) e^{- x} + \frac{e^{- x}}{- 1} + C \Rightarrow (1 - x - 1) e^{- x} + C = - x e^{- x} + C$

Page No 18.204:

Question 10:

If $\int \frac{2^{1 / x}}{x^{2}} d x = k 2^{1 / x} + C,$ then k is equal to

(a) $- \frac{1}{\log_{e} 2}$

(b) − log_e 2

(c) − 1

(d) $\frac{1}{2}$

Answer:

(a) $- \frac{1}{\log_{e} 2}$

$If \int \frac{2^{\frac{1}{x}}}{x^{2}} d x = k \cdot 2^{\frac{1}{x}} + C . . . . (1) Let \frac{1}{x} = t \Rightarrow \frac{- 1}{x^{2}} d x = d t \Rightarrow \frac{d x}{x^{2}} = - d t Putting \frac{1}{x} = t and \frac{d x}{x^{2}} = - d t in LHS of eq (1), we get - \int 2^{t} \cdot d t \Rightarrow - \frac{2^{t}}{\ln 2} + C \Rightarrow - \frac{2^{\frac{1}{x}}}{\ln 2} + C . . . (2) Comparing RHS of eq (1) with eq (2) we get, ∴ k = - \frac{1}{\ln 2} or - \frac{1}{\log_{e} 2}$

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Question 11:

$\int \frac{1}{1 + \tan x} d x =$

(a) log_e (x + sin x) + C

(b) log_e (sin x + cos x) + C

(c) $2 \sec^{2} \frac{x}{2} + C$

(d) $\frac{1}{2}$ [x + log (sin x + cos x)] + C

Answer:

Disclaimer : Generally here book is taking log_e x as log x . So we are writing ln x or log_e x instead log x only .
(d) $\frac{1}{2}$ [x + ln (sin x + cos x)] + C

$Let I = \int \frac{1}{1 + \tan x} d x = \int \frac{1}{1 + \frac{\sin x}{\cos x}} d x = \int \frac{\cos x}{\cos x + \sin x} d x = \frac{1}{2} \int \frac{2 \cos x}{\cos x + \sin x} d x = \frac{1}{2} \int [\frac{(\cos x + \sin x) + (\cos x - \sin x)}{(\cos x + \sin x)}] d x = \frac{1}{2} \int (\frac{\cos x + \sin x}{\cos x + \sin x}) d x + \frac{1}{2} \int (\frac{\cos x - \sin x}{\cos x + \sin x}) d x = \frac{1}{2} \int d x + \frac{1}{2} \int (\frac{\cos x - \sin x}{\cos x + \sin x}) d x Putting \sin x + \cos x = t \Rightarrow (\cos x - \sin x) d x = d t ∴ I = \frac{1}{2} \int d x + \frac{1}{2} \int \frac{d t}{t} = \frac{x}{2} + \frac{1}{2} \ln |t| + C = \frac{x}{2} + \frac{1}{2} \ln |\cos x + \sin x| + C (∵ t = \sin x + \cos x) = \frac{1}{2} [x + \ln (\sin x + \cos x)] + C$

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Question 12:

$\int {|x|}^{3} d x$ is equal to

(a) $\frac{- x^{4}}{4} + C$

(b) $\frac{{|x|}^{4}}{4} + C$

(c) $\frac{x^{4}}{4} + C$

(d) none of these

Answer:

(d) none of these

$\int {|x|}^{3} d x |x| = \{\begin{cases} x, x \geq 0 \\ - x, x < 0 \end{cases} Case 1 : When x \geq 0 ∴ \int {|x|}^{3} d x = \int x^{3} d x = \frac{x^{4}}{4} + C Case 2 : x < 0 \int {|x|}^{3} d x = - \int x^{3} d x = \frac{- x^{4}}{4} + C$

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Question 13:

The value of $\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$ is
(a) 2 cos $\sqrt{x}$ + C

(b) $\sqrt{\frac{\cos x}{x}} + C$

(c) sin $\sqrt{x} + C$

(d) 2 sin $\sqrt{x} + C$

Answer:

(d) 2 sin $\sqrt{x} + C$

$Let I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x Putting \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t ∴ I = 2 \int \cos t \cdot d t = 2 \sin t + C = 2 \sin \sqrt{x} + C (∵ t = \sqrt{x})$

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Question 14:

$\int e^{x} (1 - \cot x + \cot^{2} x) d x =$
(a) e^x cot x + C
(b) −e^x cot x + C
(c) e^x cosec x + C
(d) −e^x cosec x + C

Answer:

(b) −e^x cot x + C

$Let I = \int e^{x} (1 - \cot x + \cot^{2} x) d x = \int e^{x} ({cosec}^{2} x - \cot x) d x As we know that \int e \{f (x) + f'^{x} (x)\} = e^{x} f (x) + C ∴ I = - e^{x} \cot x + C$

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Question 15:

$\int \frac{\sin^{6} x}{\cos^{8} x} d x =$

(a) tan 7x + C

(b) $\frac{\tan^{7} x}{7} + C$

(c) $\frac{\tan 7 x}{7} + C$

(d) sec⁷ x + C

Answer:

(b) $\frac{\tan^{7} x}{7} + C$

$Let I = \int \frac{\sin^{6} x}{\cos^{8} x} d x = \int \frac{\sin^{6} x}{\cos^{6} x} \times \frac{1}{\cos^{2} x} d x = \int \tan^{6} x \cdot \sec^{2} x d x Putting \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int t^{6} \cdot dt = \frac{t^{7}}{7} + C = \frac{\tan^{7} x}{7} + C (∵ t = \tan x)$

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Question 16:

$\int \frac{1}{7 + 5 \cos x} d x =$

(a) $\frac{1}{\sqrt{6}} \tan^{- 1} (\frac{1}{\sqrt{6}} \tan \frac{x}{2}) + C$

(b) $\frac{1}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}} \tan \frac{x}{2}) + C$

(c) $\frac{1}{4} \tan^{- 1} (\tan \frac{x}{2}) + C$

(d) $\frac{1}{7} \tan^{- 1} (\tan \frac{x}{2}) + C$

Answer:

(a) $\frac{1}{\sqrt{6}} \tan^{- 1} (\frac{1}{\sqrt{6}} \tan \frac{x}{2}) + C$

$Let I = \int \frac{d x}{7 + 5 \cos x} Putting \cos x = \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} ∴ I = \int \frac{d x}{7 + 5 \times (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} = \int \frac{(1 + \tan^{2} \frac{x}{2}) d x}{7 (1 + \tan^{2} \frac{x}{2}) + 5 - 5 \tan^{2} \frac{x}{2}} = \int \frac{\sec^{2} \frac{x}{2} dx}{2 \tan^{2} \frac{x}{2} + 12} = \frac{1}{2} \int \frac{\sec^{2} \frac{x}{2} d x}{\tan^{2} \frac{x}{2} + {(\sqrt{6})}^{2}} Let \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \frac{1}{2} \int \frac{2 d t}{t^{2} + {(\sqrt{6})}^{2}} = \frac{1}{\sqrt{6}} \tan^{- 1} (\frac{t}{\sqrt{6}}) + C (∵ \int \frac{1}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C) = \frac{1}{\sqrt{6}} \tan^{- 1} (\frac{\tan \frac{x}{2}}{\sqrt{6}}) + C (∵ t = \tan \frac{x}{2})$

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Question 17:

$\int \frac{1}{1 - \cos x - \sin x} d x =$

(a) $\log |1 + \cot \frac{x}{2}| + C$

(b) $\log |1 - \tan \frac{x}{2}| + C$

(c) $\log |1 - \cot \frac{x}{2}| + C$

(d) $\log |1 + \tan \frac{x}{2}| + C$

Answer:

(c) $\log |1 - \cot \frac{x}{2}| + C$
Disclaimer : Here in answer $\log |1 - \cot \frac{x}{2}| + C$ refers to $\log_{e} |1 - \cot \frac{x}{2}| + C or \ln |1 - \cot \frac{x}{2}| + C$

$Let I = \int \frac{d x}{1 - \cos x - \sin x} = \int \frac{d x}{1 - (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} = \int \frac{(1 + \tan^{2} \frac{x}{2}) d x}{(1 + \tan^{2} \frac{x}{2}) - (1 - \tan^{2} \frac{x}{2}) - 2 \tan \frac{x}{2}} = \int \frac{\sec^{2} \frac{x}{2} d x}{2 \tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2}} = \frac{1}{2} \int \frac{\sec^{2} \frac{x}{2} d x}{\tan^{2} \frac{x}{2} - \tan \frac{x}{2}} Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t \Rightarrow \sec^{2} (\frac{x}{2}) d x = 2 d t ∴ I = \frac{1}{2} \int \frac{2 d t}{t^{2} - t} = \int \frac{d t}{t^{2} - t + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} = \int \frac{d t}{{(t - \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} = \frac{1}{2 \times \frac{1}{2}} \ln |\frac{t - \frac{1}{2} - \frac{1}{2}}{t - \frac{1}{2} + \frac{1}{2}}| + C (∵ \int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} \ln |\frac{x - a}{x + a}| + C) = \ln |\frac{t - 1}{t}| + C = \ln |1 - \frac{1}{t}| + C = \ln |1 - \frac{1}{\tan \frac{x}{2}}| + C (∵ t = \tan \frac{x}{2}) = \ln |1 - \cot \frac{x}{2}| + C$

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Question 18:

$\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x =$

(a) $\frac{e^{x}}{x + 4} + C$

(b) $\frac{e^{x}}{x + 3} + C$

(c) $\frac{1}{{(x + 4)}^{2}} + C$

(d) $\frac{e^{x}}{{(x + 4)}^{2}} + C$

Answer:

(a) $\frac{e^{x}}{x + 4} + C$

$Let I = \int \frac{(x + 3)}{{(x + 4)}^{2}} e^{x} d x \Rightarrow \int [\frac{x + 4 - 1}{{(x + 4)}^{2}}] e^{x} d x \Rightarrow \int [\frac{1}{(x + 4)} - \frac{1}{{(x + 4)}^{2}}] e^{x} d x As, we know that \int e^{x} \{f (x) + f' (x)\} d x = e^{x} f (x) + C ∴ I = \frac{e^{x}}{x + 4} + C$

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Question 19:

$\int \frac{\sin x}{3 + 4 \cos^{2} x} d x$

(a) log (3 + 4 cos² x) + C

(b) $\frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{\cos x}{\sqrt{3}}) + C$

(c) $- \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C$

(d) $\frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C$

Answer:

(c) $- \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C$

$Let I = \int \frac{\sin x}{3 + 4 \cos^{2} x} d x Putting \cos x = t \Rightarrow - \sin x d x = d t ∴ I = \int \frac{- d t}{3 + 4 t^{2}} = \frac{1}{4} \int \frac{- d t}{t^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{- 1}{4} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} (\frac{t \times 2}{\sqrt{3}}) + C (∵ \int \frac{1}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C) = - \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 t}{\sqrt{3}}) + C = - \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C (∵ t = \cos x)$

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Question 20:

$\int e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x$

(a) $- e^{x} \tan \frac{x}{2} + C$

(c) $- \frac{1}{2} e^{x} \tan \frac{x}{2} + C$

(d) $- \frac{1}{2} e^{x} \cot \frac{x}{2} + C$

Answer:

(b) $- e^{x} \cot \frac{x}{2} + C$

$Let I = \int e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x \Rightarrow \int e^{x} (\frac{1}{1 - \cos x} - \frac{\sin x}{1 - \cos x}) d x \Rightarrow \int e^{x} (\frac{1}{2 \sin^{2} \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^{2} \frac{x}{2}}) d x \Rightarrow \int e^{x} (\frac{1}{2} {cosec}^{2} \frac{x}{2} - \cot \frac{x}{2}) d x As, we know that \int e^{x} \{f (x) + f' (x)\} d x = e^{x} f (x) + C ∴ I = - e^{x} \cot (\frac{x}{2}) + C$

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Question 21:

$\int \frac{2}{{(e^{x} + e^{- x})}^{2}} d x$

(a) $\frac{- e^{- x}}{e^{x} + e^{- x}} + C$

(b) $- \frac{1}{e^{x} + e^{- x}} + C$

(c) $\frac{- 1}{{(e^{x} + 1)}^{2}} + C$

(d) $\frac{1}{e^{x} - e^{- x}} + C$

Answer:

(a) $\frac{- e^{- x}}{e^{x} + e^{- x}} + C$

$Let I = \int \frac{2 d x}{{(e^{x} + e^{- x})}^{2}} = \int \frac{2 d x}{{(e^{x} + \frac{1}{e^{x}})}^{2}} = 2 \int \frac{e^{2 x} d x}{{(e^{2 x} + 1)}^{2}} Let e^{2 x} + 1 = t \Rightarrow e^{2 x} \cdot 2 d x = d t \Rightarrow e^{2 x} \cdot d x = \frac{d t}{2} ∴ I = 2 \times \frac{1}{2} \int \frac{d t}{t^{2}} = - \frac{1}{t} + C = - \frac{1}{e^{2 x} + 1} + C (∵ t = e^{2 x} + 1)$

Dividing numerator and denominator by e^x
$\Rightarrow I = \frac{- \frac{1}{e^{x}}}{e^{x} + \frac{1}{e^{x}}} = \frac{- e^{- x}}{e^{x} + e^{- x}} + C$

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Question 22:

$\int \frac{e^{x} (1 + x)}{\cos^{2} (x e^{x})} d x =$

(a) 2 log_e cos (xe^x) + C

(b) sec (xe^x) + C

(c) tan (xe^x) + C

(d) tan (x + e^x) + C

Answer:

(c) tan (xe^x) + C

$Let I = \int \frac{e^{x} (1 + x)}{\cos^{2} (x e^{x})} d x Putting x e^{x} = t \Rightarrow (1 \cdot e^{x} + x e^{x}) d x = d t \Rightarrow e^{x} (1 + x) d x = d t ∴ I = \int \frac{d t}{\cos^{2} t} = \int \sec^{2} t d t = \tan t + C = \tan (x e^{x}) + C (∵ t = x e^{x})$

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Question 23:

$\int \frac{\sin^{2} x}{\cos^{4} x} d x =$

(a) $\frac{1}{3} \tan^{2} x + C$

(b) $\frac{1}{2} \tan^{2} x + C$

(c) $\frac{1}{3} \tan^{3} x + C$

(d) none of these

Answer:

(c) $\frac{1}{3} \tan^{3} x + C$

$Let I = \int \frac{\sin^{2} x d x}{\cos^{4} x} = \int \frac{\sin^{2} x}{\cos^{2} x} \times \frac{1}{\cos^{2} x} d x = \int \tan^{2} x \cdot \sec^{2} x d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int t^{2} \cdot d t = \frac{t^{3}}{3} + C = \frac{\tan^{3} x}{3} + C (∵ t = \tan x)$

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Question 24:

The primitive of the function
$f (x) = (1 - \frac{1}{x^{2}}) a^{x + \frac{1}{x}}, a > 0 is$

(a) $\frac{a^{x + \frac{1}{x}}}{\log_{e} a}$

(b) $\log_{e} a \cdot a^{x + \frac{1}{x}}$

(c) $\frac{a^{x + \frac{1}{x}}}{x} \log_{e} a$

(d) $x \frac{a^{x + \frac{1}{x}}}{\log_{e} a}$

Answer:

(a) $\frac{a^{x + \frac{1}{x}}}{\log_{e} a}$

$f (x) = (1 - \frac{1}{x^{2}}) \cdot a^{x + \frac{1}{x}} ∴ \int f (x) d x = \int (1 - \frac{1}{x^{2}}) \cdot a^{x + \frac{1}{x}} d x Let (x + \frac{1}{x}) = t \Rightarrow (1 - \frac{1}{x^{2}}) d x = d t ∴ \int f (x) d x = \int a^{t} \cdot d t = \frac{a^{t}}{\log_{e} a} + C = \frac{a^{x + \frac{1}{x}}}{\log_{e} a} + C (∵ t = x + \frac{1}{x})$

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Question 25:

The value of $\int \frac{1}{x + x \log x} d x$ is
(a) 1 + log x
(b) x + log x
(c) x log (1 + log x)
(d) log (1 + log x)

Answer:

(d) log (1 + log x)

$Let I = \int \frac{d x}{x + x \log x} \Rightarrow \int \frac{d x}{x (1 + \log x)} Putting 1 + \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = \int \frac{d t}{t} = \ln |t| + C = \ln |1 + \log x| + C$

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Question 26:

$\int \sqrt{\frac{x}{1 - x}} d x$ is equal to
(a) $\sin^{- 1} \sqrt{x} + C$

(b) $\sin^{- 1} \{\sqrt{x} - \sqrt{x (1 - x)}\} + C$

(c) $\sin^{- 1} \{\sqrt{x (1 - x)}\} + C$

(d) $\sin^{- 1} \sqrt{x} - \sqrt{x (1 - x)} + C$

Answer:

(d) $\sin^{- 1} \sqrt{x} - \sqrt{x (1 - x)} + C$

$Let I = \int \sqrt{\frac{x}{1 - x}} d x Putting \sqrt{x} = \sin θ \Rightarrow x = \sin^{2} θ \Rightarrow d x = 2 \sin θ \cos θ d θ \Rightarrow d x = \sin (2 θ) d θ ∴ I = \int \sqrt{\frac{\sin^{2} θ}{1 - \sin^{2} θ}} \times \sin (2 θ) \cdot d θ = \int \frac{\sin θ}{\cos θ} \times 2 \sin θ \cdot \cos θ d θ = \int 2 \sin^{2} θ \cdot d θ = \int (1 - \cos 2 θ) d θ = θ - \frac{\sin (2 θ)}{2} + C = θ - \frac{2 \sin θ \cos θ}{2} + C = θ - \sin θ \sqrt{1 - \sin^{2} θ} + C = \sin^{- 1} \sqrt{x} - \sqrt{x} \sqrt{1 - x} + C (∵ θ = \sin^{- 1} \sqrt{x}) = \sin^{- 1} \sqrt{x} - \sqrt{x (1 - x)} + C$

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Question 27:

$\int e^{x} \{f (x) + f' (x)\} d x =$

(a) e^x f (x) + C
(b) e^x + f (x)
(c) 2e^x f (x)
(d) e^x − f (x)

Answer:

(a) e^x f (x) + C

$Let I = \int e^{x} \{f (x) + f' (x)\} d x Putting e^{x} f (x) = t \Rightarrow [e^{x} \cdot f (x) + e^{x} f' (x)] d x = d t ∴ I = \int d t = t + C = e^{x} f (x) + C [∵ t = e^{x} f (x)]$

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Question 28:

The value of $\int \frac{\sin x + \cos x}{\sqrt{1 - \sin 2 x}} d x$ is equal to
(a) $\sqrt{\sin 2 x} + C$
(b) $\sqrt{\cos 2 x} + C$
(c) ± (sin x − cos x) + C
(d) ± log (sin x − cos x) + C

Answer:

(d) ± log (sin x − cos x) + C

$Let I = \int \frac{(\sin x + \cos x) d x}{\sqrt{1 - \sin 2 x}} = \int \frac{(\sin x + \cos x) d x}{\sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cos x}} = \int \frac{(\sin x + \cos x) d x}{\sqrt{{(\sin x - \cos x)}^{2}}} = \int \frac{(\sin x + \cos x) d x}{|\sin x - \cos x|} = \pm \int (\frac{\sin x + \cos x}{\sin x - \cos x}) d x Let \sin x - \cos x = t \Rightarrow (\cos x + \sin x) d x = d t ∴ I = \pm \int \frac{d t}{t} = \pm \ln |t| + C = \pm \ln |\sin x - \cos x| + C (∵ t = \sin x - \cos x)$

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Question 29:

If $\int x \sin x d x$ = −x cos x + α, then α is equal to
(a) sin x + C
(b) cos x + C
(c) C
(d) none of these

Answer:

(a) sin x + C

$\int \underset{I}{x} \cdot \underset{II}{\sin x} d x = - x \cos x + α \Rightarrow x \int \sin x d x - \int \{\frac{d}{d x} (x) \int \sin x d x\} d x = - x \cos x + α \Rightarrow x (- \cos x) - \int 1 \cdot (- \cos x) d x = - x \cos x + α \Rightarrow - x \cos x + \int \cos x d x = - x \cos x + α \Rightarrow - x \cos x + \sin x + C = - x \cos x + α ∴ α = \sin x + C$

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Question 30:

$\int \frac{\cos 2 x - 1}{\cos 2 x + 1} d x =$
(a) tan x − x + C
(b) x + tan x + C
(c) x − tan x + C
(d) − x − cot x + C

Answer:

(c) x − tan x + C

$\int (\frac{\cos 2 x - 1}{\cos 2 x + 1}) d x = \int (\frac{1 - 2 \sin^{2} x - 1}{2 \cos^{2} x - 1 + 1}) d x = - \int \tan^{2} x d x = - \int (\sec^{2} x - 1) d x = \int (1 - \sec^{2} x) d x = x - \tan x + C$

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Question 31:

$\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to

(a) $2 (\sin x + x \cos θ) + C$
(b) $2 (\sin x - x \cos θ) + C$
(c) $2 (\sin x + 2 x \cos θ) + C$
(d) $2 (\sin x - 2 x \cos θ) + C$

Answer:

$Let I = \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x = \int \frac{(2 \cos^{2} x - 1) - (2 \cos^{2} θ - 1)}{\cos x - \cos θ} d x = \int \frac{2 \cos^{2} x - 1 - 2 \cos^{2} θ + 1}{\cos x - \cos θ} d x = \int \frac{2 (\cos x - \cos θ) (\cos x + \cos θ)}{\cos x - \cos θ} d x = \int 2 (\cos x + \cos θ) d x = 2 (\sin x + x \cos θ) + C Therefore, \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x = 2 (\sin x + x \cos θ) + C Hence, the correct option is (a) .$

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Question 32:

$\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$ is equal to

$(a) \frac{1}{5 x} {(4 + \frac{1}{x^{2}})}^{- 5} + C (b) \frac{1}{5} {(4 + \frac{1}{x^{2}})}^{- 5} + C (c) \frac{1}{10 x} {(\frac{1}{x^{2}} + 4)}^{- 5} + C (d) \frac{1}{10} {(\frac{1}{x^{2}} + 4)}^{- 5} + C$

Answer:

$Let I = \int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x = \int \frac{x^{9}}{x^{12} {(4 + \frac{1}{x^{2}})}^{6}} d x = \int \frac{\frac{1}{x^{3}}}{{(4 + \frac{1}{x^{2}})}^{6}} d x Let (4 + \frac{1}{x^{2}}) = t On differentiating both sides, we get - \frac{2}{x^{3}} d x = d t ∴ I = - \frac{1}{2} \int \frac{1}{{(t)}^{6}} d t = - \frac{1}{2} (- \frac{1}{5}) t^{- 5} + C = \frac{1}{10} {(4 + \frac{1}{x^{2}})}^{- 5} + C Therefore, \int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x = \frac{1}{10} {(4 + \frac{1}{x^{2}})}^{- 5} + C Hence, the correct option is (d) .$

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Question 33:

$\int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x = a {(1 + x^{2})}^{\frac{3}{2}} + b \sqrt{1 + x^{2}} + C$ , then

$(a) a = \frac{1}{3}, b = 1 (b) a = - \frac{1}{3}, b = 1 (c) a = - \frac{1}{3}, b = - 1 (d) a = \frac{1}{3}, b = - 1$

Answer:

$Let I = \int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x = \int \frac{x . x^{2}}{\sqrt{1 + x^{2}}} d x Let (1 + x^{2}) = t On differentiating both sides, we get 2 x d x = d t ∴ I = \frac{1}{2} \int \frac{t - 1}{\sqrt{t}} d t = \frac{1}{2} \int (\frac{t}{\sqrt{t}} - \frac{1}{\sqrt{t}}) d t = \frac{1}{2} \int (t^{\frac{1}{2}} - t^{\frac{- 1}{2}}) d t = \frac{1}{2} (\frac{2}{3} t^{\frac{3}{2}} - \frac{2}{1} t^{\frac{1}{2}}) + C = (\frac{1}{3} {(1 + x^{2})}^{\frac{3}{2}} - \sqrt{1 + x^{2}}) + C Since, \int \frac{x^{3}}{\sqrt{1 + x^{2}}} d x = a {(1 + x^{2})}^{\frac{3}{2}} + b \sqrt{1 + x^{2}} + C Therefore, a = \frac{1}{3}, b = - 1 . Hence, the correct option is (d) .$

Page No 18.206:

Question 34:

$\int \frac{x^{3}}{x + 1} d x$ is equal to

$(a) x + \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log |1 - x| + C (b) x + \frac{x^{2}}{2} - \frac{x^{3}}{3} - \log |1 - x| + C (c) x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \log |1 + x| + C (d) x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log |1 + x| + C$

Answer:

$Let I = \int \frac{x^{3}}{x + 1} d x = \int \frac{x^{3} + 1 - 1}{x + 1} d x = \int (\frac{x^{3} + 1}{x + 1} - \frac{1}{x + 1}) d x = \int (\frac{(x + 1) (x^{2} - x + 1)}{x + 1} - \frac{1}{x + 1}) d x = \int (x^{2} - x + 1 - \frac{1}{x + 1}) d x = (\frac{x^{3}}{3} - \frac{x^{2}}{2} + x - \log |x + 1|) + C = \frac{x^{3}}{3} - \frac{x^{2}}{2} + x - \log |x + 1| + C Therefore, \int \frac{x^{3}}{x + 1} d x = \frac{x^{3}}{3} - \frac{x^{2}}{2} + x - \log |x + 1| + C Hence, the correct option is (d) .$

Page No 18.206:

Question 35:

If $\int \frac{1}{(x + 2) (x^{2} + 1)} d x = a \log |1 + x^{2}| + b \tan^{- 1} x + \frac{1}{5} \log |x + 2| + C$ , then

$(a) a = - \frac{1}{10}, b = - \frac{2}{5} (b) a = \frac{1}{10}, b = - \frac{2}{5} (c) a = - \frac{1}{10}, b = \frac{2}{5} (d) a = \frac{1}{10}, b = \frac{2}{5}$

Answer:

$Let I = \int \frac{1}{(x + 2) (x^{2} + 1)} d x We express, \frac{1}{(x + 2) (x^{2} + 1)} = \frac{A}{x + 2} + \frac{B x + C}{x^{2} + 1} \Rightarrow 1 = A (x^{2} + 1) + (B x + C) (x + 2) On comparing the coefficients of x^{2}, x and constants, we get 0 = A + B and 0 = 2 B + C and 1 = A + 2 C or A = \frac{1}{5} and B = - \frac{1}{5} and C = \frac{2}{5} ∴ I = \int (\frac{\frac{1}{5}}{x + 2} + \frac{- \frac{1}{5} x + \frac{2}{5}}{x^{2} + 1}) d x = \frac{1}{5} \int \frac{1}{x + 2} d x - \frac{1}{5} \int \frac{x}{x^{2} + 1} d x + \frac{2}{5} \int \frac{1}{x^{2} + 1} d x = \frac{1}{5} \log |x + 2| - \frac{1}{10} \log |x^{2} + 1| + \frac{2}{5} \tan^{- 1} x + C Since, \int \frac{1}{(x + 2) (x^{2} + 1)} d x = a \log |1 + x^{2}| + b \tan^{- 1} x + \frac{1}{5} \log |x + 2| + C Therefore, a = - \frac{1}{10} and b = \frac{2}{5} . Hence, the correct option is (c) .$

Page No 18.206:

Question 36:

$\int e^{x} (\cos x - \sin x) d x$ is equal to
(a) e^x cos x + C (b) e^x sin x + C
(c) -e^x cos x + C (d) -e^x sin x + C

Answer:

$I = \int e^{x} (\cos x - \sin x) d x$

$\Rightarrow I = \int e^{x} [\cos x + (- \sin x)] d x$

Let $f (x) = \cos x$

$\Rightarrow f' (x) = - \sin x$

$∴ I = e^{x} \cos x + C [\int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + C]$

Hence, the correct answer is option (a).

Page No 18.207:

Question 37:

$\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x = a x + b \log_{e} |4 e^{x} + 5 e^{- x}| + C, t h e n (a) a = - \frac{1}{8}, b = \frac{7}{8} (b) a = \frac{1}{8}, b = \frac{7}{8} (c) a = - \frac{1}{8}, b = - \frac{7}{8} (d) a = \frac{1}{8}, b = \frac{7}{8}$

Answer:

$I = \int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x$

Let $3 e^{x} - 5 e^{- x} = A \times (4 e^{x} + 5 e^{- x}) + B \times \frac{d}{d x} (4 e^{x} + 5 e^{- x})$

$\Rightarrow 3 e^{x} - 5 e^{- x} = A \times (4 e^{x} + 5 e^{- x}) + B \times (4 e^{x} - 5 e^{- x})$

$\Rightarrow 3 e^{x} - 5 e^{- x} = (4 A + 4 B) e^{x} + (5 A - 5 B) e^{- x}$

$\Rightarrow 4 A + 4 B = 3$ and $5 A - 5 B = - 5$

$4 A + 4 B = 3 \Rightarrow A + B = \frac{3}{4}$ .....(1)

$5 A - 5 B = - 5 \Rightarrow A - B = - 1$ .....(2)

Solving (1) and (2), we get

$A = - \frac{1}{8}$ and $B = \frac{7}{8}$

$∴ I = \int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x$

$\Rightarrow I = \int \frac{- \frac{1}{8} (4 e^{x} + 5 e^{- x}) + \frac{7}{8} (4 e^{x} - 5 e^{- x})}{4 e^{x} + 5 e^{- x}} d x$

$\Rightarrow I = - \frac{1}{8} \int d x + \frac{7}{8} \int \frac{4 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x$

$\Rightarrow I = - \frac{1}{8} x + \frac{7}{8} \log |4 e^{x} + 5 e^{- x}| + C [\begin{matrix} \int \frac{f' (x)}{f (x)} d x = \log |f (x)| + C \\ f (x) = 4 e^{x} + 5 e^{- x} \Rightarrow f' (x) = 4 e^{x} - 5 e^{- x} \end{matrix}]$

Comparing with the given expression, we get

$a = - \frac{1}{8}$ and $b = \frac{7}{8}$

Hence, the correct answer is option (a).

Page No 18.207:

Question 38:

$\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x is equal to (a) \frac{e^{x}}{1 + x} + C (b) - \frac{e^{x}}{1 + x^{2}} + C (c) \frac{e^{x}}{(1 + x}$ ²)2+C (d) -ex(1+x²)2+C

Answer:

$I = \int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x$

$\Rightarrow I = \int e^{x} [\frac{1 + x^{2} - 2 x}{{(1 + x^{2})}^{2}}] d x$

$\Rightarrow I = \int e^{x} [\frac{1}{1 + x^{2}} + \frac{- 2 x}{{(1 + x^{2})}^{2}}] d x$

Let $f (x) = \frac{1}{1 + x^{2}}$

$\Rightarrow f' (x) = \frac{- 2 x}{{(1 + x^{2})}^{2}}$

$∴ I = e^{x} (\frac{1}{1 + x^{2}}) + C [\int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + C]$

Hence, the correct answer is option (c).

Page No 18.207:

Question 39:

$\int \frac{x + \sin x}{1 + c o x x} d x i s e q u a l t o (a) \log |1 + \cos x| + C (b) \log |x + \sin x| + C (c) x - \tan \frac{x}{2} + C (d) x \tan \frac{x}{2} + C$

Answer:

$\int \frac{x + \sin x}{1 + \cos x} d x$

$= \int \frac{x + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} d x$
$= \int \frac{x}{2 \cos^{2} \frac{x}{2}} d x + \int \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} d x$

$= \frac{1}{2} \int x \sec^{2} \frac{x}{2} d x + \int \tan \frac{x}{2} d x$

$= \frac{1}{2} [x \int \sec^{2} \frac{x}{2} d x - \int (\frac{d}{d x} x \int \sec^{2} \frac{x}{2} d x) d x] + \int \tan \frac{x}{2} d x$

$= \frac{1}{2} [x \times \frac{\tan \frac{x}{2}}{\frac{1}{2}} - \int (\frac{\tan \frac{x}{2}}{\frac{1}{2}}) d x] + \int \tan \frac{x}{2} d x$

$= x \tan \frac{x}{2} - \int \tan \frac{x}{2} d x + \int \tan \frac{x}{2} d x + C$

$= x \tan \frac{x}{2} + C$

Hence, the correct answer is option (d).

Page No 18.207:

Question 40:

$\int \tan^{- 1} \sqrt{x} d x i s e q u a l t o (a) (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C (b) x \tan^{- 1} \sqrt{x} - \sqrt{x} + C (c) \sqrt{x} - x \tan^{- 1} \sqrt{x} + C (d) \sqrt{x} - (x + 1) \tan^{- 1} \sqrt{x} + C$

Answer:

$I = \int \tan^{- 1} \sqrt{x} d x$

Put $x = \tan^{2} θ$

$\Rightarrow d x = 2 \tan θ \sec^{2} θ d θ$

$∴ I = \int \tan^{- 1} \sqrt{\tan^{2} θ} \times 2 \tan θ \sec^{2} θ d θ$

$\Rightarrow I = 2 \int θ \times \tan θ \sec^{2} θ d θ$

$\Rightarrow I = 2 [θ \int \tan θ \sec^{2} θ d θ - \int (\frac{d}{d θ} θ \int \tan θ \sec^{2} θ d θ) d θ]$

$\Rightarrow I = 2 [θ \times \frac{\tan^{2} θ}{2} - \int \frac{\tan^{2} θ}{2} d θ]$

$\Rightarrow I = θ \tan^{2} θ - \int (\sec^{2} θ - 1) d θ$

$\Rightarrow I = θ \tan^{2} θ - \int \sec^{2} θ d θ + \int d θ$

$\Rightarrow I = θ \tan^{2} θ - \tan θ + θ + C$

$\Rightarrow I = \tan^{- 1} \sqrt{x} \times x - \sqrt{x} + \tan^{- 1} \sqrt{x} + C$ $(x = \tan^{2} θ \Rightarrow \tan θ = \sqrt{x} \Rightarrow θ = \tan^{- 1} \sqrt{x})$

$\Rightarrow I = (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C$

Hence, the correct answer is option (a).

Page No 18.207:

Question 41:

$\int \frac{1}{\sin (x - a) \sin (x - b)} d x i s e q u a l t o (a) \sin (b - a) \log |\frac{\sin (x - a)}{\sin (x - b)}| + C (b) \cos e c (b - a) \log |\frac{\sin (x - a)}{\sin (x - b)}| + C (c) \cos e c (b - a) \log |\frac{\sin (x - b)}{\sin (x - a)}| + C (d) \sin (b - a) \log |\frac{\sin (x - a)}{\sin (x - b)}| + C$

Answer:

$I = \int \frac{1}{\sin (x - a) \sin (x - b)} d x$

$\Rightarrow I = \frac{1}{\sin (b - a)} \int \frac{\sin (b - a)}{\sin (x - a) \sin (x - b)} d x$

$\Rightarrow I = \frac{1}{\sin (b - a)} \int \frac{\sin (x - a - x + b)}{\sin (x - a) \sin (x - b)} d x$

$\Rightarrow I = \frac{1}{\sin (b - a)} \int \frac{\sin [(x - a) - (x - b)]}{\sin (x - a) \sin (x - b)} d x$

$\Rightarrow I = \frac{1}{\sin (b - a)} \int \frac{\sin (x - a) \cos (x - b) - \cos (x - a) \sin (x - b)}{\sin (x - a) \sin (x - b)} d x$

$\Rightarrow I = \frac{1}{\sin (b - a)} \int \frac{\sin (x - a) \cos (x - b)}{\sin (x - a) \sin (x - b)} d x - \frac{1}{\sin (b - a)} \int \frac{\cos (x - a) \sin (x - b)}{\sin (x - a) \sin (x - b)} d x$

$\Rightarrow I = \frac{1}{\sin (b - a)} \int \cot (x - b) d x - \frac{1}{\sin (b - a)} \int \cot (x - a) d x$

$\Rightarrow I = \frac{1}{\sin (b - a)} \log |\sin (x - b)| - \frac{1}{\sin (b - a)} \log |\sin (x - a)| + C$

$\Rightarrow I = \frac{1}{\sin (b - a)} (\log |\sin (x - b)| - \log |\sin (x - a)|) + C$

$\Rightarrow I = cosec (b - a) \log |\frac{\sin (x - b)}{\sin (x - a)}| + C$

Hence, the correct answer is option (c).

Page No 18.207:

Question 1:

If 0 < x < $\frac{π}{4}$ , then $\int \sqrt{1 - \sin 2 x}$ dx is equal to ______________.

Answer:

$\sqrt{1 - \sin 2 x}$

$= \sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cos x}$

$= \sqrt{{(\sin x - \cos x)}^{2}}$

$= |\sin x - \cos x|$

When $0 < x < \frac{π}{4},$

$\sin x < \cos x$

$\Rightarrow \sin x - \cos x < 0$

$\Rightarrow \sqrt{1 - \sin 2 x} = |\sin x - \cos x| = - (\sin x - \cos x)$

$∴ \int \sqrt{1 - \sin 2 x} d x$

$= \int - (\sin x - \cos x) d x$

$= - \int \sin x d x + \int \cos x d x$

$= - (- \cos x) + \sin x + C$

$= \cos x + \sin x + C$

If 0 < x < $\frac{π}{4}$ , then $\int \sqrt{1 - \sin 2 x} d x$ is equal to $\cos x + \sin x + C$ .

Page No 18.207:

Question 2:

If $\frac{π}{4} < x < \frac{π}{2},$ then $\int \sqrt{1 - \sin 2 x}$ dx is equal to ______________

Answer:

$\sqrt{1 - \sin 2 x}$

$= \sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cos x}$

$= \sqrt{{(\sin x - \cos x)}^{2}}$

$= |\sin x - \cos x|$

When $\frac{π}{4} < x < \frac{π}{2},$

$\sin x > \cos x$

$\Rightarrow \sin x - \cos x > 0$

$\Rightarrow \sqrt{1 - \sin 2 x} = |\sin x - \cos x| = \sin x - \cos x$

$∴ \int \sqrt{1 - \sin 2 x} d x$

$= \int (\sin x - \cos x) d x$

$= \int \sin x d x - \int \cos x d x$

$= - \cos x - \sin x + C$

If $\frac{π}{4} < x < \frac{π}{2},$ then $\int \sqrt{1 - \sin 2 x} d x$ is equal to $- \cos x - \sin x + C$ .

Page No 18.207:

Question 3:

If $\int \frac{1}{\tan x + co t x}$ dx = k cos 2x + C, then k = ___________________.

Answer:

$\int \frac{1}{\tan x + \cot x} d x$

$= \int \frac{1}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}} d x$

$= \int \frac{1}{\frac{\sin^{2} x + \cos^{2} x}{\sin x \cos x}} d x$

$= \int \sin x \cos x d x$

$= \frac{1}{2} \int \sin 2 x d x (\sin 2 A = 2 \sin A \cos A)$

$= \frac{1}{2} \times (- \frac{\cos 2 x}{2}) + C$

$= - \frac{1}{4} \cos 2 x + C$

Comparing with the given expression, we get

$k = - \frac{1}{4}$

If $\int \frac{1}{\tan x + \cot x} d x$ = k cos 2x + C, then k = $- \frac{1}{4}$ .

Page No 18.207:

Question 4:

The value of the integral $\int e^{2 \log x} + e^{x \log 2}$ dx is ________________.

Answer:

$\int (e^{2 \log x} + e^{x \log 2}) d x$

$= \int (e^{\log x^{2}} + e^{\log 2^{x}}) d x (b \log a = \log a^{b})$

$= \int (x^{2} + 2^{x}) d x [a^{\log_{a} f (x)} = f (x)]$

$= \int x^{2} d x + \int 2^{x} d x$

$= \frac{x^{3}}{3} + \frac{2^{x}}{\log_{e} 2} + C$

The value of the integral $\int (e^{2 \log x} + e^{x \log 2}) d x$ is $\frac{x^{3}}{3} + \frac{2^{x}}{\log_{e} 2} + C$ .

Page No 18.208:

Question 5:

If the value of integral $\int \frac{\sin^{6} x}{\cos^{8} x} d x i s k \tan^{7} x + c,$ then k = _________________.

Answer:

$I = \int \frac{\sin^{6} x}{\cos^{8} x} d x$

$\Rightarrow I = \int \frac{\sin^{6} x}{\cos^{6} x \times \cos^{2} x} d x$

$\Rightarrow I = \int \tan^{6} x \sec^{2} x d x$

Let $\tan x = z$

$\Rightarrow \sec^{2} x d x = d z$

$∴ I = \int z^{6} d z$

$\Rightarrow I = \frac{z^{7}}{7} + C$

$\Rightarrow I = \frac{1}{7} \tan^{7} x + C$

Comparing with the given expression, we get

k = $\frac{1}{7}$

If the value of integral $\int \frac{\sin^{6} x}{\cos^{8} x} d x is k \tan^{7} x + c,$ then k = $\frac{1}{7}$ .

Page No 18.208:

Question 6:

$\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x =_____________________.$

Answer:

$I = \int e^{x} \frac{x + 3}{{(x + 4)}^{2}} d x$

$\Rightarrow I = \int e^{x} [\frac{x + 4 - 1}{{(x + 4)}^{2}}] d x$

$\Rightarrow I = \int e^{x} [\frac{x + 4}{{(x + 4)}^{2}} + \frac{- 1}{{(x + 4)}^{2}}] d x$

$\Rightarrow I = \int e^{x} [\frac{1}{x + 4} + \frac{- 1}{{(x + 4)}^{2}}] d x$

Let $f (x) = \frac{1}{x + 4}$

$\Rightarrow f' (x) = \frac{- 1}{{(x + 4)}^{2}}$

$∴ I = e^{x} (\frac{1}{x + 4}) + C [\int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + C]$

$\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x =$ $e^{x} (\frac{1}{x + 4}) + C$ .

Page No 18.208:

Question 7:

$\int \cos x e^{2 \sin x} d x =____________________.$

Answer:

$I = \int \cos x e^{2 \sin x} d x$

Let $2 \sin x = z$

$\Rightarrow 2 \cos x d x = d z$

$\Rightarrow \cos x d x = \frac{d z}{2}$

$∴ I = \frac{1}{2} \int e^{z} d z$

$\Rightarrow I = \frac{1}{2} e^{z} + C$

$\Rightarrow I = \frac{1}{2} e^{2 \sin x} + C$

$\int \cos x e^{2 \sin x} d x =$ $\frac{1}{2} e^{2 \sin x} + C$

Page No 18.208:

Question 8:

$\int e^{x} \frac{x}{{(x + 1)}^{2}} d x =_____________________.$

Answer:

$I = \int e^{x} \frac{x}{{(x + 1)}^{2}} d x$

$\Rightarrow I = \int e^{x} [\frac{x + 1 - 1}{{(x + 1)}^{2}}] d x$

$\Rightarrow I = \int e^{x} [\frac{x + 1}{{(x + 1)}^{2}} + \frac{- 1}{{(x + 1)}^{2}}] d x$

$\Rightarrow I = \int e^{x} [\frac{1}{(x + 1)} + \frac{- 1}{{(x + 1)}^{2}}] d x$

Let $f (x) = \frac{1}{x + 1}$

$\Rightarrow f' (x) = \frac{- 1}{{(x + 1)}^{2}}$

$∴ I = e^{x} (\frac{1}{x + 1}) + C [\int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + C]$

$\int e^{x} \frac{x}{{(x + 1)}^{2}} d x =$ $e^{x} (\frac{1}{x + 1}) + C$ .

Page No 18.208:

Question 9:

$\int \frac{\sin x}{3 + 4 \cos^{2} x} d x =__________________.$

Answer:

$I = \int \frac{\sin x}{3 + 4 \cos^{2} x} d x$

Let $\cos x = z$

$\Rightarrow - \sin x d x = d z$

$∴ I = - \int \frac{d z}{3 + 4 z^{2}}$

$\Rightarrow I = - \frac{1}{4} \int \frac{d z}{z^{2} + \frac{3}{4}}$

$\Rightarrow I = - \frac{1}{4} \int \frac{d z}{z^{2} + {(\frac{\sqrt{3}}{2})}^{2}}$

$\Rightarrow I = - \frac{1}{4} \times \frac{1}{(\frac{\sqrt{3}}{2})} \tan^{- 1} (\frac{z}{\frac{\sqrt{3}}{2}}) + C [\int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + C]$

$\Rightarrow I = - \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C$

$\int \frac{\sin x}{3 + 4 \cos^{2} x} d x =$ $- \frac{1}{2 \sqrt{3}} \tan^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C$

Page No 18.208:

Question 10:

$\int \frac{x^{2} + 1}{x^{2} - 1} d x =___________________.$

Answer:

$\int \frac{x^{2} + 1}{x^{2} - 1} d x$

$= \int \frac{x^{2} - 1 + 2}{x^{2} - 1} d x$

$= \int \frac{x^{2} - 1}{x^{2} - 1} d x + \int \frac{2}{x^{2} - 1} d x$

$= \int d x + 2 \int \frac{1}{x^{2} - 1} d x$

$= x + 2 \times \frac{1}{2} \log |\frac{x - 1}{x + 1}| + C (\int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \log |\frac{x - a}{x + a}| + C)$

$= x + \log |\frac{x - 1}{x + 1}| + C$

$\int \frac{x^{2} + 1}{x^{2} - 1} d x = x + \log |\frac{x - 1}{x + 1}| + C .$

Page No 18.208:

Question 11:

$\int \frac{e^{5 \log_{e} x} - e^{4 \log_{e} x}}{e^{3 \log_{e} x} - e^{2 \log_{e} x}} d x =____________________.$

Answer:

$\int \frac{e^{5 \log_{e} x} - e^{4 \log_{e} x}}{e^{3 \log_{e} x} - e^{2 \log_{e} x}} d x$

$= \int \frac{e^{\log_{e} x^{5}} - e^{\log_{e} x^{4}}}{e^{\log_{e} x^{3}} - e^{\log_{e} x^{2}}} d x (b \log a = \log a^{b})$

$= \int \frac{x^{5} - x^{4}}{x^{3} - x^{2}} d x [a^{\log_{a} f (x)} = f (x)]$

$= \int \frac{x^{4} (x - 1)}{x^{2} (x - 1)} d x$

$= \int x^{2} d x$

$= \frac{x^{3}}{3} + C$

$\int \frac{e^{5 \log_{e} x} - e^{4 \log_{e} x}}{e^{3 \log_{e} x} - e^{2 \log_{e} x}} d x = \frac{x^{3}}{3} + C .$

Page No 18.208:

Question 12:

$\int \frac{x^{4} + x^{2} + 1}{x^{2} - x + 1} d x =______________________.$

Answer:

$\int \frac{x^{4} + x^{2} + 1}{x^{2} - x + 1} d x$

$= \int \frac{x^{4} + 2 x^{2} + 1 - x^{2}}{x^{2} - x + 1} d x$

$= \int \frac{{(x^{2} + 1)}^{2} - x^{2}}{x^{2} - x + 1} d x$

$= \int \frac{(x^{2} + 1 + x) (x^{2} + 1 - x)}{x^{2} - x + 1} d x$

$= \int (x^{2} + 1 + x) d x$

$= \int x^{2} d x + \int x d x + \int d x$

$= \frac{x^{3}}{3} + \frac{x^{2}}{2} + x + C$

$\int \frac{x^{4} + x^{2} + 1}{x^{2} - x + 1} d x = \frac{x^{3}}{3} + \frac{x^{2}}{2} + x + C .$

Page No 18.208:

Question 13:

$\int \frac{\cos x - \sin x}{1 + \sin 2 x} d x =_______________________.$

Answer:

$I = \int \frac{\cos x - \sin x}{1 + \sin 2 x} d x$

$\Rightarrow I = \int \frac{\cos x - \sin x}{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x} d x$

$\Rightarrow I = \int \frac{\cos x - \sin x}{{(\sin x + \cos x)}^{2}} d x$

Let $\sin x + \cos x = z$

$\Rightarrow (\cos x - \sin x) d x = d z$

$∴ I = \int \frac{d z}{z^{2}}$

$\Rightarrow I = \int z^{- 2} d z$

$\Rightarrow I = \frac{z^{- 1}}{(- 1)} + C$

$\Rightarrow I = - \frac{1}{z} + C$

$\Rightarrow I = - \frac{1}{\sin x + \cos x} + C$

$\int \frac{\cos x - \sin x}{1 + \sin 2 x} d x = - \frac{1}{\sin x + \cos x} + C .$

Page No 18.208:

Question 14:

$\int$ e^x sine^x dx = ___________________.

Answer:

$I = \int e^{x} \sin e^{x} d x$

Let $e^{x} = z$

$\Rightarrow e^{x} d x = d z$

$∴ I = \int \sin z d z$

$\Rightarrow I = - \cos z + C$

$\Rightarrow I = - \cos e^{x} + C$

$\int$ e^x sine^x dx = $- \cos e^{x} + C$ .

Page No 18.208:

Question 15:

$\int \frac{x}{e^{3 x^{2}}} d x =_____________________.$

Answer:

$I = \int \frac{x}{e^{3 x^{2}}} d x$

Let $3 x^{2} = z$

$\Rightarrow 6 x d x = d z$

$\Rightarrow x d x = \frac{d z}{6}$

$∴ I = \int \frac{1}{e^{z}} \times \frac{d z}{6}$

$\Rightarrow I = \frac{1}{6} \int e^{- z} d z$

$\Rightarrow I = \frac{1}{6} \times \frac{e^{- z}}{(- 1)} + C$

$\Rightarrow I = - \frac{1}{6 e^{z}} + C$

$\Rightarrow I = - \frac{1}{6 e^{3 x^{2}}} + C$

$\int \frac{x}{e^{3 x^{2}}} d x = - \frac{1}{6 e^{3 x^{2}}} + C .$

Page No 18.208:

Question 16:

$\int$ x^x(1 + log x) dx = ________________.

Answer:

$I = \int x^{x} (1 + \log x) d x$

Let $x^{x} = z$

$\Rightarrow \log z = \log x^{x}$

$\Rightarrow \log z = x \log x$

$\Rightarrow \frac{1}{z} d z = (x \times \frac{1}{x} + \log x \times 1) d x$

$\Rightarrow \frac{d z}{z} = (1 + \log x) d x$

$∴ I = \int z \times \frac{d z}{z}$

$\Rightarrow I = \int d z$

$\Rightarrow I = z + C$

$\Rightarrow I = x^{x} + C$

$\int$ x^x(1 + log x)dx = ____x^x + C____.

Page No 18.208:

Question 17:

$\int \frac{e^{x}}{e^{x} + 1} d x =____________________.$

Answer:

$I = \int \frac{e^{x}}{e^{x} + 1} d x$

Let $e^{x} + 1 = z$

$\Rightarrow e^{x} d x = d z$

$∴ I = \int \frac{d z}{z}$

$\Rightarrow I = \log |z| + C$

$\Rightarrow I = \log |e^{x} + 1| + C$

$\int \frac{e^{x}}{e^{x} + 1} d x = \log |e^{x} + 1| + C .$

Page No 18.208:

Question 18:

$\int$ (x + 3) (x² + 6x + 10)⁹ dx = ______________________.

Answer:

$I = \int (x + 3) {(x^{2} + 6 x + 10)}^{9} d x$

Let $x^{2} + 6 x + 10 = z$

$\Rightarrow (2 x + 6) d x = d z$

$\Rightarrow (x + 3) d x = \frac{d z}{2}$

$∴ I = \int {(z)}^{9} \frac{d z}{2}$

$\Rightarrow I = \frac{1}{2} \int z^{9} d z$

$\Rightarrow I = \frac{1}{2} \times \frac{z^{10}}{10} + C$

$\Rightarrow I = \frac{1}{20} {(x^{2} + 6 x + 10)}^{10} + C$

$\int$ (x + 3)(x² + 6x + 10)⁹ dx = $\frac{1}{20} {(x^{2} + 6 x + 10)}^{10} + C$ .

Page No 18.208:

Question 19:

$\int \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x =________________________.$

Answer:

$I = \int \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x$

Let $\tan^{- 1} x = z$

$\Rightarrow \frac{1}{1 + x^{2}} d x = d z$

$∴ I = \int z^{3} d z$

$\Rightarrow I = \frac{z^{4}}{4} + C$

$\Rightarrow I = \frac{{(\tan^{- 1} x)}^{4}}{4} + C$

$\int \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x = \frac{{(\tan^{- 1} x)}^{4}}{4} + C .$

Page No 18.208:

Question 20:

$\int$ e^x(1 - cot x + cosec^₂ x) dx = ______________.

Answer:

$I = \int e^{x} (1 - \cot x + {cosec}^{2} x) d x$

$\Rightarrow I = \int e^{x} d x + \int e^{x} (- \cot x + {cosec}^{2} x) d x$

$\Rightarrow I = e^{x} + \int e^{x} (- \cot x + {cosec}^{2} x) d x$

Let $f (x) = - \cot x$

$\Rightarrow f' (x) = - (- {cosec}^{2} x) = {cosec}^{2} x$

$∴ I = e^{x} + e^{x} (- \cot x) + C$ $[\int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + C]$

$\Rightarrow I = e^{x} (1 - \cot x) + C$

$\int$ e^x(1 − cot x + cosec^₂ x) dx = $e^{x} (1 - \cot x) + C$ .

Page No 18.208:

Question 1:

Write a value of $\int \frac{1 + \cot x}{x + \log \sin x} d x$ .

Answer:

$Let I = \int \frac{1 + \cot x}{x + \log \sin x} d x Let x + \log \sin x = t \Rightarrow (1 + \frac{1}{\sin x} \times \cos x) d x = d t \Rightarrow (1 + \cot x) d x = d t ∴ I = \int \frac{d t}{t} = \log |t| + C = \log |x + \log \sin x| + C$

Page No 18.208:

Question 2:

Write a value of $\int e^{3 \log x} x^{4} d x$ .

Answer:

$\int e^{3 \log x} . x^{4} d x = \int e^{\log x^{3}} \cdot x^{4} d x (∵ a \log x = \log x^{a}) = \int x^{3} \cdot x^{4} d x (∵ e^{\log m} = m) = \int x^{7} \cdot d x = \frac{x^{8}}{8} + C$

Page No 18.208:

Question 3:

Write a value of $\int x^{2} \sin x^{3} d x$ .

Answer:

Let I= $\int$ x².sin x³ dx
Let x³ = t
⇒ 3x²dx = dt
$\Rightarrow x^{2} d x = \frac{d t}{3} ∴ I = \frac{1}{3} \int \sin t d t = \frac{1}{3} [- \cos t] + C = - \frac{1}{3} \cos x^{3} + C (∵ t = x^{3})$

Page No 18.208:

Question 4:

Write a value of $\int \tan^{3} x \sec^{2} x d x$ .

Answer:

Let I= $\int$ tan³ x . sec² x dx
Let tan x = t
⇒ sec²x dx = dt
$∴$ I= $\int$ t³ . dt
$= \frac{t^{4}}{4} + C = \frac{\tan^{4} x}{4} + C (∵ t = \tan x)$

Page No 18.208:

Question 5:

Write a value of $\int e^{x} (\sin x + \cos x) d x$ .

Answer:

Let I= $\int$ e^x (sin x + cos x) dx
Let e^x sin x = t
⇒ (e^x . sin x + e^x cos x) dx = dt
$∴ I = \int d t = t + C = e^{x} \sin x + C (∵ t = e^{x} \sin x)$

Page No 18.208:

Question 6:

Write a value of $\int \tan^{6} x \sec^{2} x d x$ .

Answer:

Let I= $\int$ tan⁶ x . sec² x dx
Let tan x = t
sec²x dx = dt
$∴ I =$ $\int$ t⁶ . dt
$= \frac{t^{7}}{7} + C = \frac{\tan^{7} x}{7} + C (∵ t = \tan x)$

Page No 18.209:

Question 7:

Write a value of $\int \frac{\cos x}{3 + 2 \sin x} d x$ .

Answer:

$Let I = \int \frac{\cos x}{3 + 2 \sin x} d x Let 3 + 2 \sin x = t \Rightarrow 2 \cos x d x = d t \Rightarrow \cos x d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{d t}{t} = \int \frac{1}{2} \log |t| + C = \frac{1}{2} \log |3 + 2 \sin x| + C (∵ t = 3 + 2 \sin x)$

Page No 18.209:

Question 8:

Write a value of $\int e^{x} \sec x (1 + \tan x) d x$ .

Answer:

Let I= $\int$ e^x sec x(1 + tan x) dx
       = $\int$ e^x (sec x + sec x tan x) dx
Let e^x sec x = t
⇒ (e^x sec x + e^x sec x tan x)dx = dt
⇒ e^x sec x (1 + tan x) dx = dt
$∴ I =$ $\int$ dt
     = t + C
     = e^x sec x + C                 $(∵ t = e^{x} \sec x)$

Page No 18.209:

Question 9:

Write a value of $\int \frac{\log x^{n}}{x} d x$ .

Answer:

$Let I = \int (\frac{\log x^{n}}{x}) d x = \int \frac{n \log x}{x} d x (∵ \log x^{a} = a \log x) Let \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = n \int t d t = n . \frac{t^{2}}{2} + C = \frac{n . {(\log x)}^{2}}{2} + C (∵ t = \log x)$

Page No 18.209:

Question 10:

Write a value of $\int \frac{{(\log x)}^{n}}{x} d x$ .

Answer:

$Let I = \int \frac{{(\log x)}^{n}}{x} d x Let \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = \int t^{n} d t = \frac{t^{n + 1}}{n + 1} + C = \frac{{(\log x)}^{n + 1}}{n + 1} + C (∵ t = \log x)$

Page No 18.209:

Question 11:

Write a value of $\int e^{\log \sin x} \cos x d x$ .

Answer:

Let I= $\int$ e^log^{sin x} . cos x dx
= $\int$ sin x × cos x dx $(∵ e^{\log a} = a)$
Let sin x = t
⇒ cos x dx = dt
$∴ I$ = $\int$ t . dt
$= \frac{t^{2}}{2} + C = \frac{\sin^{2} x}{2} + C (∵ t = \sin x)$

Page No 18.209:

Question 12:

Write a value of $\int \sin^{3} x \cos x d x$ .

Answer:

Let I= $\int$ sin³ x . cos x dx
Let sin x = t
⇒ cos x dx = dt
$∴ I =$ $\int$ t³ . dt
$= \frac{t^{4}}{4} + C = \frac{\sin^{4} x}{4} + C (∵ t = \sin x)$

Page No 18.209:

Question 13:

Write a value of $\int \cos^{4} x \sin x d x$ .

Answer:

Let I= $\int$ cos⁴ x .sin x dx
Let cos x = t
⇒ –sin x dx = dt
⇒ sin x dx = –dt
$∴ I =$ – $\int$ t⁴ dt
$= \frac{- t^{5}}{5} + C = - \frac{\cos^{5} x}{5} + C (∵ t = \cos x)$

Page No 18.209:

Question 14:

Write a value of $\int \tan x \sec^{3} x d x$ .

Answer:

Let I= $\int$ tan x . sec³x dx
= $\int$ sec² x . sec x tan x dx
Let sec x = t
⇒ sec x tan x dx = dt
$∴ I =$ ∫ t² dt
$= \frac{t^{3}}{3} + C = \frac{\sec^{3} x}{3} + C (∵ x = \sec x)$

Page No 18.209:

Question 15:

Write a value of $\int \frac{1}{1 + e^{x}} d x$ .

Answer:

$Let I = \int \frac{d x}{1 + e^{x}} Dividing numerator & denominator by e^{x} \Rightarrow I = \int \frac{\frac{1}{e^{x}} d x}{\frac{1}{e^{x}} + 1} = \int \frac{e^{- x} d x}{e^{- x} + 1} Let e^{- x} + 1 = t - e^{- x} d x = d t \Rightarrow e^{- x} d x = - d t ∴ I = \int - \frac{d t}{t} = - \log |t| + C = - \log |1 + e^{x}| + C (∵ t = 1 + e^{x})$

Page No 18.209:

Question 16:

Write a value of $\int \frac{1}{1 + 2 e^{x}} d x$ .

Answer:

$Let I = \int \frac{d x}{1 + 2 e^{x}} Dividing numerator & denominator by e^{x} \Rightarrow I = \int \frac{\frac{1}{e^{x}} d x}{\frac{1}{e^{x}} + 2} = \int \frac{e^{- x} d x}{e^{- x} + 2} Let e^{- x} + 2 = t \Rightarrow - e^{- x} d x = d t \Rightarrow e^{- x} d x = - d t ∴ I = - \int \frac{d t}{t} = - \log |t| + C = - \log |e^{- x} + 2| + C (∵ t = e^{- x} + 2)$

Page No 18.209:

Question 17:

Write a value of $\int \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x$ .

Answer:

$Let I = \int \frac{{(\tan^{- 1} x)}^{3}}{1 + x^{2}} d x Let \tan^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} d x = d t ∴ I = \int t^{3} . d t = \frac{t^{4}}{4} + C = \frac{{(\tan^{- 1} x)}^{4}}{4} + C (∵ t = \tan^{- 1} x)$

Page No 18.209:

Question 18:

Write a value of $\int \frac{\sec^{2} x}{{(5 + \tan x)}^{4}} d x$ .

Answer:

$Let I = \int \frac{\sec^{2} x d x}{{(5 + \tan x)}^{4}} Let 5 + \tan x = t \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{d t}{t^{4}} = \int t^{- 4} d t = [\frac{t^{- 4 + 1}}{- 4 + 1}] + C = - \frac{1}{3 t^{3}} + C = - \frac{1}{3 {(5 t + \tan x)}^{3}} + C (∵ t = 5 + \tan x)$

Page No 18.209:

Question 19:

Write a value of $\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x$ .

Answer:

$\int (\frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}}) d x \Rightarrow \int \frac{(\sin x + \cos x) d x}{\sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x}} \Rightarrow \int \frac{(\sin x + \cos x) d x}{\sqrt{{(\sin x + \cos x)}^{2}}} \Rightarrow \int d x \Rightarrow x + C$

Page No 18.209:

Question 20:

Write a value of $\int \log_{e} x d x$ .

Answer:

$\int$ log_e x dx
= $\int$ $\underset{II}{1} . \underset{I}{\log_{e} x}$ dx
= $\log_{e} x \int 1 d x - \int \{\frac{d}{d x} (\log_{e} x) \int 1 d x\} d x$
= log_e x $\int$ 1 . dx – $\int$ $\frac{1}{x} \times x$ .dx
= log_e x × x – $\int$ dx
= x log_e x – x + C
= x log_e x – x + C
= x (log_e x – 1) + C

Page No 18.209:

Question 21:

Write a value of $\int a^{x} e^{x} d x$ .

Answer:

∫ a^x . e^x dx
= ∫ (ae)^x dx
$= \frac{{(a e)}^{x}}{\ln (a e)} + C$

Page No 18.209:

Question 22:

Write a value of $\int e^{2 x^{2} + \ln x} d x$ .

Answer:

$Let I = \int (e^{2 x^{2} + \ln x}) d x = \int (e^{2 x^{2}} \times e^{\ln x}) d x = \int e^{2 x^{2}} . x d x Let 2 x^{2} = t \Rightarrow 4 x d x = d t \Rightarrow x d x = \frac{d t}{4} ∴ I = \frac{1}{4} \int e^{t} d t = \frac{1}{4} e^{t} + C = \frac{1}{4} e^{2 x^{2}} + C (∵ t = 2 x^{2})$

Page No 18.209:

Question 23:

Write a value of $\int (e^{x \log_{e} a} + e^{a \log_{e} x}) d x$ .

Answer:

$(e^{x \log_{e} a} + e^{a \log_{e} x}) d x \Rightarrow \int (e^{\log a^{x}} + e^{\log x^{a}}) d x = \int (a^{x} + x^{a}) d x = \frac{a^{x}}{\log a} + \frac{x^{a + 1}}{a + 1} + C$

Page No 18.209:

Question 24:

Write a value of $\int \frac{\cos x}{\sin x \log \sin x} d x$ .

Answer:

$Let I = \int \frac{\cos x}{\sin x \cdot \log \sin x} d x \Rightarrow \int \frac{\cot x}{\log \sin x} d x Let \log \sin x = t \Rightarrow \cot x d x = d t ∴ I = \int \frac{d t}{t} = \log t + C = \log (\log \sin x) + C$

Page No 18.209:

Question 25:

Write a value of $\int \frac{\sin 2 x}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$ .

Answer:

$Let I = \int \frac{\sin 2 x d x}{a^{2} \sin^{2} + b^{2} \cos^{2} x} Let a^{2} \sin^{2} x + b^{2} \cos^{2} x = t \Rightarrow [a^{2} (2 \sin x \cos x) + b^{2} (2 \cos x \times - \sin x)] d x = d t \Rightarrow (a^{2} - b^{2}) \sin 2 x . d x = d t \Rightarrow \sin 2 x d x = \frac{d t}{a^{2} - b^{2}} ∴ I = \frac{1}{a^{2} - b^{2}} \int \frac{d t}{t} = \frac{1}{a^{2} - b^{2}} \log t + C = \frac{1}{a^{2} - b^{2}} \log (a^{2} \sin^{2} x + b^{2} \cos^{2} x) + C (∵ t = a^{2} \sin^{2} x + b^{2} \cos^{2} x)$

Page No 18.209:

Question 26:

Write a value of $\int \frac{a^{x}}{3 + a^{x}} d x$ .

Answer:

$Let I = \int \frac{a^{x} d x}{3 + a^{x}} Let 3 + a^{x} = t \Rightarrow a^{x} . \log a d x = d t \Rightarrow a^{x} d x = \frac{d t}{\log a} ∴ I = \frac{1}{\log a} \int \frac{d t}{t} = \frac{1}{\log a} \log t + C = \frac{1}{\log a} \log (3 + a^{x}) + C (∵ t = 3 + a^{x})$

Page No 18.209:

Question 27:

Write a value of $\int \frac{1 + \log x}{3 + x \log x} d x$ .

Answer:

$Let I = \int (\frac{1 + \log x}{3 + x \log x}) d x Let 3 + x \log x = t \Rightarrow 0 + (x . \frac{1}{x} + \log x) d x = d t \Rightarrow (1 + \log x) d x = d t ∴ I = \int \frac{d t}{t} = \log t + C = \log (3 + x \log x) + C (∵ t = 3 + x \log x)$

Page No 18.209:

Question 28:

Write a value of $\int \frac{\sin x}{\cos^{3} x} d x$ .

Answer:

$Let I = \int \frac{\sin x}{\cos^{3} x} d x Let \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t ∴ I = - \int \frac{d t}{t^{3}} = - \int t^{- 3} d t = - [\frac{t^{- 3 + 1}}{- 3 + 1}] + C = \frac{1}{2 t^{2}} + C = \frac{1}{2 \cos^{2} x} + C (∵ t = \cos x) = \frac{1}{2} \sec^{2} x + C$

Page No 18.209:

Question 29:

Write a value of $\int \frac{\sin x - \cos x}{\sqrt{1 + \sin 2 x}} d x$ .

Answer:

$Let I = \int \frac{(\sin x + \cos x) d x}{\sqrt{1 - \sin 2 x}} = \int \frac{(\sin x + \cos x) d x}{\sqrt{\sin^{2} x + \cos^{2} x - 2 \sin x \cos x}} = \int \frac{(\sin x + \cos x) d x}{\sqrt{{(\sin x - \cos x)}^{2}}} = \int \frac{(\sin x + \cos x) d x}{|\sin x - \cos x|} = \pm \int (\frac{\sin x + \cos x}{\sin x - \cos x}) d x Let \sin x - \cos x = t \Rightarrow (\cos x + \sin x) d x = d t ∴ I = \pm \int \frac{d t}{t} = \pm \ln |t| + C = \pm \ln |\sin x - \cos x| + C (∵ t = \sin x - \cos x)$

Page No 18.209:

Question 30:

Write a value of $\int \frac{1}{x {(\log x)}^{n}} d x$ .

Answer:

$Let I = \int \frac{d x}{x {(\log x)}^{n}} = \int \frac{{(\log x)}^{- n} d x}{x} Let \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = \int t^{- n} . d t = \frac{t^{- n + 1}}{- n + 1} + C = \frac{{(\log x)}^{- n + 1}}{- n + 1} + C (∵ t = \log x)$

Page No 18.209:

Question 31:

Write a value of $\int e^{a x} \sin b x d x$ .

Answer:

$Let I = \int e^{a x} . \sin b x d x = \sin b x \int e^{a x} d x - \int \{\frac{d}{d x} (\sin b x) \int e^{a x} d x\} d x = \sin b x \times \frac{e^{a x}}{a} - \int \cos b x \times b . \frac{e^{a x}}{a} = \sin b x \times \frac{e^{a x}}{a} - \frac{b}{a} \int e^{a x} . \cos b x d x = \sin b x \times \frac{e^{a x}}{a} - \frac{b}{a} I_{1} . . . (1) ∴ I_{1} = \int e^{a x} \times \cos b x d x = \cos b x \int e^{a x} d x - \int \{\frac{d}{d x} (\cos b x) \int e^{a x} d x\} d x = \cos b x \times \frac{e^{a x}}{a} + \int b . \sin b x \times \frac{e^{a x}}{a} d x = \cos b x . \frac{e^{a x}}{a} + \frac{b}{a} I . . . . (2) From (1) & (2) ∴ I = \sin b x . \frac{e^{a x}}{a} - \frac{b}{a} \{\cos b x . \frac{e^{a x}}{a} + \frac{b}{a} I\} \Rightarrow I = \sin b x . \frac{e^{a x}}{a} - \frac{b}{a^{2}} \cos b x e^{a x} - \frac{b^{2}}{a^{2}} I \Rightarrow I + \frac{b^{2}}{a^{2}} I = \sin b x . \frac{e^{a x}}{a} - \frac{b \cos b x e^{a x}}{a^{2}} \Rightarrow (a^{2} + b^{2}) I = (a \sin b x - b \cos b x) e^{a x} \Rightarrow I = \frac{(a \sin b x - b \cos b x) e^{a x}}{a^{2} + b^{2}} + C$

Page No 18.209:

Question 32:

Write a value of $\int e^{a x} \cos b x d x$ .

Answer:

$Let I = \int e^{a x} . \cos b x d x = \cos b x \int e^{a x} d x - \int \{\frac{d}{d x} (\cos b x) \int e^{a x} d x\} d x = \cos b x \times \frac{e^{a x}}{a} - \int - \sin b x \times b . \frac{e^{a x}}{a} = \cos b x \times \frac{e^{a x}}{a} + \frac{b}{a} \int e^{a x} . \sin b x d x = \cos b x \times \frac{e^{a x}}{a} + \frac{b}{a} I_{1} . . . (1) ∴ I_{1} = \int e^{a x} \times \sin b x d x = \sin b x \int e^{a x} d x - \int \{\frac{d}{d x} (\sin b x) \int e^{a x} d x\} d x = \sin b x \times \frac{e^{a x}}{a} - \int b . \cos b x \times \frac{e^{a x}}{a} d x = \sin b x . \frac{e^{a x}}{a} - \frac{b}{a} I . . . . (2) From (1) & (2) ∴ I = \cos b x . \frac{e^{a x}}{a} + \frac{b}{a} \{\sin b x . \frac{e^{a x}}{a} - \frac{b}{a} I\} \Rightarrow I = \cos b x . \frac{e^{a x}}{a} + \frac{b}{a^{2}} \sin b x e^{a x} - \frac{b^{2}}{a^{2}} I \Rightarrow I + \frac{b^{2}}{a^{2}} I = \cos b x . \frac{e^{a x}}{a} + \frac{b \sin b x e^{a x}}{a^{2}} \Rightarrow (a^{2} + b^{2}) I = (a \cos b x + b \sin b x) e^{a x} \Rightarrow I = \frac{(a \cos b x + b \sin b x) e^{a x}}{a^{2} + b^{2}} + C$

Page No 18.209:

Question 33:

Write a value of $\int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x$ .

Answer:

$Let I = \int e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x As we know that \int e^{x} \{f (x) + f' (x)\} d x = e^{x} f (x) + C ∴ I = \frac{e^{x}}{x} + C$

Page No 18.209:

Question 34:

Write a value of $\int e^{a x} \{a f (x) + f' (x)\} d x$ .

Answer:

$Let I = \int e^{a x} [a f (x) + f' (x)] d x Let e^{a x} . f (x) = t \Rightarrow [e^{a x} . a f (x) + e^{a x} . f' (x)] d x = d t ∴ I = \int d t = t + C = e^{a x} . f (x) + C (∵ t = e^{a x} . f (x))$

Page No 18.209:

Question 35:

Write a value of $\int \sqrt{4 - x^{2}} d x$ .

Answer:

$\int \sqrt{4 - x^{2}} d x = \int \sqrt{2^{2} - x^{2}} d x = \frac{x}{2} \sqrt{2^{2} - x^{2}} + \frac{2^{2}}{2} \sin^{- 1} (\frac{x}{2}) + C (∵ \sqrt{a^{2} - x^{2}} = \frac{x}{2} \sqrt{a^{2} - x^{2}} - \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} + C) = \frac{x}{2} \sqrt{4 - x^{2}} + 2 \sin^{- 1} (\frac{x}{2}) + C$

Page No 18.209:

Question 36:

Write a value of $\int \sqrt{9 + x^{2}} d x$ .

Answer:

$\int \sqrt{9 + x^{2}} d x = \int \sqrt{3^{2} + x^{2}} d x (∵ \sqrt{a^{2} + x^{2}} = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \ln |x + \sqrt{x^{2} + a^{2}}|) = \frac{x}{2} \sqrt{9 + x^{2}} + \frac{9}{2} \ln |x + \sqrt{9 + x^{2}}| + C$

Page No 18.209:

Question 37:

Write a value of $\int \sqrt{x^{2} - 9} d x$ .

Answer:

$\int \sqrt{x^{2} - 9} d x = \int \sqrt{x^{2} - 3^{2}} d x = \frac{x}{2} \sqrt{x^{2} - 3^{2}} - \frac{3^{2}}{2} \ln |x + \sqrt{x^{2} - 3}| + C (∵ \sqrt{x^{2} - a^{2}} = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \ln |x + \sqrt{x^{2} + a^{2}}| + C) = \frac{x}{2} \sqrt{x^{2} - 9} - \frac{9}{2} \ln |x + \sqrt{x^{2} - 9}| + C$

Page No 18.209:

Question 38:

Evaluate: $\int \frac{x^{2}}{1 + x^{3}} d x$

Answer:

$Let I = \int \frac{x^{2} d x}{1 + x^{3}} Putting 1 + x^{3} = t \Rightarrow 3 x^{2} d x = d t \Rightarrow x^{2} d x = \frac{d t}{3} ∴ I = \frac{1}{3} \int \frac{d t}{t} = \frac{1}{3} \ln |t| + C = \frac{1}{3} \ln |1 + x^{3}| + C (∵ t = 1 + x^{3})$

Page No 18.209:

Question 39:

Evaluate: $\int \frac{x^{2} + 4 x}{x^{3} + 6 x^{2} + 5} d x$

Answer:

$Let I = \int (\frac{x^{2} + 4 x}{x^{3} + 6 x^{2} + 5}) d x Let x^{3} + 6 x^{2} + 5 = t \Rightarrow (3 x^{2} + 12 x) d x = d t \Rightarrow (x^{2} + 4 x) d x = \frac{d t}{3} Putting x^{3} + 6 x^{2} + 5 = t and (x^{2} + 4 x) d x = \frac{d t}{3} ∴ I = \frac{1}{3} \int \frac{d t}{t} = \frac{1}{3} \ln |t| + C = \frac{1}{3} \ln |x^{3} + 6 x^{2} + 5| + C$

Page No 18.209:

Question 40:

Evaluate: $\int \frac{\sec^{2} \sqrt{x}}{\sqrt{x}} d x$

Answer:

$Let I = \int \frac{\sec^{2} \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{d x}{2 \sqrt{x}} = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t Putting \sqrt{x} = t and \frac{d x}{\sqrt{x}} = 2 d t ∴ I = 2 \int \sec^{2} + d t = 2 \tan t + C = 2 \tan (\sqrt{x}) + C (∵ t = \sqrt{x})$

Page No 18.209:

Question 41:

Evaluate: $\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x$

Answer:

$Let I = \int \frac{\sin \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t Putting \sqrt{x} = t and \frac{d x}{\sqrt{x}} = 2 d t, we get ∴ I = 2 \int \sin t d t = - 2 \cos t + C (∵ t = \sqrt{x}) = - 2 \cos \sqrt{x} + C$

Page No 18.209:

Question 42:

Evaluate: $\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$

Answer:

$Let I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x Putting \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t ∴ I = 2 \int \cos t d t = 2 \sin t + C, where t = \sqrt{x} = 2 \sin \sqrt{x} + C$

Page No 18.209:

Question 43:

Evaluate: $\int \frac{{(1 + \log x)}^{2}}{x} d x$

Answer:

$Let \int \frac{{(1 + \log x)}^{2}}{x} d x Putting 1 + \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = \int t^{2} \cdot d t = \frac{t^{3}}{3} + C = \frac{{(1 + \log x)}^{3}}{3} + C [∵ t = (1 + \log x)]$

Page No 18.209:

Question 44:

Evaluate: $\int \sec^{2} (7 - 4 x) d x$

Answer:

$\int \sec^{2} (7 - 4 x) d x = \frac{\tan (7 - 4 x)}{- 4} + C (∵ \int \sec^{2} x = \tan x + C)$

Page No 18.209:

Question 45:

Evaluate: $\int \frac{\log x}{x} d x$

Answer:

$Let I = \int \frac{\log x}{x} d x & let \log x = t \Rightarrow \frac{1}{x} d x = d t ∴ I = \int t \cdot d t = \frac{t^{2}}{2} + C = \frac{{(\log x)}^{2}}{2} + C (∵ t = \log x)$

Page No 18.209:

Question 46:

Evaluate: $\int 2^{x} d x$

Answer:

$\int 2^{x} d x = \frac{2^{x}}{\ln 2} + C (∵ \int a^{x} d x = \frac{a^{x}}{\ln a} + C)$

Page No 18.209:

Question 47:

Write a value of $\int \frac{1 - \sin x}{\cos^{2} x} d x$

Answer:

$\int (\frac{1 - \sin x}{\cos^{2} x}) d x = \int (\frac{1}{\cos^{2} x} - \frac{\sin x}{\cos^{2} x}) d x \int (\frac{1}{\cos^{2} x} - \frac{\sin x}{\cos x} \times \frac{1}{\cos x}) d x = \int (\sec^{2} x - \sec x \tan x) d x = \tan x - \sec x + C$

Page No 18.209:

Question 48:

Evaluate: $\int \frac{x^{3} - 1}{x^{2}} d x$

Answer:

$\int (\frac{x^{3} - 1}{x^{2}}) d x = \int (\frac{x^{3}}{x^{2}} - \frac{1}{x^{2}}) d x = \int (x - x^{- 2}) d x = \frac{x^{2}}{2} - \frac{x^{- 2 + 1}}{- 2 + 1} + C = \frac{x^{2}}{2} + \frac{1}{x} + C$

Page No 18.209:

Question 49:

Evaluate: $\int \frac{x^{3} - x^{2} + x - 1}{x - 1} d x$

Answer:

$\int (\frac{x^{3} - x^{2} + x - 1}{x - 1}) d x = \int (\frac{x^{2} (x - 1) + 1 (x - 1)}{(x - 1)}) d x = \int \frac{(x^{2} + 1) (x - 1)}{(x - 1)} d x = \int (x^{2} + 1) d x = \frac{x^{3}}{3} + x + C$

Page No 18.209:

Question 50:

Evaluate: $\int \frac{e^{\tan^{- 1} x}}{1 + x^{2}} d x$

Answer:

$Let I = \int \frac{e^{\tan^{- 1} x}}{1 + x^{2}} d x Let \tan^{- 1} x = t \Rightarrow \frac{d x}{1 + x^{2}} = d t ∴ I = \int e^{t} d t = e^{t} + C = e^{\tan^{- 1} x} + C$

Page No 18.209:

Question 51:

Evaluate: $\int \frac{1}{\sqrt{1 - x^{2}}} d x$

Answer:

$Let I = \int \frac{d x}{\sqrt{1 - x^{2}}} Let x = \sin θ \Rightarrow d x = \cos θ ∴ I = \int \frac{\cos θ}{\cos θ} d θ = \int d θ = θ + C = \sin^{- 1} x + C (∵ x = \sin θ)$

Page No 18.209:

Question 52:

Write the value of $\int \sec x (\sec x + \tan x) d x$

Answer:

$\int \sec x (\sec x + \tan x) d x = \int (\sec^{2} x + \sec x \tan x) d x = \tan x + \sec x + C$

Page No 18.21:

Question 53:

Evaluate: $\int \frac{1}{x^{2} + 16} d x$

Answer:

$\int \frac{1}{x^{2} + 16} d x = \int \frac{1}{x^{2} + 4^{2}} d x = \frac{1}{4} \tan^{- 1} \frac{x}{4} + c Hence, \int \frac{1}{x^{2} + 16} d x = \frac{1}{4} \tan^{- 1} \frac{x}{4} + c$

Page No 18.21:

Question 54:

Evaluate: $\int (1 - x) \sqrt{x} d x$

Answer:

$\int (1 - x) \sqrt{x} d x = \int (\sqrt{x} - x \sqrt{x}) d x = \int (x^{\frac{1}{2}} - x^{\frac{3}{2}}) d x = \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} + c = \frac{2}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} + c Hence, \int (1 - x) \sqrt{x} d x = \frac{2}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}} + c$

Page No 18.21:

Question 55:

Evaluate: $\int \frac{x + \cos 6 x}{3 x^{2} + \sin 6 x} d x$

Answer:

$\int \frac{x + \cos 6 x}{3 x^{2} + \sin 6 x} d x Let (3 x^{2} + \sin 6 x) = t On differentiating both sides, we get (6 x + 6 \cos 6 x) d x = d t ∴ \int \frac{x + \cos 6 x}{3 x^{2} + \sin 6 x} d x = \frac{1}{6} \int \frac{1}{t} d t = \frac{1}{6} \log |t| + c = \frac{1}{6} \log |3 x^{2} + \sin 6 x| + c Hence, \int \frac{x + \cos 6 x}{3 x^{2} + \sin 6 x} d x = \frac{1}{6} \log |3 x^{2} + \sin 6 x| + c$

Page No 18.21:

Question 56:

$If \int (\frac{x - 1}{x^{2}}) e^{x} d x = f (x) e^{x} + C, then write the value of f (x) .$

Answer:

$\int (\frac{x - 1}{x^{2}}) e^{x} d x = \int (\frac{x}{x^{2}} - \frac{1}{x^{2}}) e^{x} d x = \int (\frac{1}{x} - \frac{1}{x^{2}}) e^{x} d x Consider, f (x) = \frac{1}{x}, then f^{'} (x) = - \frac{1}{x^{2}} Thus, the given integrand is of the form e^{x} [f (x) + f^{'} (x)] . Therefore, \int (\frac{x - 1}{x^{2}}) e^{x} d x = \frac{1}{x} e^{x} + C Hence, f (x) = \frac{1}{x} .$

Page No 18.21:

Question 57:

$If \int e^{x} (\tan x + 1) \sec x d x = e^{x} f (x) + C, then write the value of f (x) .$

Answer:

$\int e^{x} (\tan x + 1) \sec x d x = \int e^{x} (\tan x \sec x + \sec x) d x = \int e^{x} (\sec x + \tan x \sec x) d x Consider, f (x) = \sec x, then f^{'} (x) = \tan x \sec x Thus, the given integrand is of the form e^{x} [f (x) + f^{'} (x)] . Therefore, \int e^{x} (\tan x + 1) \sec x d x = \sec x e^{x} + C Hence, f (x) = \sec x .$

Page No 18.21:

Question 58:

Evaluate: $\int \frac{2}{1 - \cos 2 x} d x$

Answer:

$\int \frac{2}{1 - \cos 2 x} d x = \int \frac{2}{2 \sin^{2} x} d x = \int {cosec}^{2} x d x = - \cot x + c Hence, \int \frac{2}{1 - \cos 2 x} d x = - \cot x + c .$

Page No 18.21:

Question 59:

Write the anti-derivative of $(3 \sqrt{x} + \frac{1}{\sqrt{x}}) .$

Answer:

$\int (3 \sqrt{x} + \frac{1}{\sqrt{x}}) d x = 3 \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + c = 2 x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + c = 2 (x^{\frac{3}{2}} + x^{\frac{1}{2}}) + c Hence, the anti - derivative of (3 \sqrt{x} + \frac{1}{\sqrt{x}}) is 2 (x^{\frac{3}{2}} + x^{\frac{1}{2}}) + c .$

Page No 18.21:

Question 60:

Evaluate: $\int \cos^{- 1} (\sin x) d x$

Answer:

$\int \cos^{- 1} (\sin x) d x = \int \cos^{- 1} (\cos (\frac{π}{2} - x)) d x = \int (\frac{π}{2} - x) d x = \frac{π}{2} x - \frac{1}{2} x^{2} + c Hence, \int \cos^{- 1} (\sin x) d x = \frac{π}{2} x - \frac{1}{2} x^{2} + c$

Page No 18.21:

Question 61:

Evaluate: $\int \frac{1}{\sin^{2} x \cos^{2} x} d x$

Answer:

$\int \frac{1}{\sin^{2} x \cos^{2} x} d x = 4 \int \frac{1}{4 \sin^{2} x \cos^{2} x} d x = 4 \int \frac{1}{\sin^{2} (2 x)} d x = 4 \int {cosec}^{2} (2 x) d x = - \frac{4}{2} \cot (2 x) + c = - 2 \cot (2 x) + c Hence, \int \frac{1}{\sin^{2} x \cos^{2} x} d x = - 2 \cot (2 x) + c$

Page No 18.21:

Question 62:

Evaluate : $\int \frac{1}{x (1 + \log x)} d x$

Answer:

I = $\int \frac{1}{x (1 + \log x)} d x$

Let (1 + log x) = t

$or, \frac{1}{x} d x = d t \Rightarrow I = \int \frac{1}{t} d t$

$\Rightarrow I = \log |t| + C ∴ I = \log |1 + \log x| + C$

Page No 18.23:

Question 1:

$\int {(2 x - 3)}^{5} + \sqrt{3 x + 2} d x$

Answer:

$\int [{(2 x - 3)}^{5} + \sqrt{3 x + 2}] d x = \int {(2 x - 3)}^{5} d x + \int {(3 x + 2)}^{\frac{1}{2}} d x = \frac{{(2 x - 3)}^{5 + 1}}{2 (5 + 1)} + \frac{{(3 x + 2)}^{\frac{1}{2} + 1}}{3 (\frac{1}{2} + 1)} + C = \frac{{(2 x - 3)}^{6}}{12} + \frac{2}{9} {(3 x + 2)}^{\frac{3}{2}} + C$

Page No 18.23:

Question 2:

$\int \frac{1}{{(7 x - 5)}^{3}} + \frac{1}{\sqrt{5 x - 4}} d x$

Answer:

$\int [\frac{1}{{(7 x - 5)}^{3}} + \frac{1}{\sqrt{5 x - 4}}] d x = \int [{(7 x - 5)}^{- 3} + {(5 x - 4)}^{- \frac{1}{2}}] d x = \frac{{(7 x - 5)}^{- 3 + 1}}{7 (- 3 + 1)} + \frac{{(5 x - 4)}^{- \frac{1}{2} + 1}}{5 (- \frac{1}{2} + 1)} + C = \frac{{(7 x - 5)}^{- 2}}{- 14} + \frac{2}{5} {(5 x - 4)}^{\frac{1}{2}} + C$

Page No 18.23:

Question 3:

$\int \frac{1}{2 - 3 x} + \frac{1}{\sqrt{3 x - 2}} d x$

Answer:

$\int (\frac{1}{2 - 3 x} + \frac{1}{\sqrt{3 x - 2}}) d x = \int \frac{d x}{2 - 3 x} + \int {(3 x - 2)}^{- \frac{1}{2}} d x = \frac{\ln (2 - 3 x)}{- 3} + [\frac{{(3 x - 2)}^{- \frac{1}{2} + 1}}{3 (- \frac{1}{2} + 1)}] + C = \frac{\ln (2 - 3 x)}{- 3} + \frac{2}{3} {(3 x - 2)}^{\frac{1}{2}} + C = - \frac{1}{3} \ln (2 - 3 x) + \frac{2}{3} \sqrt{3 x - 2} + C$

Page No 18.23:

Question 4:

$\int \frac{x + 3}{{(x + 1)}^{4}} d x$

Answer:

$\int [\frac{x + 3}{{(x + 1)}^{4}}] d x = \int [\frac{x + 1 + 2}{{(x + 1)}^{4}}] d x = \int [\frac{(x + 1)}{{(x + 1)}^{4}} + \frac{2}{{(x + 1)}^{4}}] d x = \int \frac{d x}{{(x + 1)}^{3}} + 2 \int \frac{d x}{{(x + 1)}^{4}} = \int {(x + 1)}^{- 3} d x + 2 \int {(x + 1)}^{- 4} d x = [\frac{{(x + 1)}^{- 3 + 1}}{- 3 + 1}] + 2 [\frac{{(x + 1)}^{- 4 + 1}}{- 4 + 1}] + C = - \frac{1}{2} {(x + 1)}^{- 2} - \frac{2}{3} {(x + 1)}^{- 3} + C = - \frac{1}{2 {(x + 1)}^{2}} - \frac{2}{3 {(x + 1)}^{3}} + C$

Page No 18.23:

Question 5:

$\int \frac{1}{\sqrt{x + 1} + \sqrt{x}} d x$

Answer:

$\int \frac{d x}{\sqrt{x + 1} + \sqrt{x}}$

Rationalise the denominator

$= \int \frac{(\sqrt{x + 1} - \sqrt{x})}{(\sqrt{x + 1} + \sqrt{x}) (\sqrt{x + 1} - \sqrt{x})} d x = \int \frac{(\sqrt{x + 1} - \sqrt{x})}{(x + 1) - x} d x = \int {(x + 1)}^{\frac{1}{2}} d x - \int x^{\frac{1}{2}} d x = \frac{{(x + 1)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} = \frac{2}{3} {(x + 1)}^{\frac{3}{2}} - \frac{2}{3} x^{\frac{3}{2}} + C$

Page No 18.23:

Question 6:

$\int \frac{1}{\sqrt{2 x + 3} + \sqrt{2 x - 3}} d x$

Answer:

$\int \frac{d x}{(\sqrt{2 x + 3} + \sqrt{2 x} - 3)}$
Rationalise the denominator
$= \int \frac{(\sqrt{2 x + 3} - \sqrt{2 x - 3})}{(\sqrt{2 x + 3} + \sqrt{2 x - 3}) (\sqrt{2 x + 3} - \sqrt{2 x - 3})} d x = \int \frac{(\sqrt{2 x + 3} - \sqrt{2 x - 3})}{(2 x + 3) - (2 x - 3)} d x = \frac{1}{6} \int {(2 x + 3)}^{\frac{1}{2}} d x - \frac{1}{6} \int {(2 x - 3)}^{\frac{1}{2}} d x = \frac{1}{6} [\frac{{(2 x + 3)}^{\frac{1}{2} + 1}}{2 (\frac{1}{2} + 1)}] - \frac{1}{6} [\frac{{(2 x - 3)}^{\frac{1}{2} + 1}}{2 (\frac{1}{2} + 1)}] + C = \frac{1}{18} \{{(2 x + 3)}^{\frac{3}{2}} - {(2 x - 3)}^{\frac{3}{2}}\} + C$

Page No 18.23:

Question 7:

$\int \frac{2 x}{{(2 x + 1)}^{2}} d x$

Answer:

$\int \frac{2 x}{{(2 x + 1)}^{2}} d x = \int (\frac{2 x + 1 - 1}{{(2 x + 1)}^{2}}) d x = \int [\frac{2 x + 1}{{(2 x + 1)}^{2}} - \frac{1}{{(2 x + 1)}^{2}}] d x = \int \frac{d x}{2 x + 1} - \int {(2 x + 1)}^{- 2} d x = \frac{\log (2 x + 1)}{2} - [\frac{{(2 x + 1)}^{- 2 + 1}}{2 (- 2 + 1)}] + C = \frac{\log (2 x + 1)}{2} + \frac{{(2 x + 1)}^{- 1}}{2} + C = \frac{\log (2 x + 1)}{2} + \frac{1}{2 (2 x + 1)} + C$

Page No 18.23:

Question 8:

$\int \frac{1}{\sqrt{x + a} + \sqrt{x + b}} d x$

Answer:

$\int \frac{d x}{(\sqrt{x + a} + \sqrt{x + b})} = \int \frac{(\sqrt{x + a} - \sqrt{x - b})}{(\sqrt{x + a} + \sqrt{x + b}) (\sqrt{x + a} - \sqrt{x + b})} d x = \int \frac{(\sqrt{x + a} - \sqrt{x + b})}{(x + a) - (x + b)} d x = \frac{1}{a - b} \int {(x + a)}^{\frac{1}{2}} - \frac{1}{a - b} \int {(x + b)}^{\frac{1}{2}} d x = \frac{1}{(a - b)} [\frac{{(x + a)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] - \frac{1}{a - b} [\frac{{(x + b)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{2}{3 (a - b)} [{(x + a)}^{\frac{3}{2}} - {(x + b)}^{\frac{3}{2}}] + C$

Page No 18.23:

Question 9:

$\int \sin x \sqrt{1 + \cos 2 x} d x$

Answer:

$\int \sin x \sqrt{1 + \cos 2 x} d x = \int \sin x . \sqrt{2 \cos^{2} x} d x [∴ 1 + \cos 2 x = 2 \cos^{2} x] = \sqrt{2} \int \sin x \cos x d x = \frac{\sqrt{2}}{2} \int 2 \sin x \cos x d x = \frac{1}{\sqrt{2}} \int \sin 2 x d x = \frac{1}{\sqrt{2}} [\frac{- \cos 2 x}{2}] + C = \frac{- 1}{2 \sqrt{2}} \cos 2 x + C$

Page No 18.23:

Question 10:

$\int \frac{1 + \cos x}{1 - \cos x} d x$

Answer:

$\int (\frac{1 + \cos x}{1 - \cos x}) d x = \int (\frac{2 \cos^{2} \frac{x}{2}}{2 \sin^{2} \frac{x}{2}}) d x [∴ 1 + \cos x = 2 \cos^{2} \frac{x}{2} & 1 - \cos x = 2 \sin^{2} \frac{x}{2}] = \int \cot^{2} \frac{x}{2} d x = \int ({cosec}^{2} \frac{x}{2} - 1) d x = \frac{- \cot (\frac{x}{2})}{\frac{1}{2}} - x + C = - 2 \cot (\frac{x}{2}) - x + C$

Page No 18.23:

Question 11:

$\int \frac{1 - \cos x}{1 + \cos x} d x$

Answer:

$\int (\frac{1 - \cos x}{1 + \cos x}) d x = \int (\frac{2 \sin^{2} \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}) d x [1 - \cos x = 2 \sin^{2} \frac{x}{2} & 1 + \cos x = 2 \cos^{2} \frac{x}{2}] = \int \tan^{2} \frac{x}{2} d x = \int (\sec^{2} \frac{x}{2} - 1) d x = \frac{\tan \frac{x}{2}}{\frac{1}{2}} - x + C = 2 \tan \frac{x}{2} - x + C$

Page No 18.23:

Question 12:

$\int \frac{1}{1 - \sin \frac{x}{2}} d x$

Answer:

$\int \frac{d x}{1 - \sin (\frac{x}{2})} = \int \frac{(1 + \sin \frac{x}{2})}{(1 - \sin \frac{x}{2}) (1 + \sin \frac{x}{2})} d x = \int (\frac{1 + \sin \frac{x}{2}}{1 - \sin^{2} \frac{x}{2}}) d x = \int (\frac{1 + \sin \frac{x}{2}}{\cos^{2} \frac{x}{2}}) d x = \int (\sec^{2} \frac{x}{2} + \sec \frac{x}{2} \tan \frac{x}{2}) d x = \frac{\tan (\frac{x}{2})}{\frac{1}{2}} + \frac{\sec (\frac{x}{2})}{\frac{1}{2}} + C = 2 (\tan \frac{x}{2} + \sec \frac{x}{2}) + C$

Page No 18.23:

Question 13:

$\int \frac{1}{1 + \cos 3 x} d x$

Answer:

$\int \frac{d x}{1 + \cos 3 x} = \int \frac{(1 - \cos 3 x)}{(1 + \cos 3 x) (1 - \cos 3 x)} d x = \int (\frac{1 - \cos 3 x}{1 - \cos^{2} 3 x}) d x = \int (\frac{1 - \cos 3 x}{\sin^{2} 3 x}) d x = \int {cosec}^{2} 3 x d x - \int cosec 3 x \cot 3 x d x = - \frac{\cot 3 x}{3} + \frac{cosec 3 x}{3} + C = \frac{1}{3} [cosec 3 x - \cot 3 x] + C = \frac{1}{3} [\frac{1}{\sin 3 x} - \frac{\cos 3 x}{\sin 3 x}] + C = \frac{1}{3} [\frac{1 - \cos 3 x}{\sin 3 x}] + C$

Page No 18.23:

Question 14:

$\int {(e^{x} + 1)}^{2} e^{x} d x$

Answer:

$\int {(e^{x} + 1)}^{2} e^{x} d x = \int (e^{2 x} + 2 e^{x} + 1) e^{x} d x = \int (e^{3 x} + 2 e^{2 x} + e^{x}) d x = [\frac{e^{3 x}}{3} + \frac{2 e^{2 x}}{2} + e^{x}] + C$

Page No 18.23:

Question 15:

$\int {(e^{x} + \frac{1}{e^{x}})}^{2} d x$

Answer:

$\int {(e^{x} + \frac{1}{e^{x}})}^{2} d x = \int (e^{2 x} + \frac{1}{e^{2 x}} + 2 e^{x} \times \frac{1}{e^{x}}) d x = \int (e^{2 x} + e^{- 2 x} + 2) d x = \frac{e^{2 x}}{2} + \frac{e^{- 2 x}}{- 2} + 2 x + C = \frac{e^{2 x}}{2} - \frac{e^{- 2 x}}{2} + 2 x + C$

Page No 18.23:

Question 16:

$\int \frac{1 + \cos 4 x}{\cot x - \tan x} d x$

Answer:

$\int (\frac{1 + \cos 4 x}{\cot x - \tan x}) d x = \int \frac{(1 + \cos 4 x)}{(\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x})} d x = \int \frac{2 \cos^{2} 2 x \times \sin x \cos x}{(\cos^{2} x - \sin^{2} x)} d x = \int \frac{\cos^{2} 2 x \times 2 \sin x \cos x}{\cos 2 x} d x = \int \cos 2 x \sin 2 x d x = \frac{1}{2} \int 2 \sin 2 x \cos 2 x d x = \frac{1}{2} \int \sin 4 x d x = \frac{1}{2} [- \frac{\cos 4 x}{4}] + C = - \frac{1}{8} \cos 4 x + C$

Page No 18.23:

Question 17:

$\int \frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} d x$

Answer:

$\int \frac{d x}{\sqrt{x + 3} - \sqrt{x + 2}}$
Rationalising the denominator
$= \int \frac{(\sqrt{x + 3} + \sqrt{x + 2})}{(\sqrt{x + 3} - \sqrt{x + 2}) (\sqrt{x + 3} + \sqrt{x + 2})} d x = \int [\frac{{(x + 3)}^{\frac{1}{2}} + {(x + 2)}^{\frac{1}{2}}}{(x + 3) - (x + 2)}] d x = \int [{(x + 3)}^{\frac{1}{2}} + {(x + 2)}^{\frac{1}{2}}] d x = [\frac{{(x + 3)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + \frac{{(x + 2)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{2}{3} {(x + 3)}^{\frac{3}{2}} + \frac{2}{3} {(x + 2)}^{\frac{3}{2}} + C = \frac{2}{3} \{{(x + 3)}^{\frac{3}{2}} + {(x + 2)}^{\frac{3}{2}}\} + C$

Page No 18.23:

Question 18:

$\int \tan^{2} (2 x - 3) d x$

Answer:

$\int \tan^{2} (2 x - 3) d x = \int [\sec^{2} (2 x - 3) - 1] d x = \int \sec^{2} (2 x - 3) d x - \int 1 d x = \frac{\tan (2 x - 3)}{2} - x + C$

Page No 18.24:

Question 19:

$\int \frac{1}{\cos^{2} x {(1 - \tan x)}^{2}} d x$

Answer:

$Let I = \int \frac{1}{\cos^{2} x {(1 - \tan x)}^{2}} d x = \int \frac{\sec^{2} x}{{(1 - \tan x)}^{2}} d x = \int \frac{\sec^{2} x d x}{{(1 - \tan x)}^{2}}$

Let 1 $-$ tan x = t
$- \sec^{2} x d x = d t \Rightarrow \sec^{2} x d x = - d t$

$∴ I = \int \frac{- d t}{t^{2}} = - \int t^{- 2} d t = - [\frac{t^{- 2 + 1}}{- 2 + 1}] + C = \frac{1}{t} + C = \frac{1}{1 - \tan x} + C$

Page No 18.30:

Question 1:

$\int \frac{x^{2} + 5 x + 2}{x + 2} d x$

Answer:

$\int \frac{(x^{2} + 5 x + 2)}{(x + 2)} d x = \int \frac{x^{2}}{x + 2} d x + 5 \int \frac{x d x}{x + 2} + 2 \int \frac{d x}{x + 2} = \int (\frac{x^{2} - 4 + 4}{x + 2}) d x + 5 \int (\frac{x + 2 - 2}{x + 2}) d x + 2 \int \frac{d x}{x + 2} = \int \frac{(x - 2) (x + 2)}{(x + 2)} d x + \int \frac{4}{x + 2} d x + 5 \int (1 - \frac{2}{x + 2}) d x + 2 \int \frac{d x}{x + 2} = \int (x - 2) d x + 4 \int \frac{d x}{x + 2} + 5 \int d x - 10 \int \frac{d x}{x + 2} + 2 \int \frac{d x}{x + 2} = \int (x - 2) d x - 4 \int \frac{d x}{x + 2} + 5 \int d x = (\frac{x^{2}}{2} - 2 x) - 4 \ln |x + 2| + 5 x + C = \frac{x^{2}}{2} + 3 x - 4 \ln |x + 2| + C$

Page No 18.30:

Question 2:

$\int \frac{x^{3}}{x - 2} d x$

Answer:

$\int (\frac{x^{3}}{x - 2}) d x = \int (\frac{x^{3} - 8 + 8}{x - 2}) d x = \int [\frac{(x^{3} - 2^{3})}{(x - 2)} + \frac{8}{x - 2}] d x = \int [\frac{(x - 2) (x^{2} + 2 x + 4)}{(x - 2)} + \frac{8}{x - 2}] d x = \int (x^{2} + 2 x + 4) d x + 8 \int \frac{d x}{x - 2} = \frac{x^{3}}{3} + \frac{2 x^{2}}{2} + 4 x + 8 \ln |x - 2| + C = \frac{x^{3}}{3} + x^{2} + 4 x + 8 \ln |x - 2| + C$

Page No 18.30:

Question 3:

$\int \frac{x^{2} + x + 5}{3 x + 2} d x$

Answer:

$\int \frac{(x^{2} + x + 5)}{(3 x + 2)} d x = \frac{1}{9} \int \frac{9 x^{2} + 9 x + 45}{(3 x + 2)} d x = \frac{1}{9} [\int \frac{9 x^{2} - 4}{3 x + 2} d x + \int \frac{9 x + 6}{3 x + 2} d x + \int \frac{43}{3 x + 2} d x] = \frac{1}{9} [\int \frac{(3 x - 2) (3 x + 2)}{(3 x + 2)} d x + \int \frac{3 (3 x + 2)}{3 x + 2} d x + 43 \int \frac{d x}{3 x + 2}] = \frac{1}{9} [\int (3 x - 2) d x + 3 \int 1 d x + 43 \int \frac{d x}{3 x + 2}] = \frac{1}{9} [(3 \frac{x^{2}}{2} - 2 x) + 3 x + \frac{43}{3} \ln |3 x + 2| + C] = \frac{1}{9} [\frac{3}{2} x^{2} + x - \frac{43}{3} \ln |3 x + 2| + C] = \frac{1}{6} x^{2} + \frac{1}{9} x - \frac{43}{27} \ln |3 x + 2| + C$

Page No 18.30:

Question 4:

$\int \frac{2 x + 3}{{(x - 1)}^{2}} d x$

Answer:

$\int (\frac{2 x + 3}{{(x - 1)}^{2}}) d x = \int [\frac{2 x - 2 + 2 + 3}{{(x - 1)}^{2}}] d x = \int [\frac{2 (x - 1) + 5}{{(x - 1)}^{2}}] d x = 2 \int \frac{d x}{(x - 1)} + 5 \int {(x - 1)}^{- 2} d x = 2 \ln |x - 1| + 5 [\frac{{(x - 1)}^{- 2 + 1}}{- 2 + 1}] + C = 2 \ln |x - 1| - \frac{5}{x - 1} + C$

Page No 18.30:

Question 5:

$\int \frac{x^{2} + 3 x - 1}{{(x + 1)}^{2}} d x$

Answer:

$Let I = \int [\frac{x^{2} + 3 x - 1}{{(x + 1)}^{2}}] d x$
Putting x + 1 = t
⇒ x = t – 1
& dx = dt

$∴ I = \int [\frac{{(t - 1)}^{2} + 3 (t - 1) - 1}{t^{2}}] d t = \int (\frac{t^{2} - 2 t + 1 + 3 t - 3 - 1}{t^{2}}) d t = \int (\frac{t^{2} + t - 3}{t^{2}}) d t = \int (1 + \frac{1}{t} - 3 t^{- 2}) d t = t + \log |t| - 3 (\frac{t^{- 2 + 1}}{- 2 + 1}) + C = t + \log |t| + \frac{3}{t} + C = x + 1 + \log |x + 1| + \frac{3}{x + 1} + C [∵ t = x + 1]$
Let C + 1 = C′
$= x + \log (x + 1) + \frac{3}{x + 1} + C'$

Page No 18.30:

Question 6:

$\int \frac{2 x - 1}{{(x - 1)}^{2}} d x$

Answer:

$\int [\frac{2 x - 1}{{(x - 1)}^{2}}] d x = \int [\frac{2 x - 2 + 2 - 1}{{(x - 1)}^{2}}] d x = \int (\frac{2 (x - 1)}{{(x - 1)}^{2}} + \frac{1}{{(x - 1)}^{2}}) d x = 2 \int \frac{d x}{x - 1} + \int {(x - 1)}^{- 2} d x = 2 \ln |x - 1| + \frac{{(x - 1)}^{- 2 + 1}}{- 2 + 1} + C = 2 \ln |x - 1| - \frac{1}{x - 1} + C$

Page No 18.33:

Question 1:

$\int \frac{x + 1}{\sqrt{2 x + 3}} d x$

Answer:

$\int (\frac{x + 1}{\sqrt{2 x + 3}}) d x = \frac{1}{2} \int (\frac{2 x + 2}{\sqrt{2 x + 3}}) d x = \frac{1}{2} \int (\frac{2 x + 3 - 1}{\sqrt{2 x + 3}}) d x = \frac{1}{2} \int (\frac{2 x + 3}{\sqrt{2 x + 3}} - \frac{1}{\sqrt{2 x + 3}}) d x = \frac{1}{2} \int (\sqrt{2 x + 3} - \frac{1}{\sqrt{2 x + 3}}) d x = \frac{1}{2} [\int {(2 x + 3)}^{\frac{1}{2}} d x - \int {(2 x + 3)}^{- \frac{1}{2}} d x] = \frac{1}{2} [\frac{{(2 x + 3)}^{\frac{1}{2} + 1}}{2 (\frac{1}{2} + 1)} - \frac{{(2 x + 3)}^{- \frac{1}{2} + 1}}{2 (- \frac{1}{2} + 1)} + C] = \frac{1}{2} [\frac{1}{3} {(2 x + 3)}^{\frac{3}{2}} - {(2 x + 3)}^{\frac{1}{2}} + C] = \frac{1}{6} {(2 x + 3)}^{\frac{3}{2}} - \frac{1}{2} {(2 x + 3)}^{\frac{1}{2}} + C$

Page No 18.33:

Question 2:

$\int x \sqrt{x + 2} d x$

Answer:

$Let I = \int x \sqrt{x + 2} d x$
Putting x + 2 = t
Then, x = t – 2
Difference both sides
dx = dt
Now, integral becomes
$I = \int (t - 2) \sqrt{t} d t = \int (t^{\frac{3}{2}} - 2 t^{\frac{1}{2}}) d t = [\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} - 2 \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{2}{5} t^{\frac{5}{2}} - \frac{4}{3} t^{\frac{3}{2}} + C = \frac{2}{5} {(x + 2)}^{\frac{5}{2}} - \frac{4}{3} {(x + 2)}^{\frac{2}{3}} + C$

Page No 18.33:

Question 3:

$\int \frac{x - 1}{\sqrt{x + 4}} d x$

Answer:

$Let I = \int (\frac{x - 1}{\sqrt{x + 4}}) d x$

Putting x + 4 = t
Then, x = t – 4
Difference both sides
dx = dt
Now integral becomes,
$I = \int (\frac{t - 4 - 1}{\sqrt{t}}) d t = \int (\frac{t}{\sqrt{t}} - \frac{5}{\sqrt{t}}) d t = \int (t^{\frac{1}{2}} - 5 t^{- \frac{1}{2}}) d t = \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - 5 \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C = \frac{2}{3} t^{\frac{3}{2}} - 10 \sqrt{t} + C = \frac{2}{3} {(x + 4)}^{\frac{3}{2}} - 10 {(x + 4)}^{\frac{1}{2}} + C$

Page No 18.33:

Question 4:

$\int (x + 2) \sqrt{3 x + 5} d x$

Answer:

$Let I = \int (x + 2) \sqrt{3 x + 5} d x Putting 3 x + 5 = t \Rightarrow x = \frac{t - 5}{3}$

$\Rightarrow 3 d x = d t \Rightarrow d x = \frac{d t}{3}$

$∴ I = \int (\frac{t - 5}{3} + 2) \sqrt{t} \frac{d t}{3} = \frac{1}{3} \int (\frac{t - 5 + 6}{3}) \sqrt{t} d t = \frac{1}{9} \int (t + 1) \sqrt{t} d t = \frac{1}{9} \int (t^{\frac{3}{2}} + t^{\frac{1}{2}}) d t = \frac{1}{9} [\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} + \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{1}{9} [\frac{2}{5} t^{\frac{5}{2}} + \frac{2}{3} t^{\frac{3}{2}}] + C = \frac{1}{9} [\frac{2}{5} {(3 x + 5)}^{\frac{5}{2}} + \frac{2}{3} {(3 x + 5)}^{\frac{3}{2}}] + C [∵ t = 3 x + 5] = \frac{2}{9} [{(3 x + 5)}^{\frac{3}{2}} \{\frac{3 x + 5}{5} + \frac{1}{3}\}] + C = \frac{2}{9} [{(3 x + 5)}^{\frac{3}{2}} \{\frac{9 x + 15 + 5}{15}\}] + C = \frac{2}{9} [{(3 x + 5)}^{\frac{3}{2}} \{\frac{9 x + 20}{15}\}] + C = \frac{2}{135} {(3 x + 5)}^{\frac{3}{2}} (9 x + 20) + C$

Page No 18.33:

Question 5:

$\int \frac{2 x + 1}{\sqrt{3 x + 2}} d x$

Answer:

$\int (\frac{2 x + 1}{\sqrt{3 x + 2}}) d x = \frac{1}{3} \int (\frac{6 x + 3}{\sqrt{3 x + 2}}) d x = \frac{1}{3} \int (\frac{6 x + 4 - 1}{\sqrt{3 x + 2}}) d x = \frac{1}{3} \int (\frac{2 (3 x + 2)}{\sqrt{3 x + 2}} - \frac{1}{\sqrt{3 x + 2}}) d x = \frac{1}{3} \int (2 \sqrt{3 x + 2} - \frac{1}{\sqrt{3 x + 2}}) d x = \frac{1}{3} [\int 2 {(3 x + 2)}^{\frac{1}{2}} d x - \int {(3 x + 2)}^{- \frac{1}{2}} d x] = \frac{1}{3} [2 \{\frac{{(3 x + 2)}^{\frac{1}{2} + 1}}{3 (\frac{1}{2} + 1)}\} - \frac{{(3 x + 2)}^{- \frac{1}{2} + 1}}{(- \frac{1}{2} + 1) \times 3}] + C = \frac{1}{3} [\frac{4}{9} {(3 x + 2)}^{\frac{3}{2}} - \frac{2}{3} {(3 x + 2)}^{\frac{1}{2}}] + C = \frac{4}{27} {(3 x + 2)}^{\frac{3}{2}} - \frac{2}{9} {(3 x + 2)}^{\frac{1}{2}} + C = \sqrt{3 x + 2} (\frac{4}{27} (3 x + 2) - \frac{2}{9}) + C = \sqrt{3 x + 2} (\frac{4 (3 x + 2) - 6}{27}) + C = \sqrt{3 x + 2} (\frac{12 x + 8 - 6}{27}) + C = \frac{2}{27} (6 x + 1) \sqrt{3 x + 2} + C$

Page No 18.33:

Question 6:

$\int \frac{3 x + 5}{\sqrt{7 x + 9}} d x$

Answer:

$Let I = \int (\frac{3 x + 5}{\sqrt{7 x + 9}}) d x Putting 7 x + 9 = t \Rightarrow x = \frac{t - 9}{7}$

$& 7 d x = d t \Rightarrow d x = \frac{d t}{7}$

$∴ I = \int (\frac{3 (\frac{t - 9}{7}) + 5}{\sqrt{t}}) d t = \int (\frac{3}{7} \frac{t}{\sqrt{t}} - \frac{27}{7 \sqrt{t}} + \frac{5}{\sqrt{t}}) \frac{d t}{7} = \frac{3}{7 \times 7} \int t^{\frac{1}{2}} d t - \frac{27}{7 \times 7} \int t^{- \frac{1}{2}} d t + \frac{5}{7} \int t^{- \frac{1}{2}} d t = \frac{3}{7 \times 7} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] - \frac{27}{7 \times 7} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + \frac{5}{7} [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = \frac{2}{7 \times 7} t^{\frac{3}{2}} - \frac{27}{7 \times 7} \times 2 t^{\frac{1}{2}} + \frac{10 \sqrt{t}}{7} + C = \frac{2}{7 \times 7} {(7 x + 9)}^{\frac{3}{2}} - \frac{54}{7 \times 7} {(7 x + 9)}^{\frac{1}{2}} + \frac{10}{7} \sqrt{7 x + 9} + C [∵ t = 7 x + 9] = \frac{2}{7 \times 7} {(7 x + 9)}^{\frac{3}{2}} + (10 - \frac{54}{7}) \frac{\sqrt{7 x + 9}}{7} + C = \frac{2}{7} {(7 x + 9)}^{\frac{3}{2}} + (\frac{70 - 54}{7}) \frac{\sqrt{7 x + 9}}{7} + C = \frac{2}{7 \times 7} {(7 x + 9)}^{\frac{3}{2}} + \frac{16}{7 \times 7} \sqrt{7 x + 9} + C = \frac{2}{7 \times 7} [{(7 x + 9)}^{\frac{1}{2}} [7 x + 9 + 8]] + C = \frac{2}{49} [{(7 x + 9)}^{\frac{1}{2}} (7 x + 17)] + C$

Page No 18.33:

Question 7:

$\int \frac{x}{\sqrt{x + 4}} d x$

Answer:

$\int \frac{x}{\sqrt{x + 4}} d x = \int (\frac{x + 4 - 4}{\sqrt{x + 4}}) d x = \int (\sqrt{x + 4} - \frac{4}{\sqrt{x + 4}}) d x = \int {(x + 4)}^{\frac{1}{2}} d x - 4 \int {(x + 4)}^{- \frac{1}{2}} d x = \frac{{(x + 4)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - 4 \frac{{[x + 4]}^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C = \frac{2}{3} {(x + 4)}^{\frac{3}{2}} - 8 {(x + 4)}^{\frac{1}{2}} + C = {(x + 4)}^{\frac{1}{2}} [\frac{2}{3} (x + 4) - 8] + C = {(x + 4)}^{\frac{1}{2}} [\frac{2 x + 8 - 24}{3}] + C = {(x + 4)}^{\frac{1}{2}} [\frac{2 x - 16}{3}] + C = \frac{2}{3} (x - 8) \sqrt{x + 4} + C$

Page No 18.33:

Question 8:

$\int \frac{2 - 3 x}{\sqrt{1 + 3 x}} d x$

Answer:

$Let I = \int (\frac{2 - 3 x}{\sqrt{1 + 3 x}}) d x$
Putting 1 + 3x = t
⇒ 3x = t – 1

$& 3 d x = d t \Rightarrow d x = \frac{d t}{3}$

$∴ I = \int (\frac{2 - (t - 1)}{\sqrt{t}}) d t = \int (\frac{3 - t}{\sqrt{t}}) d t = \int (3 t^{- \frac{1}{2}} - t^{\frac{1}{2}}) d t = 3 \int t^{- \frac{1}{2}} d t - \int t^{\frac{1}{2}} d t = 3 [\frac{t^{\frac{- 1}{2} + 1}}{- \frac{1}{2} + 1}] - [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = 6 \sqrt{t} - \frac{2}{3} t^{\frac{3}{2}} + C = 2 \sqrt{t} (3 - \frac{t}{3}) + C = 2 \sqrt{t} (\frac{9 - t}{3}) + C [∵ t = 1 + 3 x] = \frac{2}{3} \sqrt{1 + 3 x} \{\frac{9 - (1 + 3 x)}{3}\} + C = \frac{2}{3 \times 3} \sqrt{1 + 3 x} (8 - 3 x) + C = \frac{2}{9} (8 - 3 x) \sqrt{1 + 3 x} + C$

Page No 18.33:

Question 9:

$\int (5 x + 3) \sqrt{2 x - 1} d x$

Answer:

$Let I = \int (5 x + 3) \sqrt{2 x - 1} d x Putting 2 x - 1 = t \Rightarrow 2 x = t + 1 \Rightarrow x = \frac{t + 1}{2}$

$& 2 d x = d t \Rightarrow d x = \frac{d t}{2}$

$∴ I = \int [5 (\frac{t + 1}{2}) + 3] \cdot \sqrt{t} \cdot \frac{d t}{2} = \int (\frac{5 t}{2} + \frac{5}{2} + 3) \times \frac{\sqrt{t} d t}{2} = \frac{1}{4} \int (5 t + 11) t^{\frac{1}{2}} d t = \frac{1}{4} \int (5 t^{\frac{3}{2}} + 11 t^{\frac{1}{2}}) d t = \frac{1}{4} [5 \frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} + 11 \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{1}{4} \times \frac{2}{5} \times 5 t^{\frac{5}{2}} + \frac{1}{4} \times 11 \times \frac{2}{3} t^{\frac{3}{2}} + C = \frac{1}{2} t^{\frac{5}{2}} + \frac{11}{6} t^{\frac{3}{2}} + C = \frac{t^{\frac{3}{2}}}{2} [t + \frac{11}{3}] + C = \frac{t^{\frac{3}{2}}}{2} [\frac{3 t + 11}{3}] + C = \frac{{(2 x - 1)}^{\frac{3}{2}}}{2} [\frac{3 (2 x - 1) + 11}{3}] + C [∵ t = 2 x - 1] = \frac{{(2 x - 1)}^{\frac{3}{2}}}{2} [\frac{6 x - 3 + 11}{3}] + C = {(\frac{2 x - 1}{2})}^{\frac{3}{2}} [\frac{2 (3 x + 4)}{3}] + C = \frac{{(2 x - 1)}^{\frac{3}{2}} (3 x + 4)}{3} + C$

Page No 18.33:

Question 10:

$\int \frac{x}{\sqrt{x + a} - \sqrt{x + b}} d x$

Answer:

$\int \frac{x}{\sqrt{x + a} - \sqrt{x + b}} d x = \int \frac{x}{\sqrt{x + a} - \sqrt{x + b}} \times \frac{\sqrt{x + a} + \sqrt{x + b}}{\sqrt{x + a} + \sqrt{x + b}} d x = \int \frac{x (\sqrt{x + a} + \sqrt{x + b})}{{(\sqrt{x + a})}^{2} - {(\sqrt{x + b})}^{2}} d x = \int \frac{x (\sqrt{x + a} + \sqrt{x + b})}{x + a - x - b} d x = \frac{1}{a - b} \int x (\sqrt{x + a} + \sqrt{x + b}) d x = \frac{1}{a - b} [\int x (\sqrt{x + a}) d x + \int x (\sqrt{x + b}) d x] = \frac{1}{a - b} [\int (x + a - a) (\sqrt{x + a}) d x + \int (x + b - b) (\sqrt{x + b}) d x] = \frac{1}{a - b} [\int (x + a) (\sqrt{x + a}) d x - a \int (\sqrt{x + a}) d x + \int (x + b) (\sqrt{x + b}) d x - b \int (\sqrt{x + b}) d x] = \frac{1}{a - b} [\int {(x + a)}^{\frac{3}{2}} d x - a \int {(x + a)}^{\frac{1}{2}} d x + \int {(x + b)}^{\frac{3}{2}} d x - b \int {(x + b)}^{\frac{1}{2}} d x] = \frac{1}{a - b} [\frac{{(x + a)}^{\frac{5}{2}}}{\frac{5}{2}} - a \frac{{(x + a)}^{\frac{3}{2}}}{\frac{3}{2}} + \frac{{(x + b)}^{\frac{5}{2}}}{\frac{5}{2}} - b \frac{{(x + b)}^{\frac{3}{2}}}{\frac{3}{2}}] + c where, c is an arbitrary constant = \frac{1}{a - b} [\frac{2}{5} {(x + a)}^{\frac{5}{2}} - \frac{2 a}{3} {(x + a)}^{\frac{3}{2}} + \frac{2}{5} {(x + b)}^{\frac{5}{2}} - \frac{2 b}{3} {(x + b)}^{\frac{3}{2}}] + c where, c is an arbitrary constant Hence, \int \frac{x}{\sqrt{x + a} - \sqrt{x + b}} d x = \frac{1}{a - b} [\frac{2}{5} {(x + a)}^{\frac{5}{2}} - \frac{2 a}{3} {(x + a)}^{\frac{3}{2}} + \frac{2}{5} {(x + b)}^{\frac{5}{2}} - \frac{2 b}{3} {(x + b)}^{\frac{3}{2}}] + c where, c is an arbitrary constant$

Page No 18.36:

Question 1:

$\int \sin^{2} (2 x + 5) d x$

Answer:

$\int \sin^{2} (2 x + 5) d x = \int (\frac{1 - \cos (4 x + 10)}{2}) d x [∴ \sin^{2} A = \frac{1 - \cos 2 A}{2}] = \frac{1}{2} \int (1 - \cos (4 x + 10)) d x = \frac{1}{2} [x - \frac{\sin (4 x + 10)}{4}] + C = \frac{1}{2} x - \frac{\sin (4 x + 10)}{8} + C$

Page No 18.36:

Question 2:

$\int \sin^{3} (2 x + 1) d x$

Answer:

$\int \sin^{3} (2 x + 1) d x = \frac{1}{4} \int [3 \sin (2 x + 1) - \sin (3 (2 x + 1))] d x [∴ \sin (3 θ) = 3 \sin θ - 4 \sin^{3} θ \Rightarrow \sin^{3} θ = \frac{1}{4} (3 \sin θ - \sin (3 θ))] = \frac{3}{4} \int \sin (2 x + 1) d x - \frac{1}{4} \int \sin (6 x + 3) d x = \frac{3}{4} [- \frac{\cos (2 x + 1)}{2}] - \frac{1}{4} [- \frac{\cos (6 x + 3)}{6}] + C = \frac{- 3}{8} \cos (2 x + 1) + \frac{1}{24} \cos (6 x + 3) + C$

Page No 18.36:

Question 3:

$\int \cos^{4} 2 x d x$

Answer:

$\int \cos^{4} 2 x d x = \int {(\cos^{2} 2 x)}^{2} d x = \int {(\frac{1 + \cos 4 x}{2})}^{2} d x [∴ \cos^{2} x = \frac{1 + \cos 2 x}{2}] = \frac{1}{4} \int {(1 + \cos 4 x)}^{2} d x = \frac{1}{4} \int (1 + \cos^{2} 4 x + 2 \cos 4 x) d x = \frac{1}{4} \int [1 + (\frac{1 + \cos 8 x}{2}) + 2 \cos 4 x] d x = \frac{1}{4} \int (\frac{3}{2} + \frac{\cos 8 x}{2} + 2 \cos 4 x) d x = \frac{1}{4} [\frac{3 x}{2} + \frac{\sin 8 x}{16} + \frac{2 \sin 4 x}{4}] + C = \frac{3 x}{8} + \frac{\sin 8 x}{64} + \frac{\sin 4 x}{8} + C$

Page No 18.36:

Question 4:

$\int \sin^{2} b x d x$

Answer:

$\int \sin^{2} b x d x = \int [\frac{1 - \cos 2 b x}{2}] d x [∴ \sin^{2} x = \frac{1 - \cos 2 x}{2}] = \frac{1}{2} \int (1 - \cos 2 b x) d x = \frac{1}{2} [x - \frac{\sin 2 b x}{2 b}] + C = \frac{x}{2} - \frac{\sin 2 b x}{4 b} + C$

Page No 18.36:

Question 5:

$\int \sin^{2} \frac{x}{2} d x$

Answer:

$\int \sin^{2} \frac{x}{2} d x = \int (\frac{1 - \cos x}{2}) d x [∴ \sin^{2} \frac{x}{2} = \frac{1 - \cos x}{2}] = \frac{1}{2} \int (1 - \cos x) d x = \frac{1}{2} [x - \sin x] + C$

Page No 18.36:

Question 6:

$\int \cos^{2} \frac{x}{2} d x$

Answer:

$\int \cos^{2} \frac{x}{2} d x = \int (\frac{1 + \cos x}{2}) d x [∴ \cos^{2} \frac{x}{2} = \frac{1 + \cos x}{2}] = \frac{1}{2} \int (1 + \cos x) d x = \frac{1}{2} [x + \sin x] + C$

Page No 18.36:

Question 7:

$\int \cos^{2} n x d x$

Answer:

$\int \cos^{2} n x d x = \int [\frac{1 + \cos 2 n x}{2}] d x [∴ \cos^{2} x = \frac{1 + \cos 2 x}{2}] = \frac{1}{2} \int (1 + \cos 2 n x) d x = \frac{1}{2} [x + \frac{\sin 2 n x}{2 n}] + C = \frac{x}{2} + \frac{\sin 2 n x}{4 n} + C$

Page No 18.36:

Question 8:

$\int \sin x \sqrt{1 - \cos 2 x} d x$

Answer:

$\int \sin x \cdot \sqrt{1 - \cos 2 x} d x = \int \sin x . \sqrt{2 \sin^{2} x} d x [∴ 1 - \cos 2 A = 2 \sin^{2} A] = \sqrt{2} \int \sin^{2} x d x = \sqrt{2} \int (\frac{1 - \cos 2 x}{2}) d x = \frac{1}{\sqrt{2}} \int (1 - \cos 2 x) d x = \frac{1}{\sqrt{2}} [x - \frac{\sin 2 x}{2}] + C$

Page No 18.38:

Question 1:

$\int \sin 4 x \cos 7 x d x$

Answer:

$\int \sin 4 x \cos 7 x d x = \frac{1}{2} \int 2 \cos 7 x \sin 4 x d x = \frac{1}{2} \int [\sin (7 x + 4 x) - \sin (7 x - 4 x)] d x [∴ 2 \cos A \sin B = \sin (A + B) - \sin (A - B)] = \frac{1}{2} \int (\sin (11 x) - \sin (3 x)) d x = \frac{1}{2} [- \frac{\cos (11 x)}{11} + \frac{\cos (3 x)}{3}] + C = - \frac{\cos (11 x)}{22} + \frac{\cos (3 x)}{6}$

Page No 18.38:

Question 2:

$\int \cos 3 x \cos 4 x d x$

Answer:

$\int \cos 4 x \cos 3 x d x = \frac{1}{2} \int 2 \cos 4 x \cos 3 x d x = \frac{1}{2} \int [\cos (4 x + 3 x) + \cos (4 x - 3 x)] d x [∴ 2 \cos A \cos B = \cos (A + B) + \cos (A - B)] = \frac{1}{2} \int (\cos (7 x) + \cos x) d x = \frac{1}{2} [\frac{\sin 7 x}{7} + \sin x] + C = \frac{1}{14} \sin 7 x + \frac{1}{2} \sin x + C$

Page No 18.38:

Question 3:

$\int \cos m x \cos n x d x m \neq n$

Answer:

$\int \cos m x \cos n x d x = \frac{1}{2} \int 2 \cos (m x) \cos (n x) d x = \frac{1}{2} \int [\cos (m x + n x) + \cos (m x - n x)] d x [∴ 2 \cos A \cos B = \cos (A + B) + \cos (A - B)] = \frac{1}{2} [\frac{\sin (m + n) x}{m + n} + \frac{\sin (m - n) x}{m - n}] + C$

Page No 18.38:

Question 4:

$\int \sin m x \cos n x d x m \neq n$

Answer:

$\int \sin (m x) \cdot \cos (n x) d x = \frac{1}{2} \int 2 \sin (m x) \cdot \cos (n x) d x = \frac{1}{2} \int [\sin (m x + n x) + \sin (m x - n x)] d x [∴ 2 \sin A \cdot \cos B = \sin (A + B) + \sin (A - B)] = \frac{1}{2} [- \frac{\cos (m + n) x}{m + n} - \frac{\cos (m - n) x}{m - n}] + C$

Page No 18.38:

Question 5:

Integrate the following integrals:

$\int \sin 2 x \sin 4 x \sin 6 x d x$

Answer:

$\int \sin 2 x \sin 4 x \sin 6 x d x = \frac{1}{2} \int (2 \sin 2 x \sin 4 x) \sin 6 x d x = \frac{1}{2} \int [\cos (2 x - 4 x) - \cos (2 x + 4 x)] \sin 6 x d x = \frac{1}{2} \int [\cos (2 x) - \cos (6 x)] \sin 6 x d x = \frac{1}{2} [\int \cos (2 x) \sin (6 x) d x - \int \cos (6 x) \sin (6 x) d x] = \frac{1}{4} [\int 2 \cos (2 x) \sin (6 x) d x - \int 2 \cos (6 x) \sin (6 x) d x] = \frac{1}{4} \{\int [\sin (2 x + 6 x) - \sin (2 x - 6 x)] d x - \int \sin (12 x) d x\} = \frac{1}{4} [\int \sin (8 x) d x + \int \sin (4 x) d x - \int \sin (12 x) d x] = \frac{1}{4} [\frac{- \cos (8 x)}{8} + \frac{- \cos (4 x)}{4} + \frac{\cos (12 x)}{12}] + c = - \frac{\cos (8 x)}{32} - \frac{\cos (4 x)}{16} + \frac{\cos (12 x)}{48} + c$

Hence, $\int \sin 2 x \sin 4 x \sin 6 x d x = - \frac{\cos (8 x)}{32} - \frac{\cos (4 x)}{16} + \frac{\cos (12 x)}{48} + c$ .

Page No 18.38:

Question 6:

Integrate the following integrals:

$\int \sin x \cos 2 x \sin 3 x d x$

Answer:

$\int \sin x \cos 2 x \sin 3 x d x = \frac{1}{2} \int (2 \sin x \cos 2 x) \sin 3 x d x = \frac{1}{2} \int [\sin (x + 2 x) + \sin (x - 2 x)] \sin (3 x) d x = \frac{1}{2} \int [\sin (3 x) - \sin (x)] \sin (3 x) d x = \frac{1}{2} [\int \sin^{2} (3 x) d x - \int \sin (x) \sin (3 x) d x] = \frac{1}{4} [\int 2 \sin^{2} (3 x) d x - \int 2 \sin (x) \sin (3 x) d x] = \frac{1}{4} \{\int [1 - \cos (6 x)] d x - \int [\cos (x - 3 x) - \cos (x + 3 x)] d x\} = \frac{1}{4} [\int 1 d x - \int \cos (6 x) d x - \int \cos (2 x) d x + \int \cos (4 x) d x] = \frac{1}{4} [x - \frac{\sin (6 x)}{6} - \frac{\sin (2 x)}{2} + \frac{\sin (4 x)}{4}] + c = \frac{x}{4} - \frac{\sin (6 x)}{24} - \frac{\sin (2 x)}{8} + \frac{\sin (4 x)}{16} + c$

Hence, $\int \sin x \cos 2 x \sin 3 x d x = \frac{x}{4} - \frac{\sin (6 x)}{24} - \frac{\sin (2 x)}{8} + \frac{\sin (4 x)}{16} + c$ .

Page No 18.4:

Question 1:

Evaluate each of the following integrals:

(i) $\int x^{4} d x$

(ii) $\int x^{\frac{5}{4}} d x$

(iii) $\int \frac{1}{x^{5}} d x$

(iv) $\int \frac{1}{x^{3 / 2}} d x$

(v) $\int 3^{x} d x$

(vi) $\int \frac{1}{\sqrt[3]{x^{2}}} d x$

(vii) $\int 3^{2 \log_{3}}^{x} d x$

(viii) $\int \log_{x} x d x$

Answer:

(i)
$\int x^{4} d x = \frac{x^{4 + 1}}{4 + 1} + C = \frac{x^{5}}{5} + C$

(ii)
$\int x^{\frac{5}{4}} d x = \frac{x^{\frac{5}{4} + 1}}{\frac{5}{4} + 1} + C = \frac{4}{9} x^{\frac{9}{4}} + C$

(iii)
$\int x^{- 5} d x = \frac{x^{- 5 + 1}}{- 5 + 1} + C = - \frac{1}{4} x^{- 4} + C = - \frac{1}{4 x^{4}} + C$

(iv)
$\int \frac{d x}{x^{3 / 2}} = \int x^{- 3 / 2} d x = [\frac{x^{- \frac{3}{2} + 1}}{\frac{- 3}{2} + 1}] + C = [\frac{x^{- \frac{1}{2}}}{- \frac{1}{2}}] + C = - \frac{2}{\sqrt{x}} + C$

(v)
$\int 3^{x} d x = \frac{3^{x}}{\ln 3} + C$

(vi)
$\int \frac{d x}{\sqrt[3]{x^{2}}} = \int \frac{d x}{x^{2 / 3}} = \int x^{- 2 / 3} d x = \frac{x^{- \frac{2}{3} + 1}}{- \frac{2}{3} + 1} + C = 3 x^{\frac{1}{3}} + C$

(vii)
$\int 3^{2 \log_{3^{x}}} d x = \int 3^{\log_{3} x^{2}} d x = \int x^{2} d x = \frac{x^{3}}{3} + C$

(viii)
$\int \log_{x} x d x = \int 1 \cdot d x = x + C$

Page No 18.4:

Question 2:

Evaluate: (i) $\int \sqrt{\frac{1 + \cos 2 x}{2}} d x$
(ii) $\int \sqrt{\frac{1 - \cos 2 x}{2}} d x$

Answer:

(i)
$\int \sqrt{\frac{1 + \cos 2 x}{2}} d x \int \sqrt{\frac{2 \cos^{2} x}{2}} d x [∴ 1 + \cos 2 A = 2 \cos^{2} A] = \int \cos x d x = \sin x + C$

(ii)
$\int \sqrt{\frac{1 - \cos 2 x}{2}} d x = \int \sqrt{\frac{2 \sin^{2} x}{2}} d x [∴ 1 - \cos 2 x = 2 \sin^{2} x] = \int \sin x d x = - \cos x + C$

Page No 18.4:

Question 3:

Evaluate: $\int \frac{e^{6 \log_{e} x} - e^{5 \log_{e} x}}{e^{4 \log_{e} x} - e^{3 \log_{e} x}} d x$

Answer:

$\int (\frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}}) d x = \int (\frac{e^{\log x^{6}} - e^{\log x^{5}}}{e^{\log x^{4}} - e^{\log x^{3}}}) d x = \int (\frac{x^{6} - x^{5}}{x^{4} - x^{3}}) d x = \int \frac{x^{5}}{x^{3}} d x = \int x^{2} d x = \frac{x^{3}}{3} + C$

Page No 18.4:

Question 4:

Evaluate: $\int \frac{1}{a^{x} b^{x}} d x$

Answer:

$\int \frac{d x}{a^{x} b^{x}} = \int (a^{- x} b^{- x}) d x = \frac{a^{- x} b^{- x}}{- \log_{e} (a b)} + C$

Page No 18.4:

Question 5:

Evaluate: (i) $\int \frac{\cos 2 x + 2 \sin^{2} x}{\sin^{2} x} d x$
(ii) $\int \frac{2 \cos^{2} x - \cos 2 x}{\cos^{2} x} d x$

Answer:

(i)
$\int (\frac{\cos 2 x + 2 \sin^{2} x}{\sin^{2} x}) d x = \int (\frac{1 - 2 \sin^{2} x + 2 \sin^{2} x}{\sin^{2} x}) d x [∵ \cos 2 x = 1 - 2 \sin^{2} x] = \int {cosec}^{2} x d x = - \cot x + C$

(ii)
$\int (\frac{2 \cos^{2} x - \cos 2 x}{\cos^{2} x}) d x = \int (\frac{2 \cos^{2} x - (2 \cos^{2} x - 1)}{\cos^{2} x}) d x [∵ \cos 2 x = 2 \cos^{2} x - 1] = \int \sec^{2} x d x = \tan x + C$

Page No 18.4:

Question 6:

Evaluate: $\int \frac{e^{\log \sqrt{x}}}{x} d x$

Answer:

$\int \frac{e^{\log \sqrt{x}}}{x} d x = \int \frac{\sqrt{x}}{x} d x = \int \frac{1}{\sqrt{x}} d x = \int x^{- \frac{1}{2}} d x = [\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = 2 \sqrt{x} + C$

Page No 18.47:

Question 1:

$\int \frac{1}{\sqrt{1 - \cos 2 x}} d x$

Answer:

$\int \frac{1}{\sqrt{1 - \cos 2 x}} d x = \int \frac{1}{\sqrt{2 \sin^{2} x}} d x [∵ 1 - \cos 2 x = 2 \sin^{2} x] = \frac{1}{\sqrt{2}} \int cosec x d x = \frac{1}{\sqrt{2}} \ln |cosec x - \cot x| + C = \frac{1}{\sqrt{2}} \ln |\frac{1}{\sin x} - \frac{\cos x}{\sin x}| + C = \frac{1}{\sqrt{2}} \ln |\frac{2 \sin^{2} \frac{x}{2}}{\sin x}| + C [∵ 1 - \cos x = 2 \sin^{2} \frac{x}{2}] = \frac{1}{\sqrt{2}} \ln |\frac{2 \sin^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}| + C [∵ \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}] = \frac{1}{\sqrt{2}} \ln |\tan \frac{x}{2}| + C$

Page No 18.47:

Question 2:

$\int \frac{1}{\sqrt{1 + \cos x}} d x$

Answer:

$\int \frac{1}{\sqrt{1 + \cos x}} d x = \int \frac{1}{\sqrt{2 \cos^{2} \frac{x}{2}}} d x = \frac{1}{\sqrt{2}} \int \sec \frac{x}{2} d x = \frac{1}{\sqrt{2}} \times 2 \ln |\tan \frac{x}{2} + \sec \frac{x}{2}| + C = \sqrt{2} \ln |\frac{1 + \sin \frac{x}{2}}{\cos \frac{x}{2}}| + C = \sqrt{2} \ln |\frac{{(\sin \frac{x}{4} + \cos \frac{x}{4})}^{2}}{\cos^{2} \frac{x}{4} - \sin^{2} \frac{x}{4}}| + C [\begin{matrix} ∵ 1 + \sin θ = (\sin^{2} \frac{θ}{2} + \cos^{2} \frac{θ}{2} + 2 \sin \frac{θ}{2} \cos \frac{θ}{2}) = {(\sin \frac{θ}{2} + \cos \frac{θ}{2})}^{2} \\ & \cos θ = \cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2} \end{matrix}] = \sqrt{2} \ln |\frac{{(\sin \frac{x}{4} + \cos \frac{x}{4})}^{2}}{(\cos \frac{x}{4} - \sin \frac{x}{4}) (\cos \frac{x}{4} + \sin \frac{x}{4})}| + C = \sqrt{2} \ln |\frac{\sin \frac{x}{4} + \cos \frac{x}{4}}{\cos \frac{x}{4} - \sin \frac{x}{4}}| + C = \sqrt{2} \ln |\frac{1 + \tan \frac{x}{4}}{1 - \tan \frac{x}{4}}| + C = \sqrt{2} \ln |\tan (\frac{π}{4} + \frac{x}{4})| + C$

Page No 18.47:

Question 3:

$\int \sqrt{\frac{1 + \cos 2 x}{1 - \cos 2 x}} d x$

Answer:

$\int \sqrt{\frac{1 + \cos 2 x}{1 - \cos 2 x}} d x = \int \sqrt{\frac{2 \cos^{2} x}{2 \sin^{2} x}} d x = \int \cot x d x = \ln |\sin x| + C$

Page No 18.47:

Question 4:

$\int \sqrt{\frac{1 - \cos x}{1 + \cos x}} d x$

Answer:

$\int \sqrt{\frac{1 - \cos x}{1 + \cos x}} d x = \int \sqrt{\frac{2 \sin^{2} \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}} d x [∵ 1 - \cos x = 2 \sin^{2} \frac{x}{2} & 1 + \cos x = 2 \cos^{2} \frac{x}{2}] = \int \tan \frac{x}{2} d x = - 2 \ln |\cos \frac{x}{2}| + C$

Page No 18.47:

Question 5:

Evaluate the following integrals:
$\int \frac{\sec x}{\sec 2 x} d x$

Answer:

$I = \int \frac{\sec x}{\sec 2 x} d x$

$= \int \frac{\cos 2 x}{\cos x} d x = \int \frac{(1 - 2 \sin^{2} x)}{\cos x} d x$

$= \int \frac{1}{\cos x} d x - 2 \int \frac{\sin x \cdot \sin x}{\cos x} d x = \int \sec x \cdot d x - 2 \int \tan x \cdot \sec x \cdot d x = \ln |\sec x + \tan x| - 2 \sec x + c$

Page No 18.47:

Question 6:

$\int \frac{\cos 2 x}{{(\cos x + \sin x)}^{2}} d x$

Answer:

$Let I = \int \frac{\cos 2 x}{{(\cos x + \sin x)}^{2}} d x = \int \frac{\cos^{2} x - \sin^{2} x}{{(\cos x + \sin x)}^{2}} d x = \int \frac{\cos x - \sin x}{\cos x + \sin x} d x Putting \cos x + \sin x = t \Rightarrow - \sin x + \cos x = \frac{d t}{d x} \Rightarrow (\cos x - \sin x) d x = d t ∴ I = \int \frac{1}{t} d t = \ln |t| + C = \ln |\cos x + \sin x| + C [∵ t = \cos x + \sin x]$

Page No 18.47:

Question 7:

$\int \frac{\sin (x - a)}{\sin (x - b)} d x$

Answer:

$Let I = \int \frac{\sin (x - a)}{\sin (x - b)} d x Putting x - b = t \Rightarrow x = b + t & d x = d t ∴ I = \int \frac{\sin (b + t - a)}{\sin t} d t = \int \frac{\sin \{(b - a) + t\}}{\sin t} d t = \int \frac{\sin (b - a) \cos t}{\sin t} + \int \frac{\cos (b - a) \sin t}{\sin t} d t = \int \sin (b - a) \cot t d t + \int \cos (b - a) d t = \sin (b - a) \ln |\sin t| + t \cos (b - a) + C = \sin (b - a) \ln |\sin (x - b)| + (x - b) \cos (b - a) + C [∵ t = x - b]$

Page No 18.47:

Question 8:

$\int \frac{\sin (x - α)}{\sin (x + α)} d x$

Answer:

$Let I = \int \frac{\sin (x - α)}{\sin (x + α)} d x Putting x + α = t \Rightarrow x = t - α & d x = d t ∴ I = \int \frac{\sin (t - 2 α)}{\sin t} d t = \int (\frac{\sin t \cos 2 α}{\sin t} - \frac{\cos t \sin 2 α}{\sin t}) d t = \cos 2 α \int d t - \sin 2 α \int \cot t d t = t \cos 2 α - \sin 2 α \ln |\sin t| + C = (x + α) \cos 2 α - \sin 2 α \ln |\sin (x + α)| + C [∵ t = x + α] = x \cos 2 α - \sin 2 α \ln |\sin (x + α)| + C$

Page No 18.47:

Question 9:

$\int \frac{1 + \tan x}{1 - \tan x} d x$

Answer:

$Let I = \int (\frac{1 + \tan x}{1 - \tan x}) d x = \int (\frac{1 + \frac{\sin x}{\cos x}}{1 - \frac{\sin x}{\cos x}}) d x = \int (\frac{\cos x + \sin x}{\cos x - \sin x}) d x Putting \cos x - \sin x = t \Rightarrow (- \sin x - \cos x) d x = d t \Rightarrow (\sin x + \cos x) d x = - d t ∴ I = - \int \frac{1}{t} d t = - \ln |t| + C = - \ln |\cos x - \sin x| + C [∵ t = \cos x - \sin x]$

Page No 18.47:

Question 10:

$\int \frac{\cos x}{\cos (x - a)} d x$

Answer:

$Let I = \int \frac{\cos x}{\cos (x - a)} d x Putting x - a = t \Rightarrow x = a + t \Rightarrow d x = d t ∴ I = \int \frac{\cos (a + t) d t}{\cos t} = \int \frac{\cos a \cos t}{\cos t} - \frac{\sin a \sin t}{\cos t} d t = \int (\cos a - \sin a \tan t) d t = t \cos a - \sin a \ln |\sec t| + C = (x - a) \cos a - \sin a \ln |\sec (x - a)| + C [∵ t = x - a]$

Page No 18.47:

Question 11:

$\int \sqrt{\frac{1 - \sin 2 x}{1 + \sin 2 x}} d x$

Answer:

$\int \sqrt{\frac{1 - \sin 2 x}{1 + \sin 2 x}} d x = \int \sqrt{\frac{\cos^{2} x + \sin^{2} x - 2 \sin x \cos x}{\cos^{2} x + \sin^{2} x + 2 \sin x \cos x}} d x = \sqrt{\frac{{(\cos x - \sin x)}^{2}}{{(\cos x + \sin x)}^{2}}} d x = \int \frac{\cos x - \sin x}{\cos x + \sin x} d x = \int \frac{1 - \tan x}{1 + \tan x} d x = \int \tan (\frac{π}{4} - x) d x = \frac{1}{- 1} \ln |\sec (\frac{π}{4} - x)| [∵ \int \tan (a x + b) d x = \frac{1}{a} \ln |\sec (a x + b)| + C] = \frac{- \ln |\cos (\frac{π}{4} - x)|}{- 1} + C = \ln |\cos (\frac{π}{4} - x)| + C$

Page No 18.47:

Question 12:

$\int \frac{e^{3 x}}{e^{3 x} + 1} d x$

Answer:

$Let I = \int \frac{e^{3 x}}{e^{3 x} + 1} d x Putting e^{3 x} + 1 = t \Rightarrow 3 e^{3 x} = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{3 e^{3 x}} ∴ I = \int \frac{e^{3 x}}{3 t (e^{3 x})} d t = \frac{1}{3} \int \frac{1}{t} d t = \frac{\ln |t|}{3} + C = \frac{\ln |e^{3 x} + 1|}{3} + C$

Page No 18.47:

Question 13:

$\int \frac{\sec x \tan x}{3 \sec x + 5} d x$

Answer:

$Let I = \int \frac{\sec x \tan x}{3 \sec x + 5} d x Putting \sec x = t \Rightarrow \frac{d t}{d x} = \sec x \tan x \Rightarrow d t = \sec x \tan x d x ∴ I = \int \frac{d t}{3 t + 5} = \frac{1}{3} \ln |3 t + 5| + C = \frac{1}{3} \ln |3 \sec x + 5| + C [∵ t = \sec x]$

Page No 18.47:

Question 14:

$\int \frac{1 - \cot x}{1 + \cot x} d x$

Answer:

$Let I = \int (\frac{1 - \cot x}{1 + \cot x}) d x = \int (\frac{1 - \frac{\cos x}{\sin x}}{1 + \frac{\cos x}{\sin x}}) d x = \int (\frac{\sin x - \cos x}{\sin x + \cos x}) d x Putting \sin x + \cos x = t \Rightarrow (\cos x - \sin x) d x = d t \Rightarrow (\sin x - \cos x) d x = - d t ∴ I = \int \frac{- d t}{t} = - \ln |t| + C = - \ln |\sin x + \cos x| + C$

Page No 18.47:

Question 15:

$\int \frac{\sec x cosec x}{\log (\tan x)} d x$

Answer:

$Note : Here, we are considering \log x as \log_{e} x . Let I = \int \frac{\sec x cosec x}{\log (\tan x)} d x Putting \log \tan x = t \Rightarrow \frac{\sec^{2} x}{\tan x} = \frac{d t}{d x} \Rightarrow \sec x cosec x d x = d t ∴ I = \int \frac{1}{t} d t = \log |t| + C = \log |\log (\tan x)| + C$

Page No 18.47:

Question 16:

$\int \frac{1}{x (3 + \log x)} d x$

Answer:

$Here, we are considering \log x as \log_{e} x . Let I = \int \frac{1}{x (3 + \log x)} d x Putting \log x = t \Rightarrow \frac{1}{x} = \frac{d t}{d x} \Rightarrow \frac{d x}{x} = d t ∴ I = \int \frac{d t}{3 + t} = \log |3 + t| + C = \log |3 + \log x| + C [∵ t = \log x]$

Page No 18.47:

Question 17:

$\int \frac{e^{x} + 1}{e^{x} + x} d x$

Answer:

$Let I = \int \frac{e^{x} + 1}{e^{x} + x} d x Putting e^{x} + x = t \Rightarrow e^{x} + 1 = \frac{d t}{d x} \Rightarrow (e^{x} + 1) d x = d t ∴ I = \int \frac{1}{t} d t = \ln |t| + C = \ln |e^{x} + x| + C [∵ t = e^{x} + x]$

Page No 18.47:

Question 18:

$\int \frac{1}{x \log x} d x$

Answer:

$Here, we are considering \log x as \log_{e} x . Let I = \int \frac{1}{x \log x} d x Putting \log x = t \Rightarrow \frac{1}{x} = \frac{d t}{d x} \Rightarrow \frac{1}{x} d x = d t ∴ I = \int \frac{1}{t} d t = \log |\log x| + C$

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Question 19:

$\int \frac{\sin 2 x}{a \cos^{2} x + b \sin^{2} x} d x$

Answer:

$Let I = \int \frac{\sin 2 x}{a \cos^{2} x + b \sin^{2} x} d x = \int \frac{\sin 2 x}{a (1 - \sin^{2} x) + b \sin^{2} x} d x = \int \frac{\sin 2 x}{(b - a) \sin^{2} x + a} d x Putting s {in}^{2} x = t \Rightarrow 2 \sin x . \cos x = \frac{d t}{d x} \Rightarrow \sin 2 x = \frac{d t}{d x} \Rightarrow \sin 2 x d x = d t ∴ I = \int \frac{1}{(b - a) t + a} d t = \frac{1}{(b - a)} \ln |(b - a) t + a| + C [∵ \int \frac{1}{ax + b} d x = \frac{1}{a} \ln |a x + b| + C] = \frac{1}{(b - a)} \ln |(b - a) \sin^{2} x + a| + C [∵ t = \sin^{2} x] = \frac{1}{(b - a)} \ln |b \sin^{2} x + a (1 - \sin^{2} x)| + C = \frac{1}{(b - a)} \ln |b \sin^{2} x + a \cos^{2} x| + C$

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Question 20:

$\int \frac{\cos x}{2 + 3 \sin x} d x$

Answer:

$Let I = \int \frac{\cos x}{2 + 3 \sin x} d x Putting \sin x = t \Rightarrow \cos x = \frac{d t}{d x} \Rightarrow \cos x d x = d t ∴ I = \int \frac{d t}{2 + 3 t} = \frac{1}{3} \ln |2 + 3 t| + C [∵ \int \frac{1}{a x + b} d x = \frac{1}{a} \ln |a x + b| + C] = \frac{1}{3} \ln |2 + 3 \sin x| + C [∵ t = \sin x]$

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Question 21:

$\int \frac{1 - \sin x}{x + \cos x} d x$

Answer:

$Let I = \int \frac{1 - \sin x}{x + \cos x} d x Putting x + \cos x = t \Rightarrow 1 - \sin x = \frac{d t}{d x} \Rightarrow (1 - \sin x) d x = d t ∴ I = \int \frac{1}{t} d t = \ln t + C = \ln |x + \cos x| + C [∵ t = x + \cos x]$

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Question 22:

$\int \frac{a}{b + c e^{x}} d x$

Answer:

$Let I = \int \frac{a}{b + c e^{x}} d x Dividing numerator and denominator by e^{x} \Rightarrow I = \int \frac{a e^{- x}}{b e^{- x} + c} d x Putting e^{- x} = t \Rightarrow - e^{- x} = \frac{d t}{d x} \Rightarrow e^{- x} d x = - d t ∴ I = \int \frac{- a}{b t + c} d t = \frac{- a}{b} \ln |b t + c| + C [∵ \int \frac{1}{a x + b} d x = \frac{1}{a} \ln |a x + b| + C] = \frac{- a}{b} \ln |b e^{- x} + c| + C [∵ t = e^{- x} + C]$

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Question 23:

$\int \frac{1}{e^{x} + 1} d x$

Answer:

$Let I = \int \frac{1}{e^{x} + 1} d x = \int \frac{e^{- x}}{1 + e^{- x}} d x Putting e^{- x} = t \Rightarrow - e^{- x} = \frac{d t}{d x} \Rightarrow e^{- x} d x = - d t ∴ I = \int \frac{- 1}{1 + t} d t = - \ln |1 + t| + C = - \ln |1 + e^{- x}| + C$

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Question 24:

$\int \frac{\cot x}{\log \sin x} d x$

Answer:

$Note : Here we are considering \log x as \log_{e} x Let I = \int \frac{\cot x}{\log \sin x} d x Putting \log \sin x = t \Rightarrow \cot x = \frac{d t}{d x} \Rightarrow \cot x d x = d t ∴ I = \int \frac{1}{t} d t = \log |t| + C = \log |\log \sin x| + C [∵ t = \log \sin x]$

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Question 25:

$\int \frac{e^{2 x}}{e^{2 x} - 2} d x$

Answer:

$Let I = \int \frac{e^{2 x}}{e^{2 x} - 2} d x Putting e^{2 x} = t \Rightarrow 2 e^{2 x} = \frac{d t}{d x} \Rightarrow e^{2 x} d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{1}{t - 2} d t = \frac{1}{2} \ln |t - 2| + C = \frac{1}{2} \ln |e^{2 x} - 2| + C [∵ t = e^{2 x}]$

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Question 26:

$\int \frac{2 \cos x - 3 \sin x}{6 \cos x + 4 \sin x} d x$

Answer:

$\int \frac{2 \cos x - 3 \sin x}{6 \cos x + 4 \sin x} d x \Rightarrow \int \frac{2 \cos x - 3 \sin x}{2 (3 \cos x + 2 \sin x)} d t Let, 3 \cos x + 2 \sin x = t \Rightarrow 2 \cos x - 3 \sin x = \frac{d t}{d x} \Rightarrow (2 \cos x - 3 \sin x) d x = d t Now, \int \frac{2 \cos x - 3 \sin x}{2 (3 \cos x + 2 \sin x)} d t = \int \frac{d t}{2 t} = \frac{1}{2} \log |t| + C = \frac{1}{2} \log |3 \cos x + 2 \sin x| + C$

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Question 27:

$\int \frac{\cos 2 x + x + 1}{x^{2} + \sin 2 x + 2 x} d x$

Answer:

$Let I = \int \frac{\cos 2 x + x + 1}{x^{2} + \sin 2 x + 2 x} d x Putting x^{2} + \sin 2 x + 2 x = t \Rightarrow 2 x + 2 \cos 2 x + 2 = \frac{d t}{d x} \Rightarrow (x + \cos 2 x + 1) d x = \frac{d t}{2} ∴ I = \frac{1}{2} \int \frac{1}{t} d t = \frac{1}{2} \ln |t| + C = \frac{1}{2} \ln |x^{2} + \sin 2 x + 2 x| + C [∵ t = x^{2} + \sin 2 x + 2 x]$

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Question 28:

$\int \frac{1}{\cos (x + a) \cos (x + b)} d x$

Answer:

$\int \frac{1}{\cos (x + a) \cos (x + b)} d x Multiplying and Dividing by \sin [(x + b) - (x + a)], we get = \int \frac{1}{\sin [(x + b) - (x + a)]} \times \frac{\sin [(x + b) - (x + a)]}{\cos (x + a) \cos (x + b)} d x = \int \frac{1}{\sin (b - a)} \times \frac{\sin [(x + b) - (x + a)]}{\cos (x + a) \cos (x + b)} d x = \frac{1}{\sin (b - a)} \int \frac{\sin (x + b) \cos (x + a) - \sin (x + a) \cos (x + b)}{\cos (x + a) \cos (x + b)} d x = \frac{1}{\sin (b - a)} [\int \frac{\sin (x + b)}{\cos (x + b)} d x - \int \frac{\sin (x + a)}{\cos (x + a)} d x] = \frac{1}{\sin (b - a)} [\int \tan (x + b) d x - \int \tan (x + a) d x] = \frac{1}{\sin (b - a)} [\log (\sec (x + b)) - \log (\sec (x + a))] + c = \frac{1}{\sin (b - a)} [\log (\frac{\sec (x + b)}{\sec (x + a)})] + c$

Hence, $\int \frac{1}{\cos (x + a) \cos (x + b)} d x = \frac{1}{\sin (b - a)} [\log (\frac{\sec (x + b)}{\sec (x + a)})] + c$ .

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Question 29:

$\int \frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} d x$

Answer:

$Let I = \int \frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} d x Putting 2 \sin x + \cos x = t \Rightarrow 2 \cos x - \sin x = \frac{d t}{d x} \Rightarrow (- \sin x + 2 \cos x) d x = d t ∴ I = \int \frac{1}{t} d t = \ln |t| + C = \ln |2 \sin x + \cos x| + C [∵ t = 2 \sin x + \cos x]$

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Question 30:

$\int \frac{\cos 4 x - \cos 2 x}{\sin 4 x - \sin 2 x} d x$

Answer:

$\int (\frac{\cos 4 x - \cos 2 x}{\sin 4 x - \sin 2 x}) d x = \int \frac{- 2 \sin (\frac{4 x + 2 x}{2}) \sin (\frac{4 x - 2 x}{2})}{2 \cos (\frac{4 x + 2 x}{2}) \sin (\frac{4 x - 2 x}{2})} d x [∵ \cos A - \cos B = - 2 \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2}) & \sin A - \sin B = 2 \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2})] = - \int \frac{\sin 3 x}{\cos 3 x} d x = - \int \tan 3 x d x = \frac{- \ln |\sec 3 x|}{3} + C = \frac{1}{3} \ln {(|\sec 3 x|)}^{- 1} + C = \frac{1}{3} \ln |\cos 3 x| + C$

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Question 31:

$\int \frac{se c x}{\log (\sec x + \tan x)} d x$

Answer:

$Note : Here, we are considering \log x as \log_{e} x Let I = \int \frac{\sec x}{\log (\sec x + \tan x)} d x Putting \log (\sec x + \tan x) = t \Rightarrow \frac{\sec x \tan x + \sec^{2} x}{\sec x + \tan x} = \frac{d t}{d x} \Rightarrow \sec x \frac{(\sec x + \tan x)}{\sec x + \tan x} = \frac{d t}{d x} \Rightarrow \sec x d x = d t ∴ I = \int \frac{d t}{t} = \log |t| + C = \log |\log (\sec x + \tan x)| + C$

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Question 32:

$\int \frac{cosec x}{\log \tan \frac{x}{2}} d x$

Answer:

$Note : Here, we are considering \log x as \log_{e} x Let I = \int \frac{cosec x}{\log \tan \frac{x}{2}} d x Putting \log \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \frac{\sec^{2} \frac{x}{2}}{\tan \frac{x}{2}} = \frac{d t}{d x} \Rightarrow \frac{1}{2 \sin \frac{x}{2} . \cos \frac{x}{2}} = \frac{d t}{d x} \Rightarrow \frac{1}{\sin x} = \frac{d t}{d x} \Rightarrow cosec x d x = d t ∴ I = \int \frac{d t}{t} = \log |t| + C = \log |\log \tan \frac{x}{2}| + C$

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Question 33:

$\int \frac{1}{x \log x \log (\log x)} d x$

Answer:

$Note : Here, we are considering \log x as \log_{e} x . Let I = \int \frac{1}{x \log x \log (\log x)} d x Putting \log (\log x) = t \Rightarrow \frac{1}{x \log x} = \frac{d t}{d x} \Rightarrow \frac{1}{x \log x} d x = d t ∴ I = \int \frac{d t}{t} = \log |t| + C = \log |\log (\log x)| + C [∵ t = \log (\log x)]$

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Question 34:

$\int \frac{{cosec}^{2} x}{1 + \cot x} d x$

Answer:

$Let I = \int \frac{{cosec}^{2} x}{1 + \cot x} d x Putting \cot x = t \Rightarrow - {cosec}^{2} x = \frac{d t}{d x} \Rightarrow {cosec}^{2} x d x = - d t ∴ I = \int \frac{- d t}{1 + t} = - \ln |1 + t| + C = - \ln |1 + \cot x| + C [∵ t = \cot x]$

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Question 35:

$\int \frac{10 x^{9} + 10^{x} \log_{e} 10}{10^{x} + x^{10}} d x$

Answer:

$Let I = \int \frac{10 x^{9} + 10^{x} \log_{e} 10}{10^{x} + x^{10}} d x Putting 10^{x} + x^{10} = t \Rightarrow 10^{x} \log_{e} 10 + 10 x^{9} = \frac{d t}{d x} \Rightarrow (10^{x} \log_{e} 10 + 10 x^{9}) d x = d t ∴ I = \int \frac{1}{t} d t = \ln |t| + C = \ln |10^{x} + x^{10}| + C [∵ t = 10^{x} + x^{10}]$

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Question 36:

$\int \frac{1 - \sin 2 x}{x + \cos^{2} x} d x$

Answer:

$Let I = \int \frac{1 - \sin 2 x}{x + \cos^{2} x} d x Putting x + \cos^{2} x = t \Rightarrow 1 - 2 \cos x . \sin x = \frac{d t}{d x} \Rightarrow (1 - \sin 2 x) d x = d t ∴ I = \int \frac{1}{t} d t = \ln |t| + C = \ln |x + \cos^{2} x| + C [∵ t = x + \cos^{2} x]$

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Question 37:

$\int \frac{1 + \tan x}{x + \log \sec x} d x$

Answer:

$Note : Here, we are considering \log x as \log_{e} x Let I = \int \frac{1 + \tan x}{x + \log \sec x} d x Putting x + \log \sec x = t \Rightarrow 1 + \frac{\sec x \tan x}{\sec x} = \frac{d t}{d x} \Rightarrow (1 + \tan x) d x = d t ∴ I = \int \frac{1}{t} d t = \log |t| + C = \log |x + \log \sec x| + C [∵ t = x + \log \sec x]$

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Question 38:

$\int \frac{\sin 2 x}{a^{2} + b^{2} \sin^{2} x} d x$

Answer:

$Let I = \int \frac{\sin 2 x}{a^{2} + b^{2} \sin^{2} x} d x Putting \sin^{2} x = t \Rightarrow 2 \sin x . \cos x = \frac{d t}{d x} \Rightarrow \sin 2 x = \frac{d t}{d x} \Rightarrow \sin 2 x d x = d t ∴ I = \int \frac{1}{a^{2} + b^{2} t} d t = \frac{1}{b^{2}} \ln |a^{2} + b^{2} t| + C = \frac{1}{b^{2}} \ln |a^{2} + b^{2} \sin^{2} x| + C [∵ t = \sin^{2} x]$

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Question 39:

$\int \frac{x + 1}{x (x + \log x)} d x$

Answer:

$Note : Here, we are considering \log x as \log_{e} x Let I = \int \frac{x + 1}{x (x + \log x)} d x Putting x + \log x = t \Rightarrow 1 + \frac{1}{x} = \frac{d t}{d x} \Rightarrow \frac{x + 1}{x} d x = d t ∴ I = \int \frac{1}{t} d t = \log |t| + C = \log |x + \log x| + C$

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Question 40:

$\int \frac{1}{\sqrt{1 - x^{2}} (2 + 3 \sin^{- 1} x)} d x$

Answer:

$Let I = \int \frac{1}{\sqrt{1 - x^{2}} (2 + 3 \sin^{- 1} x)} d x Putting \sin^{- 1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} = \frac{d t}{d x} \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} d x = d t ∴ I = \int \frac{1}{2 + 3 t} d t = \frac{1}{3} \ln |2 + 3 t| + C = \frac{1}{3} \ln |2 + 3 \sin^{- 1} x| + C [∵ t = \sin^{- 1} x]$

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Question 41:

$\int \frac{\sec^{2} x}{\tan x + 2} d x$

Answer:

$Let I = \int \frac{\sec^{2} x}{\tan x + 2} d x Putting \tan x = t \Rightarrow \sec^{2} x = \frac{d t}{d x} \Rightarrow \sec^{2} x d x = d t ∴ I = \int \frac{1}{t + 2} d t = \ln |t + 2| + C = \ln |\tan x + 2| + C [∵ t = \tan x]$

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Question 42:

$\int \frac{2 \cos 2 x + \sec^{2} x}{\sin 2 x + \tan x - 5} d x$

Answer:

$Let I = \int \frac{2 \cos 2 x + \sec^{2} x}{\sin 2 x + \tan x - 5} d x Putting \sin 2 x + \tan x - 5 = t \Rightarrow 2 \cos 2 x + \sec^{2} x = \frac{d t}{d x} \Rightarrow (2 \cos 2 x + \sec^{2} x) d x = d t ∴ I = \int \frac{1}{t} d t = \ln |t| + C = \ln |\sin 2 x + \tan x - 5| + C [∵ t = \sin 2 x + \tan x - 5]$

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Question 43:

$\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x$

Answer:

$\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x = \int \frac{\sin (5 x - 3 x)}{\sin 5 x \sin 3 x} d x = \int \frac{\sin 5 x \cos 3 x - \cos 5 x \sin 3 x}{\sin 5 x \sin 3 x} d x = \int \frac{\sin 5 x \cos 3 x}{\sin 5 x \sin 3 x} - \frac{\cos 5 x \sin 3 x}{\sin 5 x \sin 3 x} d x = \int [\cot 3 x - \cot 5 x] d x = \int co t 3 x d x - \int \cot 5 x d x = \frac{1}{3} \ln |\sin 3 x| - \frac{1}{5} \ln |\sin 5 x| + C$

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Question 44:

$\int \frac{1 + \cot x}{x + \log \sin x} d x$

Answer:

$Note : Here, we are considering \log x as \log_{e} x Let I = \int \frac{1 + \cot x}{x + \log \sin x} d x Putting x + \log \sin x = t \Rightarrow 1 + \cot x = \frac{d t}{d x} \Rightarrow (1 + \cot x) d x = d t ∴ I = \int \frac{1}{t} d t = \log |t| + C = \log |x + \log \sin x| + C [∵ t = x + \log \sin x]$

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Question 45:

$\int \frac{\cos (x + a)}{\sin (x + b)} d x$

Answer:

$\int \frac{\cos (x + a)}{\sin (x + b)} d x$

$= \int \frac{\cos [(x + b) + (a - b)]}{\sin (x + b)} d x$

$= \int \frac{\cos (x + b) \cos (a - b) - \sin (x + b) \sin (a - b)}{\sin (x + b)} d x$

$= \cos (a - b) \int \cot (x + b) d x - \sin (a - b) \int d x$

$= \cos (a - b) \log |\sin (x + b)| - x \sin (a - b) + C$

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Question 46:

$\int \frac{1}{\sqrt{x} (\sqrt{x} + 1)} d x$

Answer:

$Let I = \int \frac{1}{\sqrt{x} (\sqrt{x} + 1)} d x Putting \sqrt{x} + 1 = t \Rightarrow \frac{1}{2 \sqrt{x}} = \frac{d t}{d x} \Rightarrow \frac{1}{\sqrt{x}} d x = 2 d t ∴ I = 2 \int \frac{1}{t} d t = 2 \ln |t| + C = 2 \ln |\sqrt{x} + 1| + C [∵ t = \sqrt{x} + 1]$

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Question 47:

$\int \tan 2 x \tan 3 x \tan 5 x d x$

Answer:

$We know that, \tan 5 x = \tan (2 x + 3 x) \Rightarrow \tan 5 x = \frac{\tan 2 x + \tan 3 x}{1 - \tan 2 x \tan 3 x} \Rightarrow \tan 5 x - \tan 2 x \tan 3 x \tan 5 x = \tan 2 x + \tan 3 x \Rightarrow \tan 2 x \tan 3 x \tan 5 x = \tan 5 x - \tan 2 x - \tan 3 x ∴ \int \tan 2 x \tan 3 x \tan 5 x = \int (\tan 5 x - \tan 2 x - \tan 3 x) d x = \frac{1}{5} \ln |\sec 5 x| - \frac{1}{2} \ln |\sec 2 x| - \frac{1}{3} \ln |\sec 3 x| + C$

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Question 48:

$\int \{1 + \tan x \tan (x + θ)\} d x$

Answer:

$Let I = \int 1 + \tan x \tan (x + θ) d x = \int 1 + \tan x (\frac{\tan x + \tan θ}{1 - \tan x \tan θ}) d x = \int \frac{1 + \tan^{2} x}{1 - \tan x \tan θ} d x = \int \frac{\sec^{2} x d x}{1 - \tan x \tan θ} Putting \tan x = t \Rightarrow \sec^{2} x = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{\sec^{2} x} ∴ I = \int \frac{1}{1 - t \tan θ} d t = \frac{- 1}{\tan θ} \ln |1 - t \tan θ| + C [∵ \int \frac{1}{a x + b} d x = \frac{1}{a} \ln |a x + b| + C] = - \cot θ \ln |1 - \tan x \tan θ| + C = \cot θ \ln |\frac{1}{1 - \tan x \tan θ}| + C = \cot θ \ln |\frac{\cos x \cos θ}{\cos x \cos θ - \sin x \sin θ}| + C = \cot θ \ln |\frac{\cos x}{\cos (x + θ)}| + C' [Let C' = C + \cot θ \ln \cos θ]$

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Question 49:

$\int \frac{\sin 2 x}{\sin (x - \frac{π}{6}) \sin (x + \frac{π}{6})} d x$

Answer:

$Let I = \int \frac{\sin 2 x}{\sin (x - \frac{π}{6}) \sin (x + \frac{π}{6})} d x = \int \frac{\sin 2 x}{\sin^{2} x - \sin^{2} \frac{π}{6}} d x [∵ \sin (A + B) \sin (A - B) = \sin^{2} A - \sin^{2} B] = \int \frac{\sin 2 x}{\sin^{2} x - \frac{1}{4}} d x Putting \sin^{2} x - \frac{1}{4} = t \Rightarrow 2 \sin x \cos x d x = d t \Rightarrow \sin 2 x d x = d t ∴ I = \int \frac{1}{t} d t = \ln |t| + C = \ln |\sin^{2} x - \frac{1}{4}| + C [∵ t = \sin^{2} x - \frac{1}{4}]$

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Question 50:

$\int \frac{e^{x - 1} + x^{e - 1}}{e^{x} + x^{e}} d x$

Answer:

$Let I = \int \frac{e^{x - 1} + x^{e - 1}}{e^{x} + x^{e}} d x Putting e^{x} + x^{e} = t \Rightarrow e^{x} + e x^{e - 1} = \frac{d t}{d x} \Rightarrow e (e^{x - 1} + x^{e - 1}) = \frac{d t}{d x} \Rightarrow (e^{x - 1} + x^{e - 1}) d x = \frac{d t}{e} ∴ I = \frac{1}{e} \int \frac{1}{t} d t = \frac{1}{e} \ln |t| + C = \frac{1}{e} \ln |e^{x} + x^{e}| + C [∵ t = e^{x} + x^{e}]$

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Question 51:

$\int \frac{1}{\sin x \cos^{2} x} d x$

Answer:

$\int \frac{1}{\sin x \cos^{2} x} d x = \int \frac{\sin^{2} x + \cos^{2} x}{\sin x \cos^{2} x} d x = \int \tan x \sec x + cosec x d x = \sec x + \ln |cosec x - \cot x| + C = \sec x + \ln |\tan \frac{x}{2}| + C [∵ cosec x - \cot x = \frac{1 - \cos x}{\sin x} = \tan \frac{x}{2}]$

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Question 52:

$\int \frac{1}{\cos 3 x - \cos x} d x$

Answer:

$\int \frac{1}{\cos 3 x - \cos x} d x = \int \frac{1}{4 \cos^{3} x - 4 \cos x} d x [∵ \cos 3 x = 4 \cos^{3} x - 3 \cos x] = \int \frac{1}{4 \cos x (\cos^{2} x - 1)} d x = \frac{- 1}{4} \int \frac{1}{\cos x \sin^{2} x} d x = \frac{- 1}{4} \int (\frac{\sin^{2} x + \cos^{2} x}{\cos x \sin^{2} x} d x) = \frac{- 1}{4} [\int \sec x d x + \int \cot x cosec x d x] = \frac{- 1}{4} (\ln |\sec x + \tan x| - cosec x) + C = \frac{1}{4} (cosec x - \ln |\sec x + \tan x|) + C$

Page No 18.57:

Question 1:

$\int \frac{\log x}{x} d x$

Answer:

$\int \frac{\log x}{x} d x Let, \log x = t \Rightarrow \frac{1}{x} = \frac{d t}{d x} Now, \int \frac{\log x}{x} d x = \int t \cdot d t = \frac{t^{2}}{2} + C = \frac{{(\log x)}^{2}}{2} + C$

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Question 2:

$\int \frac{\log (1 + \frac{1}{x})}{x (1 + x)} d x$

Answer:

$\int \frac{\log (1 + \frac{1}{x})}{x (1 + x)} d x Let, \log (1 + \frac{1}{x}) = t \Rightarrow \frac{1}{1 + \frac{1}{x}} \times \frac{- 1}{x^{2}} = \frac{d t}{d x} \Rightarrow (\frac{x}{x + 1}) \times \frac{- 1}{x^{2}} = \frac{d t}{d x} \Rightarrow \frac{- d x}{x (x + 1)} = d t \Rightarrow \frac{d x}{x (x + 1)} = - d t Now, \int \frac{\log (1 + \frac{1}{x})}{x (1 + x)} d x = \int t \cdot (- d t) = \frac{- t^{2}}{2} + C = - \frac{1}{2} {\{\log (1 + \frac{1}{x})\}}^{2} + C$

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Question 3:

$\int \frac{{(1 + \sqrt{x})}^{2}}{\sqrt{x}} d x$

Answer:

$\int \frac{{(1 + \sqrt{x})}^{2}}{\sqrt{x}} d x Let, 1 + \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} = \frac{d t}{d x} \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t Now, \int \frac{{(1 + \sqrt{x})}^{2}}{\sqrt{x}} d x = 2 \int t^{2} d t = \frac{2}{3} t^{3} + C = \frac{2}{3} {(1 + \sqrt{x})}^{3} + C$

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Question 4:

$\int \sqrt{1 + e^{x}} e^{x} d x$

Answer:

$\int \sqrt{1 + e^{x}} \cdot e^{x} d x Let 1 + e^{x} = t \Rightarrow e^{x} = \frac{d t}{d x} \Rightarrow e^{x} d x = d t Now, \int \sqrt{1 + e^{x}} \cdot e^{x} d x = \int \sqrt{t} \cdot d t = \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C = \frac{2}{3} t^{\frac{3}{2}} + C = \frac{2}{3} {(1 + e^{x})}^{\frac{3}{2}} + C$

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Question 5:

$\int \sqrt[3]{\cos^{2} x} \sin x d x$

Answer:

$\int {(\cos^{2} x)}^{\frac{1}{3}} \sin x d x Let, \cos x = t \Rightarrow - \sin x = \frac{d t}{d x} \Rightarrow \sin x d x = - d t Now, \int {(\cos^{2} x)}^{\frac{1}{3}} \sin x d x = - \int t^{\frac{2}{3}} d t = - [\frac{t^{\frac{2}{3} + 1}}{\frac{2}{3} + 1}] + C = - \frac{3}{5} t^{\frac{5}{3}} + C = - \frac{3}{5} \cos^{\frac{5}{3}} x + C$

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Question 6:

$\int \frac{e^{x}}{{(1 + e^{x})}^{2}} d x$

Answer:

$\int \frac{e^{x} d x}{{(1 + e^{x})}^{2}} Let 1 + e^{x} = t \Rightarrow e^{x} = \frac{d t}{d x} \Rightarrow e^{x} d x = d t Now, \int \frac{e^{x} d x}{{(1 + e^{x})}^{2}} = \int \frac{d t}{t^{2}} = \int t^{- 2} d t = \frac{t^{- 2} + 1}{- 2 + 1} + C = \frac{- 1}{t} + C = - \frac{1}{1 + e^{x}} + C$

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Question 7:

$\int \cot^{3} x {cosec}^{2} x d x$

Answer:

$\int \cot^{3} x {cosec}^{2} x d x Let, \cot x = t \Rightarrow - {cosec}^{2} x = \frac{d t}{d x} \Rightarrow {cosec}^{2} x d x = - d t Now, \int \cot^{3} x {cosec}^{2} x d x = \int t^{3} (- d t) = \frac{- t^{4}}{4} + C = \frac{- \cot^{4} x}{4} + C$

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Question 8:

$\int \frac{{\{e^{\sin^{- 1} x}\}}^{2}}{\sqrt{1 - x^{2}}} d x$

Answer:

$\int \frac{{(e^{\sin^{- 1} x})}^{2}}{\sqrt{1 - x^{2}}} d x Let e^{\sin^{- 1} x} = t Differentiating both sides w . r . t . x, e^{\sin^{- 1} x} \times \frac{1}{\sqrt{1 - x^{2}}} d x = d t Now, \int \frac{{(e^{\sin^{- 1} x})}^{2}}{\sqrt{1 - x^{2}}} d x = \int e^{\sin^{- 1} x} \cdot \frac{e^{\sin^{- 1} x}}{\sqrt{1 - x^{2}}} d x = \int t \cdot d t = \frac{t^{2}}{2} + C = \frac{{(e^{\sin^{- 1} x})}^{2}}{2} + C$

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Question 9:

$\int \frac{1 + \sin x}{\sqrt{x - \cos x}} d x$

Answer:

$\int \frac{1 + \sin x}{\sqrt{x - \cos x}} d x Let, x - \cos x = t \Rightarrow (1 + \sin x) = \frac{d t}{d x} \Rightarrow (1 + \sin x) d x = d t Now, \int \frac{1 + \sin x}{\sqrt{x - \cos x}} d x = \int \frac{d t}{\sqrt{t}} = \int t^{- \frac{1}{2}} d t = \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C = 2 \sqrt{t} + C = 2 \sqrt{x - \cos x} + C$

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Question 10:

$\int \frac{1}{\sqrt{1 - x^{2}} {(\sin^{- 1} x)}^{2}} d x$

Answer:

$\int \frac{d x}{\sqrt{1 - x^{2}} {(\sin^{- 1} x)}^{2}} Let, \sin^{- 1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} = \frac{d t}{d x} \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} d x = d t Now, \int \frac{d x}{\sqrt{1 - x^{2}} {(\sin^{- 1} x)}^{2}} = \int \frac{d t}{t^{2}} = \int t^{- 2} d t = \frac{t^{- 2 + 1}}{- 2 + 1} + C = \frac{- 1}{t} + C = - \frac{1}{\sin^{- 1} x} + C$

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Question 11:

$\int \frac{\cot x}{\sqrt{\sin x}} d x$

Answer:

$\int \frac{\cot x}{\sqrt{\sin x}} d x = \int \frac{\cos x}{\sin x \sqrt{\sin x}} d x = \int \frac{\cos x}{{(\sin x)}^{\frac{3}{2}}} d x Let \sin x = t \Rightarrow \cos x = \frac{d t}{d x} \Rightarrow \cos x d x = d t Now, \int \frac{\cos x}{{(\sin x)}^{\frac{3}{2}}} d x = \int \frac{d t}{t^{\frac{3}{2}}} = \int t^{- \frac{3}{2}} d t = [\frac{t^{- \frac{3}{2} + 1}}{\frac{- 3}{2} + 1}] + C = \frac{- 2}{\sqrt{t}} + C = - \frac{2}{\sqrt{\sin x}} + C$

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Question 12:

$\int \frac{\tan x}{\sqrt{\cos x}} d x$

Answer:

$\int \frac{\tan x}{\sqrt{\cos x}} d x \Rightarrow \int \frac{\sin x}{\cos x \sqrt{\cos x}} d x \Rightarrow \int \frac{\sin x}{\cos^{\frac{3}{2}} x} d x Let \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x = - \frac{d t}{d x} Now, \int \frac{\sin x}{\cos^{\frac{3}{2}} x} d x = \int - \frac{1}{t^{\frac{3}{2}}} d t = - \int t^{- \frac{3}{2}} d t = - [\frac{t^{- \frac{3}{2} + 1}}{\frac{- 3}{2} + 1}] + C = \frac{2}{\sqrt{t}} + C = \frac{2}{\sqrt{\cos x}} + C$

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Question 13:

$\int \frac{\cos^{3} x}{\sqrt{\sin x}} d x$

Answer:

$\int \frac{\cos^{3} x}{\sqrt{\sin x}} d x = \int \frac{\cos^{2} x \cdot \cos x}{\sqrt{\sin x}} d x = \int \frac{(1 - \sin^{2} x) \cos x}{\sqrt{\sin x}} d x Let \sin x = t \Rightarrow \cos x = \frac{d t}{d x} \Rightarrow \cos x d x = d t Now, \int \frac{(1 - \sin^{2} x) \cos x}{\sqrt{\sin x}} d x = \int \frac{(1 - t^{2})}{\sqrt{t}} \cdot d t = \int (\frac{1}{\sqrt{t}} - t^{\frac{3}{2}}) d t = \int (t^{- \frac{1}{2}} - t^{\frac{3}{2}}) d t = [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} - \frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + C = 2 \sqrt{t} - \frac{2}{5} t^{\frac{5}{2}} + C = 2 \sqrt{\sin x} - \frac{2}{5} \sin^{\frac{5}{2}} x + C$

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Question 14:

$\int \frac{\sin^{3} x}{\sqrt{\cos x}} d x$

Answer:

$\int \frac{\sin^{3} x}{\sqrt{\cos x}} d x = \int (\frac{\sin^{2} x \cdot \sin x}{\sqrt{\cos x}}) d x = \int \frac{(1 - \cos^{2} x) \sin x}{\sqrt{\cos x}} d x Let \cos x = t \Rightarrow - \sin x = \frac{d t}{d x} \Rightarrow \sin x d x = - d t Now, \int \frac{(1 - \cos^{2} x) \sin x}{\sqrt{\cos x}} d x = - \int \frac{(1 - t^{2})}{\sqrt{t}} d t = \int (\frac{t^{2} - 1}{\sqrt{t}}) d t = \int (t^{\frac{3}{2}} - t^{- \frac{1}{2}}) d t = [\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} - \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = \frac{2}{5} t^{\frac{5}{2}} - 2 \sqrt{t} + C = \frac{2}{5} \cos^{\frac{5}{2}} x - 2 \sqrt{\cos x} + C$

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Question 15:

$\int \frac{1}{\sqrt{\tan^{- 1} x} . (1 + x^{2})} d x$

Answer:

$\int \frac{d x}{\sqrt{\tan^{- 1} x} (1 + x^{2})} Let \tan^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} = \frac{d t}{d x} \Rightarrow \frac{1}{1 + x^{2}} d x = d t$

$Now, \int \frac{d x}{\sqrt{\tan^{- 1} x} (1 + x^{2})} = \int \frac{d t}{\sqrt{t}} = \int t^{- \frac{1}{2}} d t = \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C = 2 \sqrt{t} + C = 2 \sqrt{\tan^{- 1} x} + C$

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Question 16:

$\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x$

Answer:

$\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x = \int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x} \times \cos^{2} x} d x = \int \frac{\sqrt{\tan x}}{\tan x} \times \sec^{2} x d x = \int \frac{1}{\sqrt{\tan x}} \times \sec^{2} x d x = \int {(\tan x)}^{- \frac{1}{2}} \sec^{2} x dx Let \tan x = t \Rightarrow \sec^{2} x = \frac{d t}{d x} \Rightarrow \sec^{2} x d x = d t Now, \int {(\tan x)}^{- \frac{1}{2}} \sec^{2} x dx = \int t^{- \frac{1}{2}} d t = [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = 2 \sqrt{t} + C = 2 \sqrt{\tan x} + C$

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Question 17:

$\int \frac{1}{x} {(\log x)}^{2} d x$

Answer:

$\int \frac{1}{x} {(\log x)}^{2} d x Let \log x = t \Rightarrow \frac{1}{x} d x = d t Now, \int \frac{1}{x} {(\log x)}^{2} d x = \int t^{2} d t = \frac{t^{3}}{3} + C = \frac{{(\log x)}^{3}}{3} + C$

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Question 18:

$\int \sin^{5} x \cos x d x$

Answer:

$\int \sin^{5} x \cos x d x Let \sin x = t \Rightarrow \cos x = \frac{d t}{d x} \Rightarrow \cos x d x = d t Now, \int \sin^{5} x \cos x d x = \int t^{5} d t = \frac{t^{6}}{6} + C = \frac{1}{6} \sin^{6} x + C$

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Question 19:

$\int \tan^{3 / 2} x \sec^{2} x d x$

Answer:

$\int \tan^{\frac{3}{2}} x \cdot \sec^{2} x d x Let \tan x = t \Rightarrow \sec^{2} x = \frac{d t}{d x} \Rightarrow \sec^{2} x d x = d t Now, \int \tan^{\frac{3}{2}} x \cdot \sec^{2} x d x = \int t^{\frac{3}{2}} d t = [\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + C = \frac{2}{5} t^{\frac{5}{2}} + C = \frac{2}{5} \tan^{\frac{5}{2}} x + C$

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Question 20:

$\int \frac{x^{3}}{{(x^{2} + 1)}^{3}} d x$

Answer:

$\int \frac{x^{3}}{{(x^{2} + 1)}^{3}} d x = \int \frac{x^{2} . x}{{(x^{2} + 1)}^{3}} d x Let x^{2} + 1 = t \Rightarrow x^{2} = t - 1 \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Now, \int \frac{x^{2} . x}{{(x^{2} + 1)}^{3}} d x = \frac{1}{2} \int \frac{(t - 1)}{t^{3}} d t = \frac{1}{2} \int (\frac{1}{t^{2}} - \frac{1}{t^{3}}) d t = \frac{1}{2} \int (t^{- 2} - t^{- 3}) d t = \frac{1}{2} [\frac{t^{- 2 + 1}}{- 2 + 1} - \frac{t^{- 3 + 1}}{- 3 + 1}] + C = \frac{1}{2} [- \frac{1}{t} + \frac{1}{2 t^{2}}] + C = \frac{1}{2} [\frac{- 1}{(x^{2} + 1)} + \frac{1}{2 {(x^{2} + 1)}^{2}}] + C = \frac{1}{2} [\frac{- 2 (x^{2} + 1) + 1}{2 {(x^{2} + 1)}^{2}}] = \frac{1}{4} [\frac{- 2 x^{2} - 2 + 1}{{(x^{2} + 1)}^{2}}] = - \frac{1}{4} \frac{(1 + 2 x^{2})}{{(x^{2} + 1)}^{2}} + C$

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Question 21:

$\int (4 x + 2) \sqrt{x^{2} + x + 1} d x$

Answer:

$\int (4 x + 2) \sqrt{x^{2} + x + 1} d x = 2 \int (2 x + 1) \sqrt{x^{2} + x + 1} d x Let x^{2} + x + 1 = t \Rightarrow (2 x + 1) = \frac{d t}{d x} \Rightarrow (2 x + 1) d x = d t Now, 2 \int (2 x + 1) \sqrt{x^{2} + x + 1} d x = 2 \int \sqrt{t} d t = 2 \int t^{\frac{1}{2}} d t = 2 [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = 2 \times \frac{2}{3} t^{\frac{3}{2}} + C = \frac{4}{3} t^{\frac{3}{2}} + C = \frac{4}{3} {(x^{2} + x + 1)}^{\frac{3}{2}} + C$

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Question 22:

$\int \frac{4 x + 3}{\sqrt{2 x^{2} + 3 x + 1}} d x$

Answer:

$\int (\frac{4 x + 3}{\sqrt{2 x^{2} + 3 x + 1}}) d x Let 2 x^{2} + 3 x + 1 = t \Rightarrow (4 x + 3) = \frac{d t}{d x} \Rightarrow (4 x + 3) d x = d t Now, \int (\frac{4 x + 3}{\sqrt{2 x^{2} + 3 x + 1}}) d x = \int \frac{d t}{\sqrt{t}} = \int t^{- \frac{1}{2}} d t = [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = 2 \sqrt{t} + C = 2 \sqrt{2 x^{2} + 3 x + 1} + C$

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Question 23:

$\int \frac{1}{1 + \sqrt{x}} d x$

Answer:

$\int \frac{d x}{1 + \sqrt{x}} = \int \frac{\sqrt{x} d x}{\sqrt{x} (1 + \sqrt{x})} Let 1 + \sqrt{x} = t \Rightarrow \sqrt{x} = t - 1 \Rightarrow \frac{1}{2 \sqrt{x}} = \frac{d t}{d x} \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t$

$Now, \int \frac{\sqrt{x}}{\sqrt{x} (1 + \sqrt{x})} d x = 2 \int (\frac{t - 1}{t}) d t = 2 \int (1 - \frac{1}{t}) d t = 2 (t - \log |t|) + C = 2 (1 + \sqrt{x}) - 2 \log |1 + \sqrt{x}| + C Let C + 2 = C' = 2 \sqrt{x} - 2 \log (1 + \sqrt{x}) + C'$

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Question 24:

$\int e^{\cos^{2} x} \sin 2 x d x$

Answer:

$Let I = \int e^{\cos^{2} x} \sin 2 x d x Let \cos^{2} x = t On differentiating both sides, we get - 2 \cos x \sin x d x = d t ∴ I = \int e^{t} 2 \sin x \cos x \frac{d t}{- 2 \sin x \cos x} = - \int e^{t} d t = - e^{t} + c = - e^{\cos^{2} x} + c$

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Question 25:

$\int \frac{1 + \cos x}{{(x + \sin x)}^{3}} d x$

Answer:

$\int \frac{(1 + \cos x)}{{(x + \sin x)}^{3}} d x Let x + \sin x = t \Rightarrow (1 + \cos x) = \frac{d t}{d x} \Rightarrow (1 + \cos x) d x = d t Now, \int \frac{(1 + \cos x)}{{(x + \sin x)}^{3}} d x = \int \frac{d t}{t^{3}} = \int t^{- 3} d t = \frac{t^{- 3 + 1}}{- 3 + 1} + C = \frac{- 1}{2 t^{2}} + C = \frac{- 1}{2 {(x + \sin x)}^{2}} + C$

Page No 18.58:

Question 26:

$\int \frac{\cos x - \sin x}{1 + \sin 2 x} d x$

Answer:

$\int (\frac{\cos x - \sin x}{1 + \sin (2 x)}) d x \Rightarrow \int (\frac{\cos x - \sin x}{\cos^{2} x + \sin^{2} x + 2 \sin x . \cos x}) d x \Rightarrow \int \frac{(\cos x - \sin x)}{{(\cos x + \sin x)}^{2}} d x Let \cos x + \sin x = t \Rightarrow (- \sin x + \cos x) = \frac{d t}{d x} \Rightarrow (- \sin x + \cos x) d x = d t Now, \int \frac{(\cos x - \sin x)}{{(\cos x + \sin x)}^{2}} d x = \int \frac{d t}{t^{2}} = \int t^{- 2} d t = \frac{t^{- 2 + 1}}{- 2 + 1} + C = \frac{- 1}{t} + C = - \frac{1}{\sin x + \cos x} + C$

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Question 27:

$\int \frac{\sin 2 x}{{(a + b \cos 2 x)}^{2}} d x$

Answer:

$\int \frac{\sin (2 x)}{{(a + b \cos 2 x)}^{2}} d x Let a + b \cos 2 x = t \Rightarrow - b \sin (2 x) d x \times 2 = d t \Rightarrow \sin (2 x) d x = \frac{- d t}{2 b} Now, \int \frac{\sin (2 x)}{{(a + b \cos 2 x)}^{2}} d x = - \frac{1}{2 b} \int \frac{d t}{t^{2}} = \frac{- 1}{2 b} \int t^{- 2} d t = \frac{- 1}{2 b} [\frac{t^{- 2 + 1}}{- 2 + 1}] + C = \frac{1}{2 b} \times \frac{1}{t} + C = \frac{1}{2 b (a + b \cos 2 x)} + C$

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Question 28:

$\int \frac{\log x^{2}}{x} d x$

Answer:

$\int \frac{\log x^{2} d x}{x} = \int \frac{2 \log x}{x} d x = 2 \int \frac{\log x}{x} d x Let \log x = t \Rightarrow \frac{1}{x} = \frac{d t}{d x} \Rightarrow \frac{1}{x} d x = d t Now, 2 \int \frac{\log x}{x} d x = 2 \int t d t = 2 [\frac{t^{2}}{2}] + C = t^{2} + C = {(\log x)}^{2} + C$

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Question 29:

$\int \frac{\sin x}{{(1 + \cos x)}^{2}} d x$

Answer:

$\int \frac{\sin x}{{(1 + \cos x)}^{2}} d x Let 1 + \cos x = t \Rightarrow - \sin x = \frac{d t}{d x} \Rightarrow \sin x d x = - d t Now, \int \frac{\sin x}{{(1 + \cos x)}^{2}} d x = \int - \frac{d t}{t^{2}} = - \int t^{- 2} d t = - [\frac{t^{- 2 + 1}}{- 2 + 1}] + C = \frac{1}{t} + C = \frac{1}{1 + \cos x} + C$

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Question 30:

$\int \cot x \cdot \log \sin x d x$

Answer:

$\int \cot x \cdot \log \sin x d x Let \log \sin x = t \Rightarrow \frac{1}{\sin x} \times \cos x = \frac{d t}{d x} \Rightarrow \cot x d x = d t Now, \int \cot x \cdot \log \sin x d x = \int t \cdot d t = \frac{t^{2}}{2} + C = \frac{{(\log |\sin x|)}^{2}}{2} + C$

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Question 31:

$\int \sec x \cdot \log (\sec x + \tan x) d x$

Answer:

$\int \sec x \cdot \log (\sec x + \tan x) d x Let \log (\sec x + \tan x) = t \Rightarrow \frac{(\sec x \tan x + \sec^{2} x)}{(\sec x + \tan x)} = \frac{d t}{d x} \Rightarrow \frac{\sec x (\sec x + \tan x)}{(\sec x + \tan x)} d x = d t Now, \int \sec x \cdot \log (\sec x + \tan x) d x = \int t \cdot d t = \frac{t^{2}}{2} + C = \frac{{[\log (\sec x + \tan x)]}^{2}}{2} + C$

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Question 32:

$\int cosec x \log (cosec x - \cot x) d x$

Answer:

$\int cosec x \cdot \log (cosec x - \cot x) d x Let \log (cosec x - \cot x) = t \Rightarrow \frac{(- cosec x \cot x + {cosec}^{2} x)}{(cosec x - \cot x)} = \frac{d t}{d x} \Rightarrow \frac{(cosec x - \cot x)}{(cosec x - \cot x)} \times cosec x d x = d t \Rightarrow cosec x d x = d t Now, \int cosec x \cdot \log (cosec x - \cot x) d x = \int t \cdot d t = \frac{t^{2}}{2} + C = \frac{{\{\log |cosec x - \cot x|\}}^{2}}{2} + C$

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Question 33:

$\int x^{3} \cos x^{4} d x$

Answer:

$\int x^{3} \cdot \cos (x^{4}) d x Let x^{4} = t \Rightarrow 4 x^{3} d x = d t \Rightarrow x^{3} d x = \frac{d t}{4} Now, \int x^{3} \cdot \cos (x^{4}) d x = \frac{1}{4} \int \cos (t) d t = \frac{1}{4} [\sin (t)] + C = \frac{1}{4} [\sin x^{4}] + C$

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Question 34:

$\int x^{3} \sin x^{4} d x$

Answer:

$\int x^{3} \cdot \sin x^{4} d x Let x^{4} = t \Rightarrow 4 x^{3} = \frac{d t}{d x} \Rightarrow x^{3} d x = \frac{d t}{4} Now, \int x^{3} \cdot \sin x^{4} d x = \frac{1}{4} \int \sin t d t = \frac{1}{4} [- \cos t] + C = \frac{1}{4} [- \cos x^{4}] + C$

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Question 35:

$\int \frac{x \sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} d x$

Answer:

$\int \frac{x \sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} d x Let \sin^{- 1} x^{2} = t \Rightarrow \frac{1 \times 2 x}{\sqrt{1 - x^{4}}} = \frac{d t}{d x} \Rightarrow \frac{x d x}{\sqrt{1 - x^{4}}} = \frac{d t}{2} Now, \int \frac{x \sin^{- 1} x^{2}}{\sqrt{1 - x^{4}}} d x = \frac{1}{2} \int t d t = \frac{t^{2}}{4} + C = \frac{{(\sin^{- 1} x^{2})}^{2}}{4} + C$

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Question 36:

$\int x^{3} \sin (x^{4} + 1) d x$

Answer:

$\int x^{3} \cdot \sin (x^{4} + 1) d x Let x^{4} + 1 = t \Rightarrow 4 x^{3} = \frac{d t}{d x} \Rightarrow x^{3} d x = \frac{d t}{4} Now, \int x^{3} \cdot \sin (x^{4} + 1) d x = \frac{1}{4} \int \sin (t) d t = \frac{1}{4} [- \cos t] + C = \frac{1}{4} [- \cos (x^{4} + 1)] + C$

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Question 37:

$\int \frac{(x + 1) e^{x}}{\cos^{2} (x e^{x})} d x$

Answer:

$\int \frac{(x + 1) e^{x}}{\cos^{2} (x \cdot e^{x})} d x Let x e^{x} = t \Rightarrow (1 \cdot e^{x} + x e^{x}) = \frac{d t}{d x} \Rightarrow (x + 1) e^{x} d x = d t Now, \int \frac{(x + 1) e^{x}}{\cos^{2} (x \cdot e^{x})} d x = \int \frac{d t}{\cos^{2} t} = \int \sec^{2} t dt = \tan (t) + C = \tan (x e^{x}) + C$

Page No 18.58:

Question 38:

$\int x^{2} e^{x^{3}} \cos (e^{x^{3}}) d x$

Answer:

$\int x^{2} \cdot e^{x^{3}} \cdot \cos (e^{x^{3}}) d x Let e^{x^{3}} = t \Rightarrow e^{x^{3}} \cdot 3 x^{2} d x = d t \Rightarrow e^{x^{3}} \cdot x^{2} d x = \frac{d t}{3} Now, \int x^{2} \cdot e^{x^{3}} \cdot \cos (e^{x^{3}}) d x = \frac{1}{3} \int \cos (t) d t = \frac{1}{3} [\sin t] + C = \frac{1}{3} [\sin (e^{x^{3}})] + C$

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Question 39:

$\int 2 x \sec^{3} (x^{2} + 3) \tan (x^{2} + 3) d x$

Answer:

$\int 2 x \sec^{3} (x^{2} + 3) \cdot \tan (x^{2} + 3) d x = \int \sec^{2} (x^{2} + 3) \cdot \sec (x^{2} + 3) \cdot \tan (x^{2} + 3) \cdot 2 x d x Let \sec (x^{2} + 3) = t \Rightarrow \sec (x^{2} + 3) \cdot \tan (x^{2} + 3) \cdot 2 x = \frac{d t}{d x} \Rightarrow \sec (x^{2} + 3) \cdot \tan (x^{2} + 3) \cdot 2 x d x = d t Now, \int \sec^{2} (x^{2} + 3) \cdot \sec (x^{2} + 3) \cdot \tan (x^{2} + 3) \cdot 2 x d x = \int t^{2} d t = \frac{t^{3}}{3} + C = \frac{\sec^{3} (x^{2} + 3)}{3} + C$

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Question 40:

$\int (\frac{x + 1}{x}) {(x + \log x)}^{2} d x$

Answer:

$\int (\frac{x + 1}{x}) \cdot {(x + \log x)}^{2} d x Let x + \log x = t \Rightarrow (1 + \frac{1}{x}) = \frac{d t}{d x} \Rightarrow (\frac{x + 1}{x}) d x = d t Now, \int (\frac{x + 1}{x}) \cdot {(x + \log x)}^{2} d x = \int t^{2} d t = \frac{t^{3}}{3} + C = \frac{{(x + \log x)}^{3}}{3} + C$

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Question 41:

$\int \tan x \sec^{2} x \sqrt{1 - \tan^{2} x} d x$

Answer:

$\int \tan x \cdot \sec^{2} x \sqrt{1 - \tan^{2} x} d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t Now, \int \tan x \cdot \sec^{2} x \sqrt{1 - \tan^{2} x} d x = \int t \cdot \sqrt{1 - t^{2}} d t Again let t^{2} = p \Rightarrow 2 t d t = d p \Rightarrow t d t = \frac{d p}{2} Again, \int t \cdot \sqrt{1 - t^{2}} d t = \frac{1}{2} \int \sqrt{1 - p} d p = \frac{1}{2} \int {(1 - p)}^{\frac{1}{2}} d p = \frac{1}{2} [\frac{{(1 - p)}^{\frac{1}{2} + 1}}{(\frac{1}{2} + 1) (- 1)}] + C = \frac{1}{2} \times \frac{- 2}{3} {(1 - p)}^{\frac{3}{2}} + C = - \frac{1}{3} {(1 - p)}^{\frac{3}{2}} + C = - \frac{1}{3} {(1 - \tan^{2} x)}^{\frac{3}{2}} + C$

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Question 42:

$\int \log x \frac{\sin \{1 + {(\log x)}^{2}\}}{x} d x$

Answer:

$\int \frac{\log x \sin \{1 + {(\log x)}^{2}\}}{x} d x Let 1 + {(\log x)}^{2} = t \Rightarrow 2 \log x \times \frac{1}{x} d x = d t \Rightarrow \frac{\log x}{x} d x = \frac{d t}{2} Now, \int \frac{\log x \sin \{1 + {(\log x)}^{2}\}}{x} d x = \frac{1}{2} \int \sin (t) d t = \frac{1}{2} [- \cos t] + C = - \frac{1}{2} \cos \{1 + {(\log x)}^{2}\} + C$

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Question 43:

$\int \frac{1}{x^{2}} \cos^{2} (\frac{1}{x}) d x$

Answer:

$\int \frac{1}{x^{2}} \cdot \cos^{2} (\frac{1}{x}) d x Let \frac{1}{x} = t \Rightarrow - \frac{1}{x^{2}} = \frac{d t}{d x} \Rightarrow \frac{1}{x^{2}} d x = - d t Now, \int \frac{1}{x^{2}} \cdot \cos^{2} (\frac{1}{x}) d x = - \int \cos^{2} t d t = - \int (\frac{1 + \cos 2 t}{2}) d t = - \frac{1}{2} \int (1 + \cos 2 t) d t = - \frac{1}{2} [t + \frac{\sin 2 t}{2}] + C = - \frac{1}{2} [\frac{1}{x} + \frac{\sin (\frac{2}{x})}{2}] + C = - \frac{1}{2} (\frac{1}{x}) - \frac{1}{4} \sin (\frac{2}{x}) + C$

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Question 44:

$\int \sec^{4} x \tan x d x$

Answer:

$\int \sec^{4} x . \tan x d x = \int \sec^{2} x . \sec^{2} x \tan x d x = \int (1 + \tan^{2} x) \sec^{2} x . \tan x d x = \int (\tan x + \tan^{3} x) \sec^{2} x d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t Now, \int (\tan x + \tan^{3} x) \sec^{2} x d x = \int (t + t^{3}) d t = \frac{t^{2}}{2} + \frac{t^{4}}{4} + C = \frac{\tan^{2} x}{2} + \frac{\tan^{4} x}{4} + C$

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Question 45:

$\int \frac{e^{\sqrt{x}} \cos (e^{\sqrt{x}})}{\sqrt{x}} d x$

Answer:

$\int \frac{e^{\sqrt{x}} \cdot \cos (e^{\sqrt{x}})}{\sqrt{x}} d x Let e^{\sqrt{x}} = t \Rightarrow e^{\sqrt{x}} \times \frac{1}{2 \sqrt{x}} = \frac{d t}{d x} \Rightarrow \frac{e^{\sqrt{x}}}{\sqrt{x}} d x = 2 d t Now, \int \frac{e^{\sqrt{x}} \cdot \cos (e^{\sqrt{x}})}{\sqrt{x}} d x = 2 \int \cos t d t = 2 \sin t + C = 2 \sin (e^{\sqrt{x}}) + C$

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Question 46:

$\int \frac{\cos^{5} x}{\sin x} d x$

Answer:

$\int \frac{\cos^{5} x}{\sin x} d x = \int \frac{\cos^{4} x . \cos x}{\sin x} d x = \int \frac{{(\cos^{2} x)}^{2} . \cos x}{\sin x} d x = \int \frac{{(1 - \sin^{2} x)}^{2} \times \cos x}{\sin x} d x = \int \frac{(1 - \sin^{4} x - 2 \sin^{2} x)}{\sin x} \cos x d x Let \sin x = t \Rightarrow \cos x d x = d t Now, \int \frac{(1 - \sin^{4} x - 2 \sin^{2} x)}{\sin x} \cos x d x = \int \frac{(1 + t^{4} - 2 t^{2})}{t} d t = \int (\frac{1}{t} + t^{3} - 2 t) d t = \log |t| + \frac{t^{4}}{4} - \frac{2 t^{2}}{2} + C = \log |t| + \frac{t^{4}}{4} - t^{2} + C = \log |\sin x| + \frac{\sin^{4} x}{4} - \sin^{2} x + C$

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Question 47:

$\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x$

Answer:

$\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} = \frac{d t}{d x} \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t Now, \int \frac{\sin \sqrt{x}}{\sqrt{x}} d x = 2 \int \sin t d t = 2 [- \cos t] + C = - 2 \cos \sqrt{x} + C$

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Question 48:

$\int \frac{(x + 1) e^{x}}{\sin^{2} (x e^{x})} d x$

Answer:

$\int \frac{(x + 1) e^{x}}{\sin^{2} (x e^{x})} d x Let x e^{x} = t \Rightarrow (1 \cdot e^{x} + x e^{x}) = \frac{d t}{d x} \Rightarrow (x + 1) e^{x} d x = d t Now, \int \frac{(x + 1) e^{x}}{\sin^{2} (x e^{x})} d x = \int \frac{d t}{\sin^{2} t} = \int {cosec}^{2} t d t = - \cot (t) + C = - \cot (x e^{x}) + C$

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Question 49:

$\int 5^{x + \tan^{- 1} x} . (\frac{x^{2} + 2}{x^{2} + 1}) d x$

Answer:

$\int 5^{x + \tan^{- 1} x} \cdot (\frac{x^{2} + 2}{x^{2} + 1}) d x Let x + \tan^{- 1} x = t (1 + \frac{1}{1 + x^{2}}) = \frac{d t}{d x} \Rightarrow (\frac{x^{2} + 1 + 1}{x^{2} + 1}) d x = d t \Rightarrow (\frac{x^{2} + 2}{x^{2} + 1}) d x = d t Now, \int 5^{x + \tan^{- 1} x} \cdot (\frac{x^{2} + 2}{x^{2} + 1}) d x = \int 5^{t} d t = \frac{5^{t}}{\log 5} + C = \frac{5^{x + \tan^{1} x}}{\log 5} + C$

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Question 50:

$\int \frac{e^{m \sin^{- 1} x}}{\sqrt{1 - x^{2}}} d x$

Answer:

$\int \frac{e^{m \sin^{- 1} x}}{\sqrt{1 - x^{2}}} d x Let \sin^{- 1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} d x = d t Now, \int \frac{e^{m \sin^{- 1} x}}{\sqrt{1 - x^{2}}} d x = \int e^{m t} \cdot d t = \frac{e^{m t}}{m} + C = \frac{e^{m \sin^{- 1} x}}{m} + C$

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Question 51:

$\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$

Answer:

$\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} = \frac{d t}{d x} \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t Now, \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x = 2 \int \cos t d t = 2 \sin t + C = 2 \sin \sqrt{x} + C$

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Question 52:

$\int \frac{\sin (\tan^{- 1} x)}{1 + x^{2}} d x$

Answer:

$\int \frac{\sin (\tan^{- 1} x)}{1 + x^{2}} d x Let \tan^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} d x = d t Now, \int \frac{\sin (\tan^{- 1} x)}{1 + x^{2}} d x = \int \sin t d t = - \cos (t) + C = - \cos (\tan^{- 1} x) + C$

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Question 53:

$\int \frac{\sin (\log x)}{x} d x$

Answer:

$\int \frac{\sin (\log x)}{x} d x Let \log x = t \Rightarrow \frac{1}{x} d x = d t Now, \int \frac{\sin (\log x)}{x} d x = \int \sin (t) d t = - \cos (t) + C = - \cos (\log x) + C$

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Question 54:

$\int \frac{e^{m \tan^{- 1} x}}{1 + x^{2}} d x$

Answer:

$\int (\frac{e^{m \tan^{- 1} x}}{1 + x^{2}}) d x Let \tan^{- 1} x = t \Rightarrow (\frac{1}{1 + x^{2}}) d x = d t Now, \int (\frac{e^{m \tan^{- 1} x}}{1 + x^{2}}) d x = \int e^{m t} d t = \frac{e^{m t}}{m} + C = \frac{e^{m \tan^{- 1} x}}{m} + C$

Page No 18.59:

Question 55:

$\int \frac{x}{\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} - a^{2}}} d x$

Answer:

$\int \frac{x d x}{\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} - a^{2}}} Let x^{2} = t \Rightarrow 2 x = \frac{d t}{d x} \Rightarrow x d x = \frac{d t}{2} Now, \int \frac{x d x}{\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} - a^{2}}} = \frac{1}{2} \int \frac{d t}{\sqrt{t + a^{2}} + \sqrt{t - a^{2}}} = \frac{1}{2} \int \frac{d t}{(\sqrt{t + a^{2}} + \sqrt{t - a^{2}})} \times \frac{(\sqrt{t + a^{2}} - \sqrt{t - a^{2}})}{(\sqrt{t + a^{2}} - \sqrt{t - a^{2}})} = \frac{1}{2} \int \frac{(\sqrt{t + a^{2}} - \sqrt{t - a^{2}})}{(t + a^{2}) - (t - a^{2})} d t = \frac{1}{4 a^{2}} \int {(t + a^{2})}^{\frac{1}{2}} d t - \frac{1}{4 a^{2}} \int {(t - a^{2})}^{\frac{1}{2}} d t = \frac{1}{4 a^{2}} [\frac{{(t + a^{2})}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] - \frac{1}{4 a^{2}} [\frac{{(t - a^{2})}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{1}{6 a^{2}} [{(t + a^{2})}^{\frac{3}{2}} - {(t - a^{2})}^{\frac{3}{2}}] + C = \frac{1}{6 a^{2}} [{(x^{2} + a^{2})}^{\frac{3}{2}} - {(x^{2} - a^{2})}^{\frac{3}{2}}] + C$

Page No 18.59:

Question 56:

$\int x \frac{\tan^{- 1} x^{2}}{1 + x^{4}} d x$

Answer:

$\int \frac{x \tan^{- 1} x^{2}}{1 + x^{4}} d x Let \tan^{- 1} x^{2} = t \Rightarrow \frac{1}{1 + {(x^{2})}^{2}} \times 2 x = \frac{d t}{d x} \Rightarrow \frac{x d x}{1 + x^{4}} = \frac{d t}{2} Now, \int \frac{x \tan^{- 1} x^{2}}{1 + x^{4}} d x = \frac{1}{2} \int t . d t = \frac{1}{2} \times \frac{t^{2}}{2} + C = \frac{{(\tan^{- 1} x^{2})}^{2}}{4} + C$

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Question 57:

$\int \frac{{(\sin^{- 1} x)}^{3}}{\sqrt{1 - x^{2}}} d x$

Answer:

$\int \frac{{(\sin^{- 1} x)}^{3}}{\sqrt{1 - x^{2}}} d x Let \sin^{- 1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} d x = d t Now, \int \frac{{(\sin^{- 1} x)}^{3}}{\sqrt{1 - x^{2}}} d x = \int t^{3} d t = \frac{t^{4}}{4} + C = \frac{{(\sin^{- 1} x)}^{4}}{4} + C$

Page No 18.59:

Question 58:

$\int \frac{\sin (2 + 3 \log x)}{x} d x$

Answer:

$\int \frac{\sin (2 + 3 \log x)}{x} d x Let 2 + 3 \log x = t \Rightarrow \frac{3}{x} = \frac{d t}{d x} \Rightarrow \frac{d x}{x} = \frac{d t}{3} Now, \int \frac{\sin (2 + 3 \log x)}{x} d x = \frac{1}{3} \int \sin t d t = \frac{1}{3} [- \cos t] + C = - \frac{1}{3} \cos (2 + 3 \log x) + C$

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Question 59:

$\int x e^{x^{2}} d x$

Answer:

$\int x . e^{x^{2}} d x Let x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Now, \int x . e^{x^{2}} d x = \frac{1}{2} \int e^{t} d t = \frac{1}{2} e^{t} + C = \frac{1}{2} e^{x^{2}} + C$

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Question 60:

$\int \frac{e^{2 x}}{1 + e^{x}} d x$

Answer:

$\int \frac{e^{2 x} d x}{1 + e^{x}} \Rightarrow \int \frac{e^{x} . e^{x}}{1 + e^{x}} d x Let 1 + e^{x} = t \Rightarrow e^{x} = t - 1 \Rightarrow e^{x} d x = d t Now, \int \frac{e^{x} . e^{x}}{1 + e^{x}} d x = \int \frac{(t - 1) . d t}{t} = (1 - \frac{1}{t}) d t = t - \log |t| + C = (1 + e^{x}) - \log (1 + e^{x}) + C Let C + 1 = C' = e^{x} - \log (1 + e^{x}) + C'$

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Question 61:

$\int \frac{\sec^{2} \sqrt{x}}{\sqrt{x}} d x$

Answer:

$\int \frac{\sec^{2} \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} = \frac{d t}{d x} \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t Now, \int \frac{\sec^{2} \sqrt{x}}{\sqrt{x}} d x = 2 \int \sec^{2} t d t = 2 \tan (t) + C = 2 \tan (\sqrt{x}) + C$

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Question 62:

$\int \tan^{3} 2 x \sec 2 x d x$

Answer:

$\int \tan^{3} x 2 x . \sec (2 x) d x = \int \tan^{2} 2 x . \sec 2 x \tan 2 x d x = \int (\sec^{2} (2 x) - 1) \sec (2 x) \tan (2 x) d x Let \sec (2 x) = t \Rightarrow \sec (2 x) \tan (2 x) \times 2 = \frac{d t}{d x} \Rightarrow \sec (2 x) \tan (2 x) d x = \frac{d t}{2} Now, \int \tan^{3} x 2 x . \sec (2 x) d x = \frac{1}{2} \int (t^{2} - 1) d t = \frac{1}{2} [\frac{t^{3}}{3} - t] + C = \frac{1}{2} [\frac{\sec^{3} (2 x)}{3} - \sec (2 x)] + C = \frac{1}{6} \sec^{3} (2 x) - \frac{\sec (2 x)}{2} + C$

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Question 63:

$\int \frac{x + \sqrt{x + 1}}{x + 2} d x$

Answer:

$We have, I = \int \frac{x + \sqrt{x + 1}}{x + 2} d x Let, x + 1 = t^{2} Differentiating both sides we get d x = 2 t d t Now, integration becomes I = \int \frac{(t^{2} - 1 + t)}{t^{2} + 1} 2 t d t = 2 \int \frac{t^{3} + t^{2} - t}{t^{2} + 1} d t = 2 \int \frac{t^{3} + t - t + t^{2} + 1 - 1 - t}{t^{2} + 1} d t = 2 \int \frac{t^{3} + t + t^{2} + 1 - t - t - 1}{t^{2} + 1} d t = 2 \int \frac{t^{3} + t}{t^{2} + 1} d t + + 2 \int \frac{t^{2} + 1}{t^{2} + 1} d t + 2 \int \frac{- 2 t - 1}{t^{2} + 1} d t = 2 \int t d t + 2 \int d t - 2 \int \frac{2 t}{t^{2} + 1} d t - 2 \int \frac{1}{t^{2} + 1} d t = t^{2} + 2 t - 2 \log |t^{2} + 1| - 2 \tan^{- 1} t + C = (x + 1) + 2 \sqrt{x + 1} - 2 \log |x + 2| - 2 \tan^{- 1} \sqrt{x + 1} + C$

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Question 64:

$\int 5^{5^{5^{x}}} 5^{5^{x}} 5^{x} d x$

Answer:

$\int 5^{5^{5^{x}}} \cdot 5^{5^{x}} \cdot 5^{x} d x Let 5^{x} = t \Rightarrow 5^{x} \log 5 = \frac{d t}{d x} \Rightarrow 5^{x} d x = \frac{d t}{\log 5} Now, \int 5^{5^{5^{x}}} \cdot 5^{5^{x}} \cdot 5^{x} d x = \int 5^{5^{t}} \cdot 5^{t} \cdot \frac{d t}{\log 5} Again let 5^{t} = p \Rightarrow 5^{t} \log 5 = \frac{dp}{dt} \Rightarrow 5^{t} dt = \frac{dp}{\log 5} Again \int 5^{5^{t}} \cdot 5^{t} \cdot \frac{d t}{\log 5} = \int 5^{p} \cdot \frac{d p}{{(\log 5)}^{2}} = \frac{5^{p}}{{(\log 5)}^{3}} + C = \frac{5^{5^{5^{x}}}}{{(\log 5)}^{3}} + C$

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Question 65:

$\int \frac{1}{x \sqrt{x^{4} - 1}} d x$

Answer:

$\int \frac{d x}{x \sqrt{x^{4} - 1}} = \int \frac{x d x}{x^{2} \sqrt{{(x^{2})}^{2} - 1}} Let x^{2} = t \Rightarrow 2 x = \frac{d t}{d x} \Rightarrow x d x = \frac{d t}{2} Now, \int \frac{x d x}{x^{2} \sqrt{{(x^{2})}^{2} - 1}} = \frac{1}{2} \int \frac{d t}{t \sqrt{t^{2} - 1}} = \frac{1}{2} \sec^{- 1} (t) + C = \frac{1}{2} \sec^{- 1} (x^{2}) + C$

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Question 66:

$\int \sqrt{e^{x} - 1} d x$

Answer:

$\int \sqrt{e^{x} - 1} d x Let e^{x} - 1 = t^{2} \Rightarrow e^{x} = t^{2} + 1 e^{x} = 2 t \frac{d t}{d x} d x = \frac{2 t d t}{e^{x}} d x = \frac{2 t d t}{t^{2} + 1} Now, \int \sqrt{e^{x} - 1} d x = \int \frac{t \cdot 2 t d t}{t^{2} + 1} = 2 \int \frac{t^{2} d t}{t^{2} + 1} = 2 \int (\frac{t^{2} + 1 - 1}{t^{2} + 1}) d t = 2 \int d t - 2 \int \frac{d t}{t^{2} + 1} = 2 t - 2 \tan^{- 1} (t) + C = 2 \sqrt{e^{x} - 1} - 2 \tan^{- 1} (\sqrt{e^{x} - 1}) + C$

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Question 67:

$\int \frac{1}{(x + 1) (x^{2} + 2 x + 2)} d x$

Answer:

$Let I = \int \frac{d x}{(x + 1) (x^{2} + 2 x + 2)} = \int \frac{d x}{(x + 1) [x^{2} + 2 x + 1 + 1]} = \int \frac{d x}{(x + 1) [{(x + 1)}^{2} + 1]} Putting x + 1 = t \Rightarrow d x = d t Now, integral becomes I = \int \frac{d t}{t [t^{2} + 1]} = \int \frac{t \cdot d t}{t^{2} (t^{2} + 1)} Again putting t^{2} = p \Rightarrow 2 t d t = d p \Rightarrow t d t = \frac{d p}{2} Now, integral becomes I = \frac{1}{2} \int \frac{d p}{p (p + 1)} = \frac{1}{2} \int \frac{d p}{p^{2} + p} = \frac{1}{2} \int \frac{d p}{p^{2} + p + \frac{1}{4} - \frac{1}{4}} = \frac{1}{2} \int \frac{d p}{{(p + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} = \frac{1}{2} [\frac{1}{2 \times \frac{1}{2}} \log |\frac{p + \frac{1}{2} - \frac{1}{2}}{p + \frac{1}{2} + \frac{1}{2}}|] + C = \frac{1}{2} \log |\frac{p}{p + 1}| + C = \frac{1}{2} \log |\frac{t^{2}}{t^{2} + 1}| + C = \frac{1}{2} \log |\frac{{(x + 1)}^{2}}{{(x + 1)}^{2} + 1}| + C = \log \sqrt{|\frac{{(x + 1)}^{2}}{{(x + 1)}^{2} + 1}|} + C = \log |\frac{x + 1}{\sqrt{x^{2} + 2 x + 2}}| + C$

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Question 68:

$\int \frac{x^{5}}{\sqrt{1 + x^{3}}} d x$

Answer:

$\int \frac{x^{5} d x}{\sqrt{1 + x^{3}}} = \int \frac{x^{3} . x^{2} d x}{\sqrt{1 + x^{3}}} Let 1 + x^{3} = t \Rightarrow x^{3} = t - 1 \Rightarrow 3 x^{2} = \frac{d t}{d x} \Rightarrow x^{2} d x = \frac{d t}{3} Now, \int \frac{x^{3} . x^{2} d x}{\sqrt{1 + x^{3}}} = \frac{1}{3} \int \frac{(t - 1)}{\sqrt{t}} d t = \frac{1}{3} \int (\sqrt{t} - \frac{1}{\sqrt{t}}) d t = \frac{1}{3} \int (t^{\frac{1}{2}} - t^{- \frac{1}{2}}) d t = \frac{1}{3} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = \frac{1}{3} [\frac{2}{3} t^{\frac{3}{2}} - 2 \sqrt{t}] + C = \frac{2}{9} {(1 + x^{3})}^{\frac{3}{2}} - \frac{2}{3} {(1 + x^{3})}^{\frac{1}{2}} + C$

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Question 69:

$\int 4 x^{3} \sqrt{5 - x^{2}} d x$

Answer:

$\int 4 x^{3} \sqrt{5 - x^{2}} d x = 4 \int x^{2} \times x \sqrt{5 - x^{2}} d x Let 5 - x^{2} = t \Rightarrow x^{2} = 5 - t \Rightarrow 2 x = - \frac{d t}{d x} \Rightarrow x d x = - \frac{d t}{2} Now, 4 \int x^{2} \times x \sqrt{5 - x^{2}} d x = \frac{4}{- 2} \int (5 - t) . \sqrt{t} d t = - 2 \int 5 t^{\frac{1}{2}} + 2 \int t^{\frac{3}{2}} d t = - 10 [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + 2 [\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + C = - \frac{20}{3} t^{\frac{3}{2}} + \frac{4}{5} t^{\frac{5}{2}} + C = - \frac{20}{3} {(5 - x^{2})}^{\frac{3}{2}} + \frac{4}{5} {(5 - x^{2})}^{\frac{5}{2}} + C$

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Question 70:

$\int \frac{1}{\sqrt{x} + x} d x$

Answer:

$\int \frac{d x}{\sqrt{x} + x} = \int \frac{d x}{\sqrt{x} (1 + \sqrt{x})} Let 1 + \sqrt{x} = t \Rightarrow \frac{1}{2 \sqrt{x}} = \frac{d t}{d x} \Rightarrow \frac{d x}{\sqrt{x}} = 2 d t Now, \int \frac{d x}{\sqrt{x} (1 + \sqrt{x})} = \int \frac{2 d t}{t} = 2 \int \frac{d t}{t} = 2 \log |t| + C = 2 \log (1 + \sqrt{x}) + C$

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Question 71:

$\int \frac{1}{x^{2} {(x^{4} + 1)}^{3 / 4}} d x$

Answer:

$\int \frac{d x}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}} = \int \frac{d x}{x^{2} {[x^{4} (1 + \frac{1}{x^{4}})]}^{\frac{3}{4}}} = \int \frac{d x}{x^{2} . x^{3} {(1 + \frac{1}{x^{4}})}^{\frac{3}{4}}} = \int \frac{{(1 + \frac{1}{x^{4}})}^{- \frac{3}{4}}}{x^{5}} d x Let 1 + \frac{1}{x^{4}} = t \Rightarrow - \frac{4}{x^{5}} d x = d t \Rightarrow \frac{d x}{x^{5}} = - \frac{d t}{4} Now, \int \frac{{(1 + \frac{1}{x^{4}})}^{- \frac{3}{4}}}{x^{5}} d x = - \frac{1}{4} \int t^{- \frac{3}{4}} d t = - \frac{1}{4} [\frac{t^{- \frac{3}{4} + 1}}{- \frac{3}{4} + 1}] + C = - t^{\frac{1}{4}} + C = - {(1 + \frac{1}{x^{4}})}^{\frac{1}{4}} + C$

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Question 72:

$\int \frac{\sin^{5} x}{\cos^{4} x} d x$

Answer:

$\int \frac{\sin^{5} x}{\cos^{4} x} d x = \int (\frac{\sin^{4} x . \sin x}{\cos^{4} x}) d x = \int \frac{{(\sin^{2} x)}^{2} . \sin x}{\cos^{4} x} d x = \int \frac{{(1 - \cos^{2} x)}^{2} \sin x}{\cos^{4} x} d x = \int (\frac{1 + \cos^{4} x - 2 \cos^{2} x}{\cos^{4} x}) \sin x d x = \int (\frac{1}{\cos^{4} x} + 1 - \frac{2}{\cos^{2} x}) \sin x d x Let \cos x = t \Rightarrow - \sin x = \frac{d t}{d x} \Rightarrow \sin x d x = - d t Now, \int (\frac{1}{\cos^{4} x} + 1 - \frac{2}{\cos^{2} x}) \sin x d x = - \int (t^{- 4} + 1 - 2 t^{- 2}) d t = - [\frac{t^{- 4 + 1}}{- 4 + 1} + t - \frac{2 t^{- 2 + 1}}{- 2 + 1}] + C = - [- \frac{1}{3 t^{3}} + t + \frac{2}{t}] + C = \frac{1}{3 t^{3}} - t - \frac{2}{t} + C = \frac{1}{3 \cos^{3} x} - \cos x - \frac{2}{\cos x} + C$

Page No 18.65:

Question 1:

$\int x^{2} \sqrt{x + 2} d x$

Answer:

$\int x^{2} \sqrt{x + 2} d x Let x + 2 = t \Rightarrow x = t - 2 \Rightarrow d x = d t Now, \int x^{2} \sqrt{x + 2} d x = \int {(t - 2)}^{2} \sqrt{t} d t = \int (4^{2} - 4 t + 4) t^{\frac{1}{2}} d t = \int (t^{2 + \frac{1}{2}} - 4 t^{1 + \frac{1}{2}} + 4 t^{\frac{1}{2}}) d t = \int (t^{\frac{5}{2}} - 4 t^{\frac{3}{2}} + 4 t^{\frac{1}{2}}) d t = [\frac{t^{\frac{5}{2} + 1}}{\frac{5}{2} + 1}] - 4 [\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}] + 4 [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = \frac{2}{7} t^{\frac{7}{2}} - \frac{8}{5} t^{\frac{5}{2}} + \frac{8}{3} t^{\frac{3}{2}} + C = \frac{2}{7} {(x + 2)}^{\frac{7}{2}} - \frac{8}{5} {(x + 2)}^{\frac{5}{2}} + \frac{8}{3} {(x + 2)}^{\frac{3}{2}} + C$

Page No 18.65:

Question 2:

$\int \frac{x^{2}}{\sqrt{x - 1}} d x$

Answer:

$\int \frac{x^{2}}{\sqrt{x - 1}} d x Let x - 1 = t^{2} \Rightarrow x = t^{2} + 1 \Rightarrow 1 = 2 t \frac{d t}{d x} \Rightarrow d x = 2 t d t Now, \int \frac{x^{2}}{\sqrt{x - 1}} d x = \int \frac{{(t^{2} + 1)}^{2}}{t} 2 t d t = 2 \int (t^{4} + 2 t^{2} + 1) d t = 2 [\frac{t^{4 + 1}}{4 + 1} + \frac{2 t^{2 + 1}}{2 + 1} + t] + C = 2 [\frac{t^{5}}{5} + \frac{2 t^{3}}{3} + t] + C = 2 [\frac{3 t^{5} + 10 t^{3} + 15 t}{15}] + C = \frac{2}{15} t [3 t^{4} + 10 t^{2} + 15] + C = \frac{2}{15} \sqrt{x - 1} [3 {(x - 1)}^{2} + 10 (x - 1) + 15] + C = \frac{2}{15} \sqrt{x - 1} [3 (x^{2} - 2 x + 1) + 10 x - 10 + 15] + C = \frac{2}{15} \sqrt{x - 1} [3 x^{2} - 6 x + 3 + 10 x - 10 + 15] + C = \frac{2}{15} \sqrt{x - 1} [3 x^{2} + 4 x + 8] + C$

Page No 18.65:

Question 3:

$\int \frac{x^{2}}{\sqrt{3 x + 4}} d x$

Answer:

$\int \frac{x^{2} d x}{\sqrt{3 x + 4}} Let 3 x + 4 = t \Rightarrow x = \frac{t - 4}{3} \Rightarrow 1 = \frac{1}{3} . \frac{d t}{d x} \Rightarrow d x = \frac{d t}{3} Now, \int \frac{x^{2} d x}{\sqrt{3 x + 4}} = \frac{1}{3} \int \frac{{(\frac{t - 4}{3})}^{2}}{\sqrt{t}} d t = \frac{1}{27} \int (\frac{t^{2}}{\sqrt{t}} - \frac{8 t}{\sqrt{t}} + \frac{16}{\sqrt{t}}) d t = \frac{1}{27} \int (t^{\frac{3}{2}} - 8 t^{\frac{1}{2}} + 16 t^{- \frac{1}{2}}) d t = \frac{1}{27} [\frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} + \frac{8 t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + \frac{16 t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}] + C = \frac{1}{27} [\frac{2}{5} t^{\frac{5}{2}} - \frac{8 \times 2}{3} t^{\frac{3}{2}} + 32 t^{\frac{1}{2}}] + C = \frac{2}{135} {(t)}^{\frac{5}{2}} - \frac{16}{81} t^{\frac{3}{2}} + \frac{32}{27} t^{\frac{1}{2}} + C = \frac{2}{135} {(3 x + 4)}^{\frac{5}{2}} - \frac{16}{81} {(3 x + 4)}^{\frac{3}{2}} + \frac{32}{27} {(3 x + 4)}^{\frac{1}{2}} + C$

Page No 18.65:

Question 4:

$\int \frac{2 x - 1}{{(x - 1)}^{2}} d x$

Answer:

$\int [\frac{2 x - 1}{{(x - 1)}^{2}}] d x Let x - 1 = t \Rightarrow x = t + 1 \Rightarrow 1 = \frac{d t}{d x} Now, \int [\frac{2 x - 1}{{(x - 1)}^{2}}] d x = \int [\frac{2 (t + 1) - t}{t^{2}}] d t = \int (\frac{2 t + 1}{t^{2}}) d t = 2 \int \frac{d t}{t} + \int t^{- 2} d t = 2 \log |t| + \frac{t^{- 2 + 1}}{- 2 + 1} + C = 2 \log (x - 1) - \frac{1}{x - 1} + C$

Page No 18.65:

Question 5:

$\int (2 x^{2} + 3) \sqrt{x + 2} d x$

Answer:

$\int (2 x^{2} + 3) \sqrt{x + 2} d x Let x + 2 = t \Rightarrow x = t - 2 \Rightarrow d x = d t \int [2 {(t - 2)}^{2} + 3] \sqrt{t} d t = \int (2 \sqrt{t} (t^{2} - 4 t + 4) + 3 \sqrt{t}) d t = 2 \int (t^{\frac{5}{2}} - 4 t^{\frac{3}{2}} + 4 t^{\frac{1}{2}}) d t + 3 \int t^{\frac{1}{2}} d t = 2 [\frac{t^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} - \frac{4 t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} + \frac{4 t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + 3 [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = 2 [\frac{2}{7} t^{\frac{7}{2}} - \frac{8}{5} t^{\frac{5}{2}} + \frac{8}{3} t^{\frac{3}{2}}] + 2 t^{\frac{3}{2}} + C = \frac{4}{7} t^{\frac{7}{2}} - \frac{16}{5} t^{\frac{5}{2}} + \frac{16}{3} t^{\frac{3}{2}} + 2 t^{\frac{3}{2}} + C = \frac{4}{7} t^{\frac{7}{2}} - \frac{16}{5} t^{\frac{5}{2}} + \frac{22}{3} t^{\frac{3}{2}} + C = \frac{4}{7} {(x + 2)}^{\frac{7}{2}} - \frac{16}{5} {(x + 2)}^{\frac{5}{2}} + \frac{22}{3} {(x + 2)}^{\frac{3}{2}} + C$

Page No 18.65:

Question 6:

$\int \frac{x^{2} + 3 x + 1}{{(x + 1)}^{2}} d x$

Answer:

$\int (\frac{x^{2} + 3 x + 1}{{(x + 1)}^{2}}) d x Let x + 1 = t \Rightarrow x = t - 1 \Rightarrow 1 = \frac{d t}{d x} \Rightarrow d x = d t Now, \int (\frac{x^{2} + 3 x + 1}{{(x + 1)}^{2}}) d x = \int [\frac{{(t - 1)}^{2} + 3 (t - 1) + 1}{t^{2}}] d t = \int (\frac{t^{2} - 2 t + 1 + 3 t - 3 + 1}{t^{2}}) d t = \int (\frac{t^{2} + t - 1}{t^{2}}) d t = \int (1 + \frac{1}{t} - t^{- 2}) d t = t + \log |t| - \frac{t^{- 2 + 1}}{- 2 + 1} + C = t + \log |t| + \frac{1}{t} + C = x + 1 + \log |x + 1| + \frac{1}{x + 1} + C Let 1 + C = C' = x + \log |x + 1| + \frac{1}{x + 1} + C'$

Page No 18.65:

Question 7:

$\int \frac{x^{2}}{\sqrt{1 - x}} d x$

Answer:

$\int \frac{x^{2}}{\sqrt{1 - x}} d x Let 1 - x = t \Rightarrow x = 1 - t \Rightarrow 1 = - \frac{d t}{d x} \Rightarrow d x = - d t Now, \int \frac{x^{2}}{\sqrt{1 - x}} d x = \int \frac{{(1 - t)}^{2}}{\sqrt{t}} d t = \int (\frac{1 - t^{2} - 2 t}{\sqrt{t}}) d t = \int (\frac{1}{\sqrt{t}} + \frac{t^{2}}{\sqrt{t}} - \frac{2 t}{\sqrt{t}}) d t = \int (t^{- \frac{1}{2}} + t^{\frac{3}{2}} - 2 t^{\frac{1}{2}}) d t = [\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + \frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} - \frac{2 t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] + C = 2 t^{\frac{1}{2}} + \frac{2}{5} t^{\frac{5}{2}} - \frac{4}{3} t^{\frac{3}{2}} + C = 2 t^{\frac{1}{2}} [1 + \frac{t^{2}}{5} - \frac{2}{3} t] + C = 2 t^{\frac{1}{2}} [\frac{15 + 3 t^{2} - 10 t}{15}] + C = 2 \sqrt{1 - x} [\frac{15 + 3 {(1 - x)}^{2} - 10 (1 - x)}{15}] + C = \frac{2}{15} \sqrt{1 - x} [15 + 3 (1^{2} + x^{2} - 2 x) - 10 + 10 x] + C = \frac{2}{15} \sqrt{1 - x} [15 + 3 + 3 x^{2} - 6 x - 10 + 10 x] + C = \frac{2}{15} \sqrt{1 - x} [3 x^{2} + 4 x + 8] + C$

Page No 18.65:

Question 8:

$\int x {(1 - x)}^{23} d x$

Answer:

$\int x {(1 - x)}^{23} d x Let 1 - x = t \Rightarrow x = 1 - t \Rightarrow 1 = - \frac{d t}{d x} \Rightarrow d x = - d t Now, \int x {(1 - x)}^{23} d x = - \int (1 - t) \cdot t^{23} d t = - \int (t^{23} - t^{24}) d t = \int (t^{24} - t^{23}) d t = \frac{t^{25}}{25} - \frac{t^{24}}{24} + C = \frac{24 t^{25} - 25 t^{24}}{600} + C = \frac{t^{24}}{600} [24 t - 25] + C = \frac{{(1 - x)}^{24}}{600} [24 (1 - x) - 25] + C = \frac{- 1}{600} {(1 - x)}^{24} (1 + 24 x) + C$

Page No 18.65:

Question 9:

$\int \frac{1}{\sqrt{x} + \sqrt[4]{x}} d x$

Answer:

$Let I = \int \frac{1}{\sqrt{x} + \sqrt[4]{x}} d x Let x = t^{4} On differentiating both sides, we get d x = 4 t^{3} d t ∴ I = \int \frac{4 t^{3}}{\sqrt{t^{4}} + \sqrt[4]{t^{4}}} d t = \int \frac{4 t^{3}}{t^{2} + t} d t = 4 \int \frac{t^{2}}{t + 1} d t = 4 \int \frac{(t - 1) (t + 1) + 1}{t + 1} d t = 4 \int [(t - 1) + \frac{1}{t + 1}] d t = 4 [\frac{t^{2}}{2} - t + \log (t + 1)] + c = 2 \sqrt{x} - 4 \sqrt[4]{x} + 4 \log (\sqrt[4]{x} + 1) + c Hence, \int \frac{1}{\sqrt{x} + \sqrt[4]{x}} d x = 2 \sqrt{x} - 4 \sqrt[4]{x} + 4 \log (\sqrt[4]{x} + 1) + c$

Page No 18.65:

Question 10:

$\int \frac{1}{x^{\frac{1}{3}} (x^{\frac{1}{3}} - 1)} d x$

Answer:

$Let I = \int \frac{1}{x^{\frac{1}{3}} (x^{\frac{1}{3}} - 1)} d x = \int \frac{1}{x^{\frac{2}{3}} - x^{\frac{1}{3}}} d x Let x = t^{3} On differentiating both sides, we get d x = 3 t^{2} d t ∴ I = \int \frac{3 t^{2}}{{(t^{3})}^{\frac{2}{3}} - {(t^{3})}^{\frac{1}{3}}} d t = \int \frac{3 t^{2}}{t^{2} - t} d t = 3 \int \frac{t}{t - 1} d t = 3 \int \frac{(t - 1) + 1}{t - 1} d t = 3 \int [(1) + \frac{1}{t - 1}] d t = 3 [t + \log (t - 1)] + c = 3 x^{\frac{1}{3}} + 3 \log (x^{\frac{1}{3}} - 1) + c Hence, \int \frac{1}{x^{\frac{1}{3}} (x^{\frac{1}{3}} - 1)} d x = 3 x^{\frac{1}{3}} + 3 \log (x^{\frac{1}{3}} - 1) + c$

Page No 18.69:

Question 1:

$\int \tan^{3} x \sec^{2} x d x$

Answer:

∫ tan³ x sec² x dx
Let tan x = t
⇒ sec² x dx = dt
Now, ∫ tan³ x sec² x dx
= ∫ t³.dt
$= \frac{t^{4}}{4} + C = \frac{\tan^{4} x}{4} + C$

Page No 18.69:

Question 2:

$\int \tan x \sec^{4} x d x$

Answer:

∫ tan x. sec⁴ x dx
= ∫ tan x. sec² x . sec² x dx
= ∫ tan x (1 + tan² x) sec² x dx
Let tan x = t
⇒ sec² x dx = dt
Now, ∫ tan x (1 + tan² x) sec² x dx
= ∫ t (1 + t²) dt
= ∫ (t + t³) dt
$= \frac{t^{2}}{2} + \frac{t^{4}}{4} + C = \frac{1}{2} \tan^{2} x + \frac{1}{4} \tan^{4} x + C$

Page No 18.69:

Question 3:

$\int \tan^{5} x \sec^{4} x d x$

Answer:

∫ tan⁵ x sec⁴ x dx
= ∫ tan⁵ x. sec² x . sec² x dx
= ∫ tan⁵ x (1 + tan² x) sec² x dx
Let tan x = t
⇒ sec²x dx = dt
Now, ∫tan⁵x (1+tan² x) sec² x dx
= ∫ t⁵ (1 + t²) dt
= ∫ (t⁵ + t⁷) dt
$= \frac{t^{6}}{6} + \frac{t^{8}}{8} + C = \frac{\tan^{6} x}{6} + \frac{\tan^{8} x}{8} + C$

Page No 18.69:

Question 4:

$\int \sec^{6} x \tan x d x$

Answer:

∫ sec⁶ x tan x dx
=∫ sec⁶ x.sec x tan x dx
Let sec x = t
⇒ sec x tan x dx = dt
Now, ∫ sec⁶ x.sec x tan x dx
= ∫ t⁶. dt
$= \frac{t^{6}}{6} + C = \frac{\sec^{6} x}{6} + C$

Page No 18.69:

Question 5:

$\int \tan^{5} x d x$

Answer:

∫ tan⁵ x dx
= ∫ tan⁴ x. tan x dx
= ∫(sec² x – 1)² . tan x dx
= ∫ (sec⁴ x – 2 sec² x + 1) tan x dx
= ∫ tan x . sec⁴ x dx – 2 ∫ sec² x . tan x dx+ ∫ tan x dx
= ∫ sec² x. sec² x . tan x dx – 2 ∫ tan x sec² x dx + ∫ tan x dx
= ∫ (1 + tan² x) . tan x . sec² x dx – 2 ∫ tan x . sec² x dx + ∫ tan x dx
Let I₁=∫ (1 + tan² x) . tan x . sec² x dx – 2 ∫ tan x . sec² x dx
And I₂=∫ tan x dx
∫ tan⁵ x dx=I₁ + I₂
Now, I₁=∫ (1 + tan² x) . tan x . sec² x dx – 2 ∫ tan x . sec² x dx
Let tan x = t
⇒ sec²x dx = dt
I₁=∫ (1 + tan² x) . tan x . sec² x dx – 2 ∫ tan x . sec² x dx
∫ (1 + t²) . t. dt – 2 ∫ t. dt
∫ (t + t³) dt – 2 ∫ t dt
$\frac{t^{2}}{2} + \frac{t^{4}}{4} - \frac{2 t^{2}}{2} + C_{1} = \frac{t^{4}}{4} - \frac{t^{2}}{2} + C_{1} = \frac{\tan^{4} x}{4} - \frac{\tan^{2} x}{2} + C_{1}$
And I₂=∫ tan x dx
$= \log |\sec x| + C_{2}$
$\int \tan^{5} x d x = \frac{\tan^{4} x}{4} - \frac{\tan^{2} x}{2} + C_{1} + \log |\sec x| + C_{2} = \frac{\tan^{4} x}{4} - \frac{\tan^{2} x}{2} + \log |\sec x| + C_{1} + C_{2} = \frac{\tan^{4} x}{4} - \frac{\tan^{2} x}{2} + \log |\sec x| + C [∴ C = C_{1} + C_{2}]$

Page No 18.69:

Question 6:

$\int \sqrt{\tan x} \sec^{4} x d x$

Answer:

$\int \sqrt{\tan x} \cdot \sec^{4} x d x = \int \sqrt{\tan x} \cdot \sec^{2} x \cdot \sec^{2} x d x = \int \sqrt{\tan x} \cdot (1 + \tan^{2} x) \sec^{2} x d x Let \tan x = t \Rightarrow \sec^{2} x d x = d t Now, \int \sqrt{\tan x} \cdot (1 + \tan^{2} x) \sec^{2} x d x = \int \sqrt{t} (1 + t^{2}) d t = \int (\sqrt{t} + t^{\frac{5}{2}}) d t = \int (t^{\frac{1}{2}} + t^{\frac{5}{2}}) d t = \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{7} t^{\frac{7}{2}} + C = \frac{2}{3} \tan^{\frac{3}{2}} x + \frac{2}{7} \tan^{\frac{7}{2}} x + C$

Page No 18.69:

Question 7:

$\int \sec^{4} 2 x d x$

Answer:

∫ sec⁴ 2x dx
= ∫ sec² 2x . sec² 2x dx
= ∫ (1 + tan² 2x) . sec² 2x dx
Let tan 2x = t
⇒sec² 2x . 2 dx = dt
$\Rightarrow \sec^{2} 2 x . d x = \frac{d t}{2} Now, \int (1 + \tan^{2} 2 x) . \sec^{2} 2 x d x = \frac{1}{2} \int (1 + t^{2}) d t = \frac{1}{2} [t + \frac{t^{3}}{3}] + C = \frac{t}{2} + \frac{t^{3}}{6} + C = \frac{\tan (2 x)}{2} + \frac{\tan^{3} (2 x)}{6} + C$

Page No 18.69:

Question 8:

$\int {cosec}^{4} 3 x d x$

Answer:

∫ cosec⁴ 3x dx
= ∫ cosec²3x . cosec² 3x dx
= ∫ (1 + cot² 3x) cosec² 3x dx
Let cot (3x) = t
⇒–cosec² (3x) × 3 dx = dt
$\Rightarrow {cosec}^{2} (3 x) d x = - \frac{d t}{3} Now, \int (1 + \cot^{2} 3 x) = \frac{- 1}{3} \int (1 + t^{2}) d t = \frac{- 1}{3} [t + \frac{t^{3}}{3}] + C = - \frac{t}{3} - \frac{t^{3}}{9} + C = \frac{- \cot (3 x)}{3} - \frac{\cot^{3} 3 x}{9} + C$

Page No 18.69:

Question 9:

$\int \cot^{n} {cosec}^{2} x d x, n \neq - 1$

Answer:

∫ cotⁿ x cosec² x dx
Let cot x = t
⇒ –cosec² x dx = dt
⇒ cosec² x dx = –dt
$Now, \int \cot^{n} x {cosec}^{2} x d x = - \int t^{n} d t = \frac{- t^{n + 1}}{n + 1} + C = - \frac{\cot^{n + 1} x}{n + 1} + C$

Page No 18.69:

Question 10:

$\int \cot^{5} x {cosec}^{4} x d x$

Answer:

∫ cot⁵x . cosec⁴ x dx
= ∫ cot⁵ x . cosec² x . cosec² x dx
= ∫ cot⁵ x . (1 + cot² x) . cosec² x dx
Let cot x = t
⇒ – cosec² x dx = dt
⇒ cosec² x dx = –dt
Now, ∫ cot⁵x . cosec⁴ x dx
= ∫ t⁵ (1 + t²) dt
= ∫(t⁵ + t⁷) dt
$= - [\frac{t^{6}}{6} + \frac{t^{8}}{8}] + C = - [\frac{\cot^{6} x}{6} + \frac{\cot^{8} x}{8}] + C$

Page No 18.69:

Question 11:

$\int \cot^{5} x d x$

Answer:

∫ cot⁵ x dx
= ∫ cot⁴ x . cot x dx
= ∫ (cosec² x – 1)² cot x dx
= ∫ (cosec⁴ x – 2 cosec² x + 1) cot x dx
= ∫ cosec⁴ x . cot x dx – 2 ∫ cot x . cosec² x dx + ∫ cot x dx
= ∫ cosec² x . cosec² x . cot x . dx – 2 ∫ cot x cosec² x dx + ∫ cot x dx
=∫ (1 + cot ² x) . cot x . cosec² x dx – 2 ∫ cot x cosec² x dx + ∫ cot x dx
= ∫ (cot x + cot³ x) cosec² x dx – 2 ∫ cot x cosec² x dx + ∫ cot x dx
Now, let I₁= ∫ (cot x + cot³ x) cosec² x dx – 2 ∫ cot x cosec² x dx
And I₂= ∫ cot x dx
First we integrate I₁
I₁= ∫ (cot x + cot³ x) cosec² x dx – 2 ∫ cot x cosec² x dx
Let cot x = t
⇒ – cosec² x dx = dt
⇒ cosec² x dx = – dt

I₁= ∫ (t + t³) (– dt) – 2∫ t (–dt)
= –∫(t + t³) + 2∫t dt
$= [- \frac{t^{2}}{2} - \frac{t^{4}}{4}] + 2 . \frac{t^{2}}{2} + C_{1} = \frac{t^{2}}{2} - \frac{t^{4}}{4} + C_{1} = \frac{\cot^{2} x}{2} - \frac{\cot^{4} x}{4} + C_{1}$
Now we integrate I₂
I₂= ∫ cot x dx
   = $\log |\sin x| + C_{2}$
Now, ∫ cot⁵ x dx=I₁ + I₂
                        = $- \frac{1}{4} \cot^{4} x + \frac{1}{2} \cot^{2} x + \log |\sin x| + C_{1} + C_{2}$
                        = $- \frac{1}{4} \cot^{4} x + \frac{1}{2} \cot^{2} x + \log |\sin x| + C [∴ C = C_{1} + C_{2}]$

Page No 18.69:

Question 12:

$\int \cot^{6} x d x$

Answer:

∫ cot⁶ x dx
= ∫ cot⁴ x . (cosec² x – 1) dx
= ∫ cot⁴ x × cosec² x dx – ∫ cot⁴ x dx
= ∫ cot⁴ x . cosec² x dx – ∫ cot² x . cot² x dx
= ∫ cot⁴ x – cosec² x dx – ∫ (cosec² x – 1) cot² x dx
= ∫ cot⁴ x . cosec² x dx – ∫ cot² x . cosec² x dx + ∫ cot² x dx
= ∫ cot⁴ x . cosec² x dx – ∫ cot² x . cosec² x dx + ∫ (cosec² x – 1) dx
Now, let I₁= ∫ cot⁴ x . cosec² x dx – ∫ cot² x . cosec² x dx
And I₂= ∫ (cosec² x – 1) dx
First we integrate I₁
I₁= ∫ cot⁴ x . cosec² x dx – ∫ cot² x . cosec² x dx
Let cot x = t
⇒ –cosec²x dx = dt
⇒ cosec² dx = – dt
I₁=– ∫ t⁴ dt + ∫ t² dt
$= \frac{- t^{5}}{5} + \frac{t^{3}}{3} + C_{1} = - \frac{\cot^{5} x}{5} + \frac{\cot^{3} x}{3} + C_{1}$
Now we integrate I₂
I₂= ∫ (cosec² x – 1) dx
   = – cot x – x + C₁
Now, ∫ cot⁶ x dx=I₁ + I₂
                        = $- \frac{1}{5} \cot^{5} x + \frac{1}{3} \cot^{3} x - \cot x - x + C_{1} + C_{2}$
                        = $- \frac{1}{5} \cot^{5} x + \frac{1}{3} \cot^{3} x - \cot x - x + C [∴ C = C_{1} + C_{2}]$ −14cot4x+12cot2x+log|sin x|+C     [∴C=C1+C2]

Page No 18.73:

Question 1:

$\int \sin^{4} x \cos^{3} x d x$

Answer:

∫ sin⁴ x cos³ x dx
= ∫ sin⁴ x . cos² x cos x dx
= ∫ sin⁴ x . (1 – sin² x ) cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ sin⁴ x . (1 – sin² x ) cos x dx
= ∫ t⁴ (1 – t²) dt
= ∫ (t⁴ – t⁶) dt
$= \frac{t^{5}}{5} - \frac{t^{7}}{7} + C = \frac{\sin^{5} x}{5} - \frac{\sin^{7} x}{7} + C$

Page No 18.73:

Question 2:

$\int \sin^{5} x d x$

Answer:

∫ sin⁵ x dx
= ∫ sin⁴ x . sin x dx
= ∫ (1 – cos² x)² sin x dx
= ∫ (1 – cos⁴ x – 2 cos² x) sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos⁴ x – 2 cos² x) sin x dx
=–∫ (1 + t⁴ – 2t²) dt
$= - [t + \frac{t^{5}}{5} - \frac{2 t^{3}}{3}] + C = - t - \frac{t^{5}}{5} + \frac{2 t^{3}}{3} + C = - \cos x + \frac{2}{3} \cos^{3} x - \frac{\cos^{5} x}{5} + C$

Page No 18.73:

Question 3:

$\int \cos^{5} x d x$

Answer:

∫ cos⁵ x dx
= ∫ cos⁴ x . cos x dx
= ∫ (1 – sin² x)² cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin² x)² cos x dx
= ∫ (1 – t²)² . dt
= ∫ (1 + t⁴ – 2t²) dt
= ∫ dt + ∫ t⁴ dt – 2 ∫t² dt
$= t + \frac{t^{5}}{5} - \frac{2 t^{3}}{3} + C = \sin x + \frac{\sin^{5} x}{5} - \frac{2}{3} \sin^{3} x + C$

Page No 18.73:

Question 4:

$\int \sin^{5} x \cos x d x$

Answer:

∫ sin⁵ x cos x dx
Let sin x = t
cos x dx = dt
Now, ∫ sin⁵ x cos x dx
= ∫ t⁵ . dt
$= \frac{t^{6}}{6} + C = \frac{\sin^{6} x}{6} + C$

Page No 18.73:

Question 5:

$\int \sin^{3} x \cos^{6} x d x$

Answer:

∫ sin³ x . cos⁶ x dx
= ∫ sin² x . cos⁶ x . sin x dx
= ∫ (1 – cos² x) . cos⁶ x . sin x dx
Let cos x = t
⇒ –sin x dx = dt
Now, ∫ (1 – cos² x) . cos⁶ x . sin x dx
= –∫ (1 – t²) . t⁶ dt
= ∫ (t² – 1) t⁶ dt
= ∫ (t⁸ – t⁶) dt
$= \frac{t^{9}}{9} - \frac{t^{7}}{7} + C = \frac{\cos^{9} x}{9} - \frac{\cos^{7} x}{7} + C$

Page No 18.73:

Question 6:

$\int \cos^{7} x d x$

Answer:

∫ cos⁷ x dx
= ∫ cos⁶ x . cos x dx
= ∫ (cos² x)³ cos x dx
= ∫ (1 – sin² x)³ . cos x dx
Let sin x = t
⇒ cos x dx = dt
Now, ∫ (1 – sin² x)³.cos x dx
= ∫ (1 – t²)³ dt
= ∫ (1 – t⁶ – 3t² + 3t⁴) dt
$= [t - \frac{t^{7}}{7} - \frac{3 t^{3}}{3} + \frac{3 t^{5}}{5}] + C = \sin x - \frac{1}{7} \sin^{7} x - \sin^{3} x + \frac{3}{5} \sin^{5} x + C$

Page No 18.73:

Question 7:

$\int x \cos^{3} x^{2} \sin x^{2} d x$

Answer:

∫ x . cos³ x² sin x² dx
Let x² = t
⇒ 2x dx = dt
$\Rightarrow x d x = \frac{d t}{2} Now, \int x . \cos^{3} x^{2} \sin x^{2} d x = \frac{1}{2} \int \cos^{3} t . \sin t . d t Again let \cos t = p \Rightarrow - \sin t d t = d p \Rightarrow \sin t d t = - d p So, \frac{1}{2} \int \cos^{3} t . \sin t . d t = - \frac{1}{2} p^{3} d p = - \frac{1}{2} (\frac{p^{4}}{4}) + C = - \frac{p^{4}}{8} + C = - \frac{\cos^{4} t}{8} + C = - \frac{\cos^{4} x^{2}}{8} + C$

Page No 18.73:

Question 8:

$\int \sin^{7} x d x$

Answer:

∫ sin⁷ x dx
= ∫ sin⁶ x . sin x dx
= ∫ (sin² x)³ sin x dx
= ∫ (1 – cos²x)³ sin x dx
Let cos x = t
⇒ –sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos²x)³ sin x dx
= ∫ (1 – t²)³ . (–dt)
= –∫ (1 – t⁶ – 3t² + 3t⁴) dt
$= - [t - \frac{t^{7}}{7} - t^{3} + \frac{3 t^{5}}{5}] + C = - [\cos x - \frac{\cos^{7} x}{7} - \cos^{3} x + \frac{3}{5} \cos^{5} x] + C = - \cos x + \frac{1}{7} \cos^{7} x + \cos^{3} x - \frac{3}{5} \cos^{5} x + C$

Page No 18.73:

Question 9:

$\int \sin^{3} x \cos^{5} x d x$

Answer:

∫ sin³ x . cos⁵ x dx
= ∫ sin² x . cos⁵ x . sin x dx
= ∫ (1 – cos² x) . cos⁵ x sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos² x) . cos⁵ x sin x dx
= –∫ (1 – t²) t⁵ dt
= –∫ (t⁵ – t⁷) dt
= ∫(t⁷ – t⁵) dt
$= \frac{t^{8}}{8} - \frac{t^{6}}{6} + C = \frac{\cos^{8} x}{8} - \frac{\cos^{6} x}{6} + C$

Page No 18.73:

Question 10:

$\int \frac{1}{\sin^{4} x \cos^{2} x} d x$

Answer:

$\int \frac{d x}{\sin^{4} x . \cos^{2} x} Dividing numerator & denominator by \sin^{2} x = \int \frac{\frac{1}{\sin^{2} x}}{\sin^{4} x . \cot^{2} x} d x = \int \frac{{cosec}^{6} x}{\cot^{2}} d x = \int \frac{{cosec}^{4} x . {cosec}^{2} x d x}{\cot^{2} x} = \int \frac{{(1 + \cot^{2} x)}^{2} . {cosec}^{2} x d x}{\cot^{2} x} Let \cot x = t \Rightarrow - {cosec}^{2} x = \frac{d t}{d x} \Rightarrow - {cosec}^{2} x d x = d t Now, \int \frac{{(1 + \cot^{2} x)}^{2} . {cosec}^{2} x d x}{\cot^{2} x} = \int {(\frac{1 + t^{2}}{t})}^{2} (- d t) = - \int \frac{(1 + t^{4} + 2 t^{2})}{t^{2}} d t = - \int (t^{- 2} + t^{2} + 2) d t = - [\frac{t^{- 2 + 1}}{- 2 + 1} + \frac{t^{3}}{3} + 2 t] + C = - [- \frac{1}{t} + \frac{t^{3}}{3} + 2 t] + C = - \frac{1}{3} t^{3} - 2 t + \frac{1}{t} + C = - \frac{1}{3} \cot^{3} x - 2 \cot x + \frac{1}{\cot x} + C = - \frac{1}{3} \cot^{3} x - 2 \cot x + \tan x + C$

Page No 18.73:

Question 11:

$\int \frac{1}{\sin^{3} x \cos^{5} x} d x$

Answer:

$\int \frac{d x}{\sin^{3} x . \cos^{5} x} d x Dividing numerator & denominator by \cos^{8} x = \int \frac{\frac{1}{\cos^{8} x} d x}{\frac{\sin^{3} x}{\cos^{3} x}} = \int \frac{\sec^{8} x}{\tan^{3} x} d x = \int \frac{\sec^{6} x . \sec^{2} x d x}{\tan^{3} x} = \int \frac{{(1 + \tan^{2} x)}^{3} . \sec^{2} x d x}{\tan^{3} x} Let \tan x = t \Rightarrow \sec^{2} x d x = d t Now, \int \frac{{(1 + \tan^{2} x)}^{3} . \sec^{2} x d x}{\tan^{3} x} = \int \frac{{(1 + t^{2})}^{3}}{t^{3}} . d t = \int \frac{1 + t^{6} + 3 t^{2} + 3 t^{4}}{t^{3}} d t = \int (\frac{1}{t^{3}} + t^{3} + \frac{3}{t} + 3 t) d t = \int t^{- 3} d t + \int t^{3} d t + 3 \int \frac{d t}{t} + 3 \int t d t = [\frac{t^{- 3 + 1}}{- 3 + 1}] + [\frac{t^{3 + 1}}{3 + 1}] + 3 \log |t| + \frac{3 t^{2}}{2} + C = - \frac{1}{2} {(\tan x)}^{- 2} + \frac{1}{4} \tan^{4} x + 3 \log |\tan x| + \frac{3}{2} \tan^{2} x + C$

Page No 18.73:

Question 12:

$\int \frac{1}{\sin^{3} x \cos x} d x$

Answer:

$\int \frac{d x}{\sin^{3} x . \cos x} Dividing numerator & denominator by \sin^{4} x = \int \frac{\frac{1}{\sin^{4} x} d x}{\frac{\sin^{3} x . \cos x}{\sin^{4} x}} = \int \frac{{cosec}^{4} x d x}{\cot x} = \int \frac{{cosec}^{2} x . {cosec}^{2} x d x}{\cot x} = \int \frac{(1 + \cot^{2} x) . {cosec}^{2} x d x}{\cot x} Let \cot x = t \Rightarrow - {cosec}^{2} x = \frac{d t}{d x} \Rightarrow {cosec}^{2} x d x = - d t Now, \int \frac{(1 + \cot^{2} x) . {cosec}^{2} x}{\cot x} d x = \int \frac{(1 + t^{2}) . (- d t)}{t} = - \int (\frac{1}{t} + t) d t = - \log |t| - \frac{t^{2}}{2} + C = - \log |\cot x| - \frac{\cot^{2} x}{2} + C = \log {|\cot x|}^{- 1} - \frac{({cosec}^{2} x - 1)}{2} + C = \log |\frac{1}{\cot x}| - \frac{{cosec}^{2} x}{2} + \frac{1}{2} + C = \log |\tan x| - \frac{1}{2 \sin^{2} x} + C' [∴ C' = C + \frac{1}{2}]$

Page No 18.73:

Question 13:

$\int \frac{1}{\sin x \cos^{3} x} d x$

Answer:

$\int \frac{d x}{\sin x . \cos^{3} x} Dividing numerator & denominaor by \cos^{4} x = \int \frac{\frac{1}{\cos^{4} x} d x}{\frac{\sin x . \cos^{3} x}{\cos^{4} x}} = \int \frac{\sec^{4} x d x}{\tan x} = \int \frac{\sec^{2} x . \sec^{2} x d x}{\tan x} = \int \frac{(1 + \tan^{2} x) . \sec^{2} x}{\tan x} d x Let \tan x = t \Rightarrow \sec^{2} x = \frac{d x}{d t} \Rightarrow \sec^{2} x d x = d t Now, \int \frac{(1 + \tan^{2} x) . \sec^{2} x}{\tan x} d x = \int \frac{(1 + t^{2})}{t} d t = \int (\frac{1}{t} + t) d t = \log |t| + \frac{t^{2}}{2} + C = \log |\tan x| + \frac{\tan^{2} x}{2} + C$

Page No 18.79:

Question 1:

Evaluate the following integrals:

$\int \frac{x^{2}}{{(a^{2} - x^{2})}^{\frac{3}{2}}} d x$

Answer:

$Let I = \int \frac{x^{2}}{{(a^{2} - x^{2})}^{\frac{3}{2}}} d x Let x = a \cos θ On differentiating both sides, we get d x = - a \sin θ d θ ∴ I = \int \frac{a^{2} \cos^{2} θ}{{(a^{2} - a^{2} \cos^{2} θ)}^{\frac{3}{2}}} \times - a \sin θ d θ = - \int \frac{a^{3} \cos^{2} θ \sin θ}{a^{3} {(1 - \cos^{2} θ)}^{\frac{3}{2}}} d θ = - \int \frac{\cos^{2} θ \sin θ}{\sin^{3} θ} d θ = - \int \cot^{2} θ d θ = - \int (\cos e c^{2} θ - 1) d θ = - (- \cot θ - θ) + c = \cot θ + θ + c = \cot (\cos^{- 1} \frac{x}{a}) + \cos^{- 1} \frac{x}{a} + c = \cot (\cot^{- 1} \frac{x}{\sqrt{a^{2} - x^{2}}}) + \cos^{- 1} \frac{x}{a} + c = \frac{x}{\sqrt{a^{2} - x^{2}}} + \cos^{- 1} \frac{x}{a} + c Hence, \int \frac{x^{2}}{{(a^{2} - x^{2})}^{\frac{3}{2}}} d x = \frac{x}{\sqrt{a^{2} - x^{2}}} + \cos^{- 1} \frac{x}{a} + c$

Page No 18.79:

Question 3:

Evaluate the following integrals:

$\int \cos \{2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}}\} d x$

Answer:

$Let I = \int \cos \{2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}}\} d x Let x = \cos 2 θ On differentiating both sides, we get d x = - 2 \sin 2 θ d θ ∴ I = - \int \cos \{2 \cot^{- 1} \sqrt{\frac{1 + \cos 2 θ}{1 - \cos 2 θ}}\} 2 \sin 2 θ d θ = - 2 \int \cos \{2 \cot^{- 1} \sqrt{\frac{2 \cos^{2} θ}{2 \sin^{2} θ}}\} \sin 2 θ d θ = - 2 \int \cos \{2 \cot^{- 1} (\cot θ)\} \sin 2 θ d θ = - 2 \int \cos 2 θ \sin 2 θ d θ = - \int \sin 4 θ d θ = \frac{\cos 4 θ}{4} + c_{1} = \frac{1}{4} (2 \cos^{2} 2 θ - 1) + c_{1} = \frac{1}{2} x^{2} - \frac{1}{4} + c_{1} = \frac{1}{2} x^{2} + c, where c = - \frac{1}{4} + c_{1} Hence, \int \cos \{2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}}\} d x = \frac{1}{2} x^{2} + c$

Page No 18.79:

Question 4:

Evaluate the following integrals:

$\int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x$

Answer:

$Let I = \int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x Let x = \tan θ On differentiating both sides, we get d x = \sec^{2} θ d θ ∴ I = \int \frac{\sqrt{1 + \tan^{2} θ}}{\tan^{4} θ} \sec^{2} θ d θ = \int \frac{\sec^{3} θ}{\tan^{4} θ} d θ = \int \frac{\cos θ}{\sin^{4} θ} d θ = \int \cot θ {cosec}^{3} θ d θ Let {cosec}^{3} θ = t On differentiating both sides, we get - 3 {cosec}^{3} θ \cot θ d θ = d t ∴ I = - \frac{1}{3} \int \cot θ {cosec}^{3} θ \frac{d t}{{cosec}^{3} θ \cot θ} = - \frac{t}{3} + c = - \frac{1}{3} ({cosec}^{3} θ) + c = - \frac{1}{3} {(cosec (\tan^{- 1} x))}^{3} + c = - \frac{1}{3} {(cosec ({cosec}^{- 1} \frac{\sqrt{1 + x^{2}}}{x}))}^{3} + c = - \frac{1}{3} {(\frac{\sqrt{1 + x^{2}}}{x})}^{3} + c Hence, \int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x = - \frac{1}{3} {(\frac{\sqrt{1 + x^{2}}}{x})}^{3} + c$

Page No 18.79:

Question 5:

Evaluate the following integrals:

$\int \frac{1}{{(x^{2} + 2 x + 10)}^{2}} d x$

Answer:

$Let I = \int \frac{1}{{(x^{2} + 2 x + 10)}^{2}} d x = \int \frac{1}{{[{(x + 1)}^{2} + 3^{2}]}^{2}} d x Let x + 1 = 3 \tan θ On differentiating both sides, we get d x = 3 \sec^{2} θ d θ ∴ I = \int \frac{1}{{[3^{2} \tan^{2} θ + 3^{2}]}^{2}} 3 \sec^{2} θ d θ = \frac{1}{27} \int \frac{\sec^{2} θ}{\sec^{4} θ} d θ = \frac{1}{27} \int \frac{1}{\sec^{2} θ} d θ = \frac{1}{27} \int \cos^{2} θ d θ = \frac{1}{54} \int (1 + \cos 2 θ) d θ = \frac{1}{54} (θ + \frac{\sin 2 θ}{2}) + c = \frac{1}{54} (θ + \frac{\tan θ}{1 + \tan^{2} θ}) + c = \frac{1}{54} (\tan^{- 1} \frac{x + 1}{3} + \frac{\tan (\tan^{- 1} \frac{x + 1}{3})}{1 + \tan^{2} (\tan^{- 1} \frac{x + 1}{3})}) + c = \frac{1}{54} (\tan^{- 1} \frac{x + 1}{3} + \frac{\frac{x + 1}{3}}{1 + {(\frac{x + 1}{3})}^{2}}) + c = \frac{1}{54} (\tan^{- 1} \frac{x + 1}{3} + \frac{\frac{x + 1}{3}}{\frac{x^{2} + 2 x + 10}{9}}) + c = \frac{1}{54} (\tan^{- 1} \frac{x + 1}{3} + \frac{3 (x + 1)}{x^{2} + 2 x + 10}) + c Hence, \int \frac{1}{{(x^{2} + 2 x + 10)}^{2}} d x = \frac{1}{54} (\tan^{- 1} \frac{x + 1}{3} + \frac{3 (x + 1)}{x^{2} + 2 x + 10}) + c$

Page No 18.83:

Question 1:

$\int \frac{1}{a^{2} - b^{2} x^{2}} d x$

Answer:

$\int \frac{d x}{a^{2} - b^{2} x^{2}} = \frac{1}{b^{2}} \int \frac{d x}{(\frac{a^{2}}{b^{2}}) - x^{2}} = \frac{1}{b^{2}} \times \frac{1}{2 \frac{a}{b}} \log |\frac{\frac{a}{b} + x}{\frac{a}{b} - x}| + C [∴ \int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} \log |\frac{a + x}{a - x}| + C] = \frac{1}{2 a b} \log |\frac{a + b x}{a - b x}| + c$

Page No 18.83:

Question 2:

$\int \frac{1}{a^{2} x^{2} - b^{2}} d x$

Answer:

$\int \frac{d x}{a^{2} x^{2} - b^{2}} = \frac{1}{a^{2}} \int \frac{d x}{x^{2} - {(\frac{b}{a})}^{2}} = \frac{1}{a^{2}} \times \frac{1}{2 \frac{b}{a}} \log |\frac{x - \frac{b}{a}}{x + \frac{b}{a}}| + C [∴ \int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} \log |\frac{x - a}{x + a}| + C] = \frac{1}{2 a b} \log |\frac{a x - b}{a x + b}| + C$

Page No 18.83:

Question 3:

$\int \frac{1}{a^{2} x^{2} + b^{2}} d x$

Answer:

$\int \frac{d x}{a^{2} x^{2} + b^{2}} = \frac{1}{a^{2}} \int \frac{d x}{x^{2} + {(\frac{b}{a})}^{2}} = \frac{1}{a^{2}} \times \frac{a}{b} \tan^{- 1} (\frac{x}{\frac{b}{a}}) + C [∴ \int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + C] = \frac{1}{a b} \tan^{- 1} (\frac{a x}{b}) + C$

Page No 18.83:

Question 4:

$\int \frac{x^{2} - 1}{x^{2} + 4} d x$

Answer:

$\int \frac{x^{2} - 1}{x^{2} + 4} d x = \int (\frac{x^{2} + 4 - 4 - 1}{x^{2} + 4}) d x = \int (\frac{x^{2} + 4}{x^{2} + 4}) d x - 5 \int \frac{d x}{x^{2} + 2^{2}} = \int d x - 5 \int \frac{d x}{x^{2} + 2^{2}} = x - \frac{5}{2} \tan^{- 1} (\frac{x}{2}) + C [∴ \int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + C]$

Page No 18.83:

Question 5:

$\int \frac{1}{\sqrt{1 + 4 x^{2}}} d x$

Answer:

$\int \frac{d x}{\sqrt{1 + 4 x^{2}}} = \int \frac{d x}{\sqrt{1 + {(2 x)}^{2}}} let 2 x = t \Rightarrow 2 d x = d t \Rightarrow d x = \frac{d t}{2} Now, \int \frac{d x}{\sqrt{1 + {(2 x)}^{2}}} = \frac{1}{2} \int \frac{d t}{\sqrt{1 + t^{2}}} = \frac{1}{2} \log |t + \sqrt{1 + t^{2}}| + C [∵ \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \log |x + \sqrt{x^{2} + a^{2}}| + C] = \frac{1}{2} \log |2 x + \sqrt{1 + 4 x^{2}}| + C$

Page No 18.83:

Question 6:

$\int \frac{1}{\sqrt{a^{2} + b^{2} x^{2}}} d x$

Answer:

$\int \frac{d x}{\sqrt{a^{2} + b^{2} x^{2}}} = \int \frac{d x}{\sqrt{b^{2} (\frac{a^{2}}{b^{2}} + x^{2})}} = \frac{1}{b} \int \frac{d x}{\sqrt{x^{2} + {(\frac{a}{b})}^{2}}} = \frac{1}{b} \log |x + \sqrt{x^{2} + \frac{a^{2}}{b^{2}}}| + C = \frac{1}{b} [\log |x + \frac{\sqrt{b^{2} x^{2} + a^{2}}}{b}|] + C = \frac{1}{b} [\log |\frac{b x + \sqrt{b^{2} x^{2} + a^{2}}}{b}|] + C = \frac{1}{b} [\log |b x + \sqrt{b^{2} x^{2} + a^{2}}| - \log b] + C = \frac{1}{b} \log |b x + \sqrt{b^{2} x^{2} + a^{2}}| - \frac{\log b}{b} + C let C - \frac{\log b}{b} = C' = \frac{1}{b} \log |b x + \sqrt{b^{2} x^{2} + a^{2}}| + C'$

Page No 18.83:

Question 7:

$\int \frac{1}{\sqrt{a^{2} - b^{2} x^{2}}} d x$

Answer:

$\int \frac{d x}{\sqrt{a^{2} - b^{2} x^{2}}} = \int \frac{d x}{\sqrt{b^{2} (\frac{a^{2}}{b^{2}} - x^{2})}} = \frac{1}{b} \int \frac{d x}{\sqrt{{(\frac{a}{b})}^{2} - x^{2}}} = \frac{1}{b} \sin^{- 1} (\frac{x b}{a}) + C$

Page No 18.83:

Question 8:

$\int \frac{1}{\sqrt{{(2 - x)}^{2} + 1}} d x$

Answer:

$\int \frac{d x}{\sqrt{{(2 - x)}^{2} + 1}} let 2 - x = t \Rightarrow - d x = d t \Rightarrow d x = - d t Now, \int \frac{d x}{\sqrt{{(2 - x)}^{2} + 1}} = - \int \frac{d t}{\sqrt{t^{2} + 1}} = - \log |t + \sqrt{t^{2} + 1}| + C = - \log |(2 - x) + \sqrt{{(2 - x)}^{2} + 1}| + C$

Page No 18.83:

Question 9:

$\int \frac{1}{\sqrt{{(2 - x)}^{2} - 1}} d x$

Answer:

$\int \frac{d x}{\sqrt{{(2 - x)}^{2} - 1}} let 2 - x = t \Rightarrow - d x = d t \Rightarrow d x = - d t Now, \int \frac{d x}{\sqrt{{(2 - x)}^{2} - 1}} = \int \frac{- d t}{\sqrt{t^{2} - 1}} = - \log |t + \sqrt{t^{2} - 1}| + C = - \log |(2 - x) + \sqrt{{(2 - x)}^{2} - 1}| + C$

Page No 18.83:

Question 10:

$\int \frac{x^{4} + 1}{x^{2} + 1} d x$

Answer:

$\int (\frac{x^{4} + 1}{x^{2} + 1}) d x = \int (\frac{x^{4} - 1 + 1 + 1}{x^{2} + 1}) d x = \int [\frac{(x^{4} - 1)}{x^{2} + 1} + \frac{2}{x^{2} + 1}] d x = \int [\frac{(x^{2} - 1) (x^{2} + 1)}{(x^{2} + 1)} + \frac{2}{x^{2} + 1}] d x = \int [(x^{2} - 1) + \frac{2}{x^{2} + 1}] d x = \frac{x^{3}}{3} - x + 2 \tan^{- 1} (x) + C$

Page No 18.86:

Question 1:

$\int \frac{1}{4 x^{2} + 12 x + 5} d x$

Answer:

$\int \frac{d x}{4 x^{2} + 12 x + 5} = \frac{1}{4} \int \frac{d x}{x^{2} + 3 x + \frac{5}{4}} = \frac{1}{4} \int \frac{d x}{x^{2} + 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{5}{4}} = \frac{1}{4} \int \frac{d x}{{(x + \frac{3}{2})}^{2} - \frac{9}{4} + \frac{5}{4}} = \frac{1}{4} \int \frac{d x}{{(x + \frac{3}{2})}^{2} - 1^{2}} let x + \frac{3}{2} = t \Rightarrow d x = d t Now, \frac{1}{4} \int \frac{d x}{{(x + \frac{3}{2})}^{2} - 1^{2}} = \frac{1}{4} \int \frac{d x}{t^{2} - 1^{2}} = \frac{1}{4} \times \frac{1}{2 \times 1} \log |\frac{t - 1}{t + 1}| + C = \frac{1}{8} \log |\frac{x + \frac{3}{2} - 1}{x + \frac{3}{2} + 1}| + C = \frac{1}{8} \log |\frac{x + \frac{1}{2}}{x + \frac{5}{2}}| + C = \frac{1}{8} \log |\frac{2 x + 1}{2 x + 5}| + C$

Page No 18.86:

Question 2:

$\int \frac{1}{x^{2} - 10 x + 34} d x$

Answer:

$\int \frac{d x}{x^{2} - 10 x + 34} = \int \frac{d x}{x^{2} - 10 x + 25 - 25 + 34} = \int \frac{d x}{{(x - 5)}^{2} + 9} = \int \frac{d x}{{(x - 5)}^{2} + 3^{2}} let x - 5 = t \Rightarrow d x = d t Now, \int \frac{d x}{{(x - 5)}^{2} + 3^{2}} = \int \frac{d t}{t^{2} + 3^{2}} = \frac{1}{3} \tan^{- 1} (\frac{t}{3}) + C = \frac{1}{3} \tan^{- 1} (\frac{x - 5}{3}) + C$

Page No 18.86:

Question 3:

$\int \frac{1}{1 + x - x^{2}} d x$

Answer:

$\int \frac{d x}{1 + x - x^{2}} = \int \frac{- d x}{x^{2} - x - 1} = \int \frac{- d x}{x^{2} - x + \frac{1}{4} - \frac{1}{4} - 1} = \int \frac{- d x}{{(x - \frac{1}{2})}^{2} - \frac{5}{4}} = \int \frac{d x}{\frac{5}{4} - {(x - \frac{1}{2})}^{2}} = \int \frac{d x}{{(\frac{\sqrt{5}}{2})}^{2} - {(x - \frac{1}{2})}^{2}} let x - \frac{1}{2} = t \Rightarrow d x = d t Now, \int \frac{d x}{{(\frac{\sqrt{5}}{2})}^{2} - {(x - \frac{1}{2})}^{2}} = \int \frac{d t}{{(\frac{\sqrt{5}}{2})}^{2} - t^{2}} = \frac{1}{2 \times \frac{\sqrt{5}}{2}} \log |\frac{\frac{\sqrt{5}}{2} + t}{\frac{\sqrt{5}}{2} - t}| + C$
$= \frac{1}{\sqrt{5}} \log |\frac{\sqrt{5} + 2 t}{\sqrt{5} - 2 t}| + C = \frac{1}{\sqrt{5}} \log |\frac{\sqrt{5} + 2 (x - \frac{1}{2})}{\sqrt{5} - 2 (x - \frac{1}{2})}| + C = \frac{1}{\sqrt{5}} \log |\frac{\sqrt{5} - 1 + 2 x}{\sqrt{5} + 1 - 2 x}| + C$

Page No 18.86:

Question 4:

$\int \frac{1}{2 x^{2} - x - 1} d x$

Answer:

$\int \frac{d x}{2 x^{2} - x - 1} = \frac{1}{2} \int \frac{d x}{x^{2} - \frac{x}{2} - \frac{1}{2}} = \frac{1}{2} \int \frac{d x}{x^{2} - \frac{x}{2} + {(\frac{1}{4})}^{2} - {(\frac{1}{4})}^{2} - \frac{1}{2}} = \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{4})}^{2} - \frac{1}{16} - \frac{1}{2}} = \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{4})}^{2} - (\frac{1 + 8}{16})} = \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{4})}^{2} - {(\frac{3}{4})}^{2}} let x - \frac{1}{4} = t \Rightarrow d x = d t$
$Now, \frac{1}{2} \int \frac{d x}{{(x - \frac{1}{2})}^{2} - {(\frac{3}{4})}^{2}} = \frac{1}{2} \int \frac{d t}{t^{2} - {(\frac{3}{4})}^{2}} = \frac{1}{2} \int \frac{d t}{t^{2} - {(\frac{3}{4})}^{2}} = \frac{1}{2 \times \frac{3}{4}} \times \frac{1}{2} \log |\frac{t - \frac{3}{4}}{t + \frac{3}{4}}| + C = \frac{2}{3} \times \frac{1}{2} \log |\frac{x - \frac{1}{4} - \frac{3}{4}}{x - \frac{1}{4} + \frac{3}{4}}| + C = \frac{2}{3} \times \frac{1}{2} \log |\frac{x - 1}{x + \frac{1}{2}}| + C = \frac{1}{3} \log |\frac{2 (x - 1)}{2 x + 1}| + C = \frac{1}{3} [\log |\frac{(x - 1)}{2 x + 1}| + \log |2|] + C = \frac{1}{3} \log |\frac{x - 1}{2 x + 1}| + \frac{1}{3} \log |2| + C = \frac{1}{3} \log |\frac{x - 1}{2 x + 1}| + C' [∵ C' = \frac{1}{3} \log |2| + C]$

Page No 18.86:

Question 5:

$\int \frac{1}{x^{2} + 6 x + 13} d x$

Answer:

$\int \frac{d x}{x^{2} + 6 x + 13} = \int \frac{d x}{x^{2} + 2 \times x \times 3 + 9 - 9 + 13} = \int \frac{d x}{{(x + 3)}^{2} + 2^{2}} = \frac{1}{2} \tan^{- 1} (\frac{x + 3}{2}) + C$

Page No 18.90:

Question 1:

$\int \frac{\sec^{2} x}{1 - \tan^{2} x} d x$

Answer:

$\int \frac{\sec^{2} x d x}{1 - \tan^{2} x} let \tan x = t \Rightarrow \sec^{2} x d x = d t Now, \int \frac{\sec^{2} x d x}{1 - \tan^{2} x} = \int \frac{d t}{1 - t^{2}} = \frac{1}{2} \log |\frac{1 + t}{1 - t}| + C = \frac{1}{2} \log |\frac{1 + \tan x}{1 - \tan x}| + C$

Page No 18.90:

Question 2:

$\int \frac{e^{x}}{1 + e^{2 x}} d x$

Answer:

$\int \frac{e^{x} d x}{1 + e^{2 x}} let e^{x} = t \Rightarrow e^{x} d x = d t Now, \int \frac{e^{x} d x}{1 + e^{2 x}} = \int \frac{d t}{1 + t^{2}} = \tan^{- 1} (t) + C = \tan^{- 1} (e^{x}) + C$

Page No 18.90:

Question 3:

$\int \frac{\cos x}{\sin^{2} x + 4 \sin x + 5} d x$

Answer:

$\int \frac{\cos x d x}{\sin^{2} x + 4 \sin x + 5} let \sin x = t \Rightarrow \cos x d x = d t Now, \int \frac{\cos x d x}{\sin^{2} x + 4 \sin x + 5} = \int \frac{d t}{t^{2} + 4 t + 5} = \int \frac{d t}{t^{2} + 2 \times t \times 2 + 4 + 1} = \int \frac{d t}{{(t + 2)}^{2} + 1^{2}} = \frac{1}{1} \tan^{- 1} (\frac{t + 2}{1}) + C = \tan^{- 1} (\sin x + 2) + C$

Page No 18.90:

Question 4:

$\int \frac{e^{x}}{e^{2 x} + 5 e^{x} + 6} d x$

Answer:

$\int \frac{e^{x} d x}{e^{2 x} + 5 e^{x} + 6} let e^{x} = t \Rightarrow e^{x} d x = d t Now, \int \frac{e^{x} d x}{e^{2 x} + 5 e^{x} + 6} = \int \frac{d t}{t^{2} + 5 t + 6} = \int \frac{d t}{t^{2} + 5 t + {(\frac{5}{2})}^{2} - {(\frac{5}{2})}^{2} + 6} = \int \frac{d t}{{(t + \frac{5}{2})}^{2} - \frac{25}{4} + 6} = \int \frac{d t}{{(t + \frac{5}{2})}^{2} - \frac{25 + 24}{4}} = \int \frac{d t}{{(t + \frac{5}{2})}^{2} - {(\frac{1}{2})}^{2}} = \frac{1}{2 \times \frac{1}{2}} \log |\frac{t + \frac{5}{2} - \frac{1}{2}}{t + \frac{5}{2} + \frac{1}{2}}| + C = \log |\frac{t + 2}{t + 3}| + C = \log |\frac{e^{x} + 2}{e^{x} + 3}| + C$

Page No 18.90:

Question 5:

$\int \frac{e^{3 x}}{4 e^{6 x} - 9} d x$

Answer:

$\int \frac{e^{3 x} d x}{4 e^{6 x} - 9} let e^{3 x} = t \Rightarrow e^{3 x} \times 3 d x = d t \Rightarrow e^{3 x} d x = \frac{d t}{3} Now, \int \frac{e^{3 x} d x}{4 e^{6 x} - 9} = \frac{1}{3} \int \frac{d t}{4 t^{2} - 9} = \frac{1}{3} \int \frac{d t}{{(2 t)}^{2} - 3^{2}} = \frac{1}{3} \times \frac{1}{2 \times 3} \log |\frac{2 t - 3}{2 t + 3}| \times \frac{1}{2} + C = \frac{1}{36} \log |\frac{2 t - 3}{2 t + 3}| + C = \frac{1}{36} \log |\frac{2 e^{3 x} - 3}{2 e^{3 x} + 3}| + C$

Page No 18.90:

Question 6:

$\int \frac{d x}{e^{x} + e^{- x}}$

Answer:

$\int \frac{d x}{e^{x} + e^{- x}} = \int \frac{d x}{e^{x} + \frac{1}{e^{x}}} = \int \frac{e^{x} d x}{e^{2 x} + 1} let e^{x} = t \Rightarrow e^{x} d x = d t Now, \int \frac{e^{x} d x}{e^{2 x} + 1} = \int \frac{d t}{1 + t^{2}} = \tan^{- 1} (t) + c = \tan^{- 1} (e^{x}) + c$

Page No 18.90:

Question 7:

$\int \frac{x}{x^{4} + 2 x^{2} + 3} d x$

Answer:

$\int \frac{x d x}{x^{4} + 2 x^{2} + 3} let x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Now, \int \frac{x d x}{x^{4} + 2 x^{2} + 3} = \frac{1}{2} \int \frac{d t}{t^{2} + 2 t + 3} = \frac{1}{2} \int \frac{d t}{t^{2} + 2 t + 1 + 2} = \frac{1}{2} \int \frac{d t}{{(t + 1)}^{2} + {(\sqrt{2})}^{2}} = \frac{1}{2} \times \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t + 1}{\sqrt{2}}) + C [∵ \int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + C] = \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x^{2} + 1}{\sqrt{2}}) + C$

Page No 18.90:

Question 8:

$\int \frac{3 x^{5}}{1 + x^{12}} d x$

Answer:

$\int \frac{3 x^{5}}{1 + x^{12}} d x let x^{6} = t \Rightarrow 6 x^{5} d x = d t \Rightarrow x^{5} d x = \frac{d t}{6} Now, \int \frac{3 x^{5}}{1 + x^{12}} d x = \frac{3}{6} \int \frac{d t}{1 + t^{2}} = \frac{1}{2} \tan^{- 1} (t) + C$

$= \frac{1}{2} \tan^{- 1} (x^{6}) + C$

Page No 18.90:

Question 9:

$\int \frac{x^{2} d x}{x^{6} - a^{6}} d x$

Answer:

$\int \frac{x^{2} d x}{x^{6} - a^{6}} let x^{3} = t \Rightarrow 3 x^{2} d x = d t \Rightarrow x^{2} d x = \frac{d t}{3} Now, \int \frac{x^{2} d x}{x^{6} - a^{6}} = \frac{1}{3} \int \frac{d t}{t^{2} - {(a^{3})}^{2}} = \frac{1}{3} \times \frac{1}{2 a^{3}} \log |\frac{t - a^{3}}{t + a^{3}}| + C = \frac{1}{6 a^{3}} \log |\frac{x^{3} - a^{3}}{x^{3} + a^{3}}| + C$

Page No 18.90:

Question 10:

$\int \frac{x^{2}}{x^{6} + a^{6}} d x$

Answer:

$\int \frac{x^{2}}{x^{6} + a^{6}} d x \Rightarrow \int \frac{x^{2} d x}{{(x^{3})}^{2} + {(a^{3})}^{2}} let x^{3} = t \Rightarrow 3 x^{2} d x = d t \Rightarrow x^{2} d x = \frac{d t}{3} Now, \int \frac{x^{2}}{x^{6} + a^{6}} d x = \frac{1}{3} \int \frac{d t}{t^{2} + {(a^{3})}^{2}} = \frac{1}{3 a^{3}} \tan^{- 1} (\frac{t}{a^{3}}) + C = \frac{1}{3 a^{3}} {\tan^{-}}^{1} (\frac{x^{3}}{a^{3}}) + C$

Page No 18.90:

Question 11:

$\int \frac{1}{x (x^{6} + 1)} d x$

Answer:

$\int \frac{d x}{x (x^{6} + 1)} = \int \frac{x^{5} d x}{x^{6} (x^{6} + 1)} let x^{6} = t \Rightarrow 6 x^{5} d x = d t \Rightarrow x^{5} d x = \frac{d t}{6} Now, \int \frac{d x}{x^{6} (x^{6} + 1)} = \frac{1}{6} \int \frac{d t}{t (t + 1)} = \frac{1}{6} \int \frac{d t}{t^{2} + t} = \frac{1}{6} \int \frac{d t}{t^{2} + t + \frac{1}{4} - \frac{1}{4}} = \frac{1}{6} \int \frac{d t}{{(t + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} = \frac{1}{6} \times \frac{1}{2 \times \frac{1}{2}} \log |\frac{t + \frac{1}{2} - \frac{1}{2}}{t + \frac{1}{2} + \frac{1}{2}}| + C = \frac{1}{6} \log |\frac{t}{t + 1}| + C = \frac{1}{6} \log |\frac{x^{6}}{x^{6} + 1}| + C$

Page No 18.90:

Question 12:

$\int \frac{x}{x^{4} - x^{2} + 1} d x$

Answer:

$\int \frac{x d x}{x^{4} - x^{2} + 1} Let x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Now, \int \frac{x d x}{x^{4} - x^{2} + 1} = \frac{1}{2} \int \frac{d t}{t^{2} - t + 1} = \frac{1}{2} \int \frac{d t}{t^{2} - t + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} = \frac{1}{2} \int \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}} = \frac{1}{2} \int \frac{d t}{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{1}{2} \times \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{t - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 t - 1}{\sqrt{3}}) + C = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x^{2} - 1}{\sqrt{3}}) + C$

Page No 18.90:

Question 13:

$\int \frac{x}{3 x^{4} - 18 x^{2} + 11} d x$

Answer:

$\int \frac{x d x}{3 x^{4} - 18 x^{2} + 11} let x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Now, \int \frac{x d x}{3 x^{4} - 18 x^{2} + 11} = \frac{1}{2} \int \frac{d t}{3 t^{2} - 18 t + 11} = \frac{1}{3 \times 2} \int \frac{d t}{t^{2} - 6 t + \frac{11}{3}} = \frac{1}{6} \int \frac{d t}{t^{2} - 6 t + 9 - 9 + \frac{11}{3}} = \frac{1}{6} \int \frac{d t}{{(t - 3)}^{2} - \frac{16}{3}} = \frac{1}{6} \int \frac{d t}{{(t - 3)}^{2} - {(\frac{4}{\sqrt{3}})}^{2}} = \frac{1}{6} \times \frac{1}{2 \times \frac{4}{\sqrt{3}}} \log |\frac{t - 3 - \frac{4}{\sqrt{3}}}{t - 3 + \frac{4}{\sqrt{3}}}| + C = \frac{\sqrt{3}}{48} \log |\frac{x^{2} - 3 - \frac{4}{\sqrt{3}}}{x^{2} - 3 + \frac{4}{\sqrt{3}}}| + C$

Page No 18.90:

Question 14:

$\int \frac{e^{x}}{(1 + e^{x}) (2 + e^{x})} d x$

Answer:

$\int \frac{e^{x} d x}{(1 + e^{x}) (2 + e^{x})} let e^{x} = t = e^{x} d x = d t = \int \frac{d t}{(1 + t) (2 + t)} = \int \frac{d t}{2 + t + 2 t + t^{2}} = \int \frac{d t}{t^{2} + 3 t + 2} = \int \frac{d t}{t^{2} + 3 t + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2} = \int \frac{d t}{{(t + \frac{3}{2})}^{2} - \frac{9}{4} + 2} = \int \frac{d t}{{(t + \frac{3}{2})}^{2} - \frac{1}{4}} = \int \frac{d t}{{(t + \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}}$

$= \frac{1}{2 \times \frac{1}{2}} \log |\frac{t + \frac{3}{2} - \frac{1}{2}}{t + \frac{3}{2} + \frac{1}{2}}| + C = \log |\frac{t + 1}{t + 2}| + C = \log |\frac{e^{x} + 1}{e^{x} + 2}| + C$

Page No 18.90:

Question 15:

$\int \frac{\tan^{2} x \sec^{2} x}{1 - \tan^{6} x} d x$

Answer:

$I = \int \frac{\tan^{2} x \sec^{2} x}{1 - \tan^{6} x} d x$

Let $\tan^{3} x = z$

$\Rightarrow 3 \tan^{2} x \sec^{2} x d x = d z$

$\Rightarrow \tan^{2} x \sec^{2} x d x = \frac{d z}{3}$

$∴ I = \frac{1}{3} \int \frac{d z}{1 - z^{2}}$

$\Rightarrow I = \frac{1}{3} \times \frac{1}{2} \log |\frac{1 + z}{1 - z}| + C (\int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} \log |\frac{a + x}{a - x}| + C)$

$\Rightarrow I = \frac{1}{6} \log |\frac{1 + \tan^{3} x}{1 - \tan^{3} x}| + C$

Page No 18.90:

Question 16:

Evaluate the following integrals:

$\int \frac{1}{\cos x + cosec x} d x$

Answer:

$Let I = \int \frac{1}{\cos x + cosec x} d x = \int \frac{\sin x}{1 + \cos x \sin x} d x = \int \frac{2 \sin x}{2 + 2 \cos x \sin x} d x = \int \frac{\sin x + \cos x + \sin x - \cos x}{2 + 2 \cos x \sin x} d x = \int \frac{\sin x + \cos x}{2 + 2 \cos x \sin x} d x + \int \frac{\sin x - \cos x}{2 + 2 \cos x \sin x} d x = \int \frac{\sin x + \cos x}{3 - \sin^{2} x - \cos^{2} x + 2 \cos x \sin x} d x + \int \frac{\sin x - \cos x}{1 + \sin^{2} x + \cos^{2} x + 2 \cos x \sin x} d x = \int \frac{\sin x + \cos x}{3 - {(\sin x - \cos x)}^{2}} d x + \int \frac{\sin x - \cos x}{1 + {(\sin x + \cos x)}^{2}} d x = I_{1} + I_{2} . . . (1) where, I_{1} = \int \frac{\sin x + \cos x}{3 - {(\sin x - \cos x)}^{2}} d x and I_{2} = \int \frac{\sin x - \cos x}{1 + {(\sin x + \cos x)}^{2}} d x Now, I_{1} = \int \frac{\sin x + \cos x}{3 - {(\sin x - \cos x)}^{2}} d x Let (\sin x - \cos x) = t On differentiating both sides, we get (\cos x + \sin x) d x = d t ∴ I_{1} = \int \frac{1}{3 - {(t)}^{2}} d t = \frac{1}{2 \sqrt{3}} \log |\frac{\sqrt{3} + t}{\sqrt{3} - t}| + c_{1} = \frac{1}{2 \sqrt{3}} \log |\frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - (\sin x - \cos x)}| + c_{1} . . . (2) Now, I_{2} = \int \frac{\sin x - \cos x}{1 + {(\sin x + \cos x)}^{2}} d x Let (\sin x + \cos x) = t On differentiating both sides, we get (\cos x - \sin x) d x = d t ∴ I_{2} = - \int \frac{1}{1 + {(t)}^{2}} d t = - \tan^{- 1} t + c_{2} = - \tan^{- 1} (\sin x + \cos x) + c_{2} . . . (3) On substituting (2) and (3) in (1), we get I = \frac{1}{2 \sqrt{3}} \log |\frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - \sin x + \cos x}| - \tan^{- 1} (\sin x + \cos x) + c Hence, \int \frac{1}{\cos x + cosec x} d x = \frac{1}{2 \sqrt{3}} \log |\frac{\sqrt{3} + \sin x - \cos x}{\sqrt{3} - \sin x + \cos x}| - \tan^{- 1} (\sin x + \cos x) + c$

Page No 18.93:

Question 1:

$\int \frac{1}{\sqrt{2 x - x^{2}}} d x$

Answer:

$\int \frac{d x}{\sqrt{2 x - x^{2}}} = \int \frac{d x}{\sqrt{2 x - x^{2} - 1 + 1}} = \int \frac{d x}{\sqrt{1 - (x^{2} - 2 x + 1)}} = \int \frac{d x}{\sqrt{1 - {(x - 1)}^{2}}} = \sin^{- 1} (x - 1) + C [∵ \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} (\frac{x}{a}) + C]$

Page No 18.93:

Question 2:

$\int \frac{1}{\sqrt{8 + 3 x - x^{2}}} d x$

Answer:

$\int \frac{d x}{\sqrt{8 + 3 x - x^{2}}} \Rightarrow \int \frac{d x}{\sqrt{8 - (x^{2} - 3 x)}} \Rightarrow \int \frac{d x}{\sqrt{8 - (x^{2} - 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2})}} \Rightarrow \int \frac{d x}{\sqrt{8 - {(x - \frac{3}{2})}^{2} + \frac{9}{4}}} \Rightarrow \int \frac{d x}{\sqrt{{(\frac{\sqrt{41}}{2})}^{2} - {(x - \frac{3}{2})}^{2}}} \Rightarrow \sin^{- 1} (\frac{x - \frac{3}{2}}{\frac{\sqrt{41}}{2}}) + C \Rightarrow \sin^{- 1} (\frac{2 x - 3}{\sqrt{41}}) + C$

Page No 18.93:

Question 3:

$\int \frac{1}{\sqrt{5 - 4 x - 2 x^{2}}} d x$

Answer:

$\int \frac{d x}{\sqrt{5 - 4 x - 2 x^{2}}} = \int \frac{d x}{\sqrt{2 [\frac{5}{2} - 2 x - x^{2}]}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{5}{2} - 2 x - x^{2}}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{5}{2} - (x^{2} + 2 x)}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{5}{2} - (x^{2} + 2 x + 1 - 1)}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{5}{2} - {(x + 1)}^{2} + 1}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{7}{2} - {(x + 1)}^{2}}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{{(\frac{\sqrt{7}}{\sqrt{2}})}^{2} - {(x + 1)}^{2}}} = \frac{1}{\sqrt{2}} \sin^{- 1} (\frac{(x + 1) \sqrt{2}}{\sqrt{7}}) + C = \frac{1}{\sqrt{2}} \sin^{- 1} (\sqrt{\frac{2}{7}} (x + 1)) + C$

Page No 18.93:

Question 4:

$\int \frac{1}{\sqrt{3 x^{2} + 5 x + 7}} d x$

Answer:

$\int \frac{d x}{\sqrt{3 x^{2} + 5 x + 7}} = \int \frac{d x}{\sqrt{3 (x^{2} + \frac{5}{3} x + \frac{7}{3})}} = \frac{1}{\sqrt{3}} \int \frac{d x}{\sqrt{x^{2} + \frac{5}{3} x + {(\frac{5}{6})}^{2} - {(\frac{5}{6})}^{2} + \frac{7}{3}}} = \frac{1}{\sqrt{3}} \int \frac{d x}{\sqrt{{(x + \frac{5}{6})}^{2} - \frac{25}{36} + \frac{7}{3}}} = \frac{1}{\sqrt{3}} \int \frac{d x}{\sqrt{{(x + \frac{5}{6})}^{2} + \frac{- 25 + 84}{36}}} = \frac{1}{\sqrt{3}} \int \frac{d x}{\sqrt{{(x + \frac{5}{6})}^{2} + \frac{59}{36}}} = \frac{1}{\sqrt{3}} \int \frac{d x}{\sqrt{{(x + \frac{5}{6})}^{2} + {(\frac{\sqrt{59}}{36})}^{2}}} = \frac{1}{\sqrt{3}} \log |x + \frac{5}{6} + \sqrt{{(x + \frac{5}{6})}^{2} + \frac{59}{36}}| + C = \frac{1}{\sqrt{3}} \log |x + \frac{5}{6} + \sqrt{x^{2} + \frac{5}{3} x + \frac{7}{3}}| + C$

Page No 18.93:

Question 5:

$\int \frac{1}{\sqrt{(x - α) (β - x)}} d x, (β > α)$

Answer:

$Let I = \int \frac{d x}{\sqrt{(x - α) (β - x)}} = \int \frac{d x}{\sqrt{β x - x^{2} - α β + α x}} = \int \frac{d x}{\sqrt{- x^{2} + (α + β) x - α β}} = \int \frac{d x}{\sqrt{- [x^{2} - (α + β) x + α β]}} = \int \frac{d x}{\sqrt{- [x^{2} - (α + β) x + {(\frac{α + β}{2})}^{2} - {(\frac{α + β}{2})}^{2} + α β]}} = \int \frac{d x}{\sqrt{- {\{x - (\frac{α + β}{2})\}}^{2} + {(\frac{α + β}{2})}^{2} - α β}} = \int \frac{d x}{\sqrt{- {[x - (\frac{α + β}{2})]}^{2} + \frac{{(α + β)}^{2} - 4 α β}{4}}} = \int \frac{d x}{\sqrt{- {[x - (\frac{α + β}{2})]}^{2} + {(\frac{α - β}{2})}^{2}}} = \int \frac{d x}{\sqrt{{(\frac{α - β}{2})}^{2} - {(x - (\frac{α + β}{2}))}^{2}}} = \sin^{- 1} [\frac{x - (\frac{α + β}{2})}{\frac{α - β}{2}}] + C = \sin^{- 1} (\frac{2 x - α - β}{α - β}) + C$

Page No 18.93:

Question 6:

$\int \frac{1}{\sqrt{7 - 3 x - 2 x^{2}}} d x$

Answer:

$\int \frac{d x}{\sqrt{7 - 3 x - 2 x^{2}}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{7}{2} - \frac{3}{2} x - x^{2}}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{7}{2} - (x^{2} - \frac{3}{2} x)}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{{(\frac{\sqrt{7}}{\sqrt{2}})}^{2} - (x^{2} + \frac{3}{2} x + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{2})}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{{(\frac{\sqrt{7}}{\sqrt{2}})}^{2} - {(x + \frac{3}{4})}^{2} + \frac{9}{16}}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{7}{2} + \frac{9}{16} - {(x + \frac{3}{4})}^{2}}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\frac{56 + 9}{16} - {(x + \frac{3}{4})}^{2}}} = \frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{{(\frac{\sqrt{65}}{4})}^{2} - {(x + \frac{3}{4})}^{2}}} = \frac{1}{\sqrt{2}} \sin^{- 1} [\frac{x + \frac{3}{4}}{\frac{\sqrt{65}}{4}}] + C = \frac{1}{\sqrt{2}} \sin^{- 1} [\frac{4 x + 3}{\sqrt{65}}] + C$

Page No 18.93:

Question 7:

$\int \frac{1}{\sqrt{16 - 6 x - x^{2}}} d x$

Answer:

$\int \frac{d x}{\sqrt{16 - 6 x - x^{2}}} = \int \frac{d x}{\sqrt{16 - (x^{2} + 6 x)}} = \int \frac{d x}{\sqrt{16 - (x^{2} + 6 x + 3^{2} - 3^{2})}} = \int \frac{d x}{\sqrt{16 + 9 - {(x + 3)}^{2}}} = \int \frac{d x}{\sqrt{5^{2} - {(x + 3)}^{2}}} = \sin^{- 1} (\frac{x + 3}{5}) + C$

Page No 18.93:

Question 8:

$\int \frac{1}{\sqrt{7 - 6 x - x^{2}}} d x$

Answer:

$\int \frac{d x}{\sqrt{7 - 6 x - x^{2}}} = \int \frac{d x}{\sqrt{7 - (x^{2} + 6 x)}} = \int \frac{d x}{\sqrt{7 - [x^{2} + 6 x + 3^{2} - 3^{2}]}} = \int \frac{d x}{\sqrt{7 + 9 - {(x + 3)}^{2}}} = \int \frac{d x}{\sqrt{4^{2} - {(x + 3)}^{2}}} = \sin^{- 1} (\frac{x + 3}{4}) + C$

Page No 18.93:

Question 9:

$\int \frac{1}{\sqrt{5 x^{2} - 2 x}} d x$

Answer:

$\int \frac{d x}{\sqrt{5 x^{2} - 2 x}} = \int \frac{d x}{\sqrt{5 (x^{2} - \frac{2}{5} x)}} = \frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{x^{2} - \frac{2}{5} x + {(\frac{1}{5})}^{2} - {(\frac{1}{5})}^{2}}} = \frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{{(x - \frac{1}{5})}^{2} - {(\frac{1}{5})}^{2}}} = \frac{1}{\sqrt{5}} \log |x - \frac{1}{5} + \sqrt{{(x - \frac{1}{5})}^{2} + {(\frac{1}{5})}^{2}}| + C = \frac{1}{\sqrt{5}} \log |\frac{5 x - 1}{5} + \frac{\sqrt{5 x^{2} - 2 x}}{\sqrt{5}}| + C$

Page No 18.98:

Question 1:

$\int \frac{x}{\sqrt{x^{4} + a^{4}}} d x$

Answer:

$\int \frac{x d x}{\sqrt{x^{4} + a^{4}}} = \int \frac{x d x}{\sqrt{{(x^{2})}^{2} + {(a^{2})}^{2}}} let x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Now, \int \frac{x d x}{\sqrt{{(x^{2})}^{2} + {(a^{2})}^{2}}} = \frac{1}{2} \int \frac{d t}{\sqrt{t^{2} + {(a^{2})}^{2}}} = \frac{1}{2} \log |t + \sqrt{t^{2} + a^{4}}| + C = \frac{1}{2} \log |x^{2} + \sqrt{x^{4} + a^{4}}| + C$

Page No 18.98:

Question 2:

$\int \frac{\sec^{2} x}{\sqrt{4 + \tan^{2} x}} d x$

Answer:

$\int \frac{\sec^{2} x d x}{\sqrt{4 + \tan^{2} x}} let \tan x = t \Rightarrow \sec^{2} x d x = d t Now, \int \frac{\sec^{2} x d x}{\sqrt{4 + \tan^{2} x}} = \int \frac{d t}{\sqrt{2^{2} + t^{2}}} = \log |t + \sqrt{4 + t^{2}}| + C = \log |\tan x + \sqrt{4 + \tan^{2} x}| + C$

Page No 18.98:

Question 3:

$\int \frac{e^{x}}{\sqrt{16 - e^{2 x}}} d x$

Answer:

$\int \frac{e^{x} d x}{\sqrt{16 - {(e^{x})}^{2}}} let e^{x} = t \Rightarrow e^{x} d x = d t Now, \int \frac{e^{x} d x}{\sqrt{16 - {(e^{x})}^{2}}} = \int \frac{d t}{\sqrt{16 - t^{2}}} = \int \frac{d t}{\sqrt{4^{2} - t^{2}}} = \sin^{- 1} (\frac{t}{4}) + C = \sin^{- 1} (\frac{e^{x}}{4}) + C$

Page No 18.99:

Question 4:

$\int \frac{\cos x}{\sqrt{4 + \sin^{2} x}} d x$

Answer:

$\int \frac{\cos x d x}{\sqrt{4 + \sin^{2} x}} let \sin x = t \Rightarrow \cos x d x = d t Now, \int \frac{\cos x d x}{\sqrt{4 + \sin^{2} x}} = \int \frac{d t}{\sqrt{2^{2} + t^{2}}} = \log |t + \sqrt{4 + t^{2}}| + C = \log |\sin x + \sqrt{4 + \sin^{2} x}| + C$

Page No 18.99:

Question 5:

$\int \frac{\sin x}{\sqrt{4 \cos^{2} x - 1}} d x$

Answer:

$\int \frac{\sin x d x}{\sqrt{4 \cos^{2} x - 1}} let \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t Now, \int \frac{\sin x d x}{\sqrt{4 \cos^{2} x - 1}} = \int \frac{- d t}{\sqrt{4 t^{2} - 1}} = \int \frac{- d t}{\sqrt{4 (t^{2} - \frac{1}{4})}} = - \frac{1}{2} \int \frac{d t}{\sqrt{t^{2} - {(\frac{1}{2})}^{2}}} = - \frac{1}{2} \log |t + \sqrt{t^{2} - \frac{1}{4}}| + C = - \frac{1}{2} \log |t + \frac{\sqrt{4 t^{2} - 1}}{2}| + C = - \frac{1}{2} \log |\frac{2 t + \sqrt{4 t^{2} - 1}}{2}| + C = - \frac{1}{2} [\log |2 t + \sqrt{4 t^{2} - 1}| - \log 2] + C = - \frac{1}{2} \log |2 t + \sqrt{4 t^{2} - 1}| + \frac{\log 2}{2} + C let C ´ = \frac{\log 2}{2} + C = - \frac{1}{2} \log |2 \cos t + \sqrt{4 \cos^{2} t - 1}| + C ´$

Page No 18.99:

Question 6:

$\int \frac{x}{\sqrt{4 - x^{4}}} d x$

Answer:

$\int \frac{x d x}{\sqrt{4 - x^{4}}} \Rightarrow \int \frac{x d x}{\sqrt{2^{2} - {(x^{2})}^{2}}} let x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} Now, \int \frac{x d x}{\sqrt{2^{2} - {(x^{2})}^{2}}} = \frac{1}{2} \int \frac{d t}{\sqrt{2^{2} - t^{2}}} = \frac{1}{2} \times \sin^{- 1} (\frac{1}{2}) + C = \frac{1}{2} \sin^{- 1} (\frac{x^{2}}{2}) + C$

Page No 18.99:

Question 7:

$\int \frac{1}{x \sqrt{4 - 9 {(\log x)}^{2}}} d x$

Answer:

$\int \frac{d x}{x \sqrt{4 - 9 {(\log x)}^{2}}} let \log x = t \Rightarrow \frac{1}{x} d x = d t Now, \int \frac{d x}{x \sqrt{4 - 9 {(\log x)}^{2}}} = \int \frac{d t}{\sqrt{4 - 9 t^{2}}} = \int \frac{d t}{\sqrt{2^{2} - {(3 t)}^{2}}} = \frac{1}{3} \sin^{- 1} (\frac{3 t}{2}) + C = \frac{1}{3} \sin^{- 1} (\frac{3 \log x}{2}) + C$

Page No 18.99:

Question 8:

$\int \frac{\sin 8 x}{\sqrt{9 + \sin^{4} 4 x}} d x$

Answer:

$\int \frac{\sin 8 x d x}{\sqrt{9 + \sin^{4} (4 x)}} \Rightarrow \int \frac{2 \sin (4 x) \cdot \cos (4 x)}{\sqrt{9 + {(\sin^{2} (4 x))}^{2}}} d x let \sin^{2} (4 x) = t \Rightarrow 2 \sin (4 x) \cdot \cos 4 x \times 4 d x = d t \Rightarrow 2 \sin (4 x) \cos (4 x) d x = \frac{d t}{4} Now, \int \frac{2 \sin (4 x) \cdot \cos (4 x)}{\sqrt{9 + {(\sin^{2} (4 x))}^{2}}} d x = \frac{1}{4} \int \frac{d t}{\sqrt{9 + t^{2}}} = \frac{1}{4} \int \frac{d t}{\sqrt{3^{2} + t^{2}}} = \frac{1}{4} \log |t + \sqrt{3^{2} + t^{2}}| + C = \frac{1}{4} \log |\sin^{2} 4 x + \sqrt{9 + \sin^{4} 4 x}| + C$

Page No 18.99:

Question 9:

$\int \frac{\cos 2 x}{\sqrt{\sin^{2} 2 x + 8}} d x$

Answer:

$\int \frac{\cos (2 x) \cdot d x}{\sqrt{\sin^{2} 2 x + 8}} let \sin (2 x) = t \Rightarrow \cos (2 x) \times 2 \cdot d x = d t \Rightarrow \cos (2 x) \cdot d x = \frac{d t}{2} Now, \int \frac{\cos (2 x) \cdot d x}{\sqrt{\sin^{2} 2 x + 8}} = \frac{1}{2} \int \frac{d t}{\sqrt{t^{2} + {(2 \sqrt{2})}^{2}}} = \frac{1}{2} \log |t + \sqrt{t^{2} + 8}| + C = \frac{1}{2} \log |\sin (2 x) + \sqrt{\sin^{2} (2 x) + 8}| + C$

Page No 18.99:

Question 10:

$\int \frac{\sin 2 x}{\sqrt{\sin^{4} x + 4 \sin^{2} x - 2}} d x$

Answer:

$\int \frac{\sin (2 x) d x}{\sqrt{\sin^{4} x + 4 \sin^{2} x - 2}} let \sin^{2} x = t \Rightarrow 2 \sin x \cos x d x = d t \Rightarrow \sin (2 x) d x = d t Now, \int \frac{\sin (2 x) d x}{\sqrt{\sin^{4} x + 4 \sin^{2} x - 2}} = \int \frac{d t}{\sqrt{t^{2} + 4 t - 2}} = \int \frac{d t}{\sqrt{t^{2} + 4 t + 4 - 4 - 2}} = \int \frac{d t}{\sqrt{{(t + 2)}^{2} - {(\sqrt{6})}^{2}}} = \log |t + 2 + \sqrt{{(t + 2)}^{2} - 6}| + C = \log |t + 2 + \sqrt{t^{2} + 4 t - 2}| + C = \log |\sin^{2} x + 2 + \sqrt{\sin^{4} x + 4 \sin^{2} x - 2}| + C$

Page No 18.99:

Question 11:

$\int \frac{\sin 2 x}{\sqrt{\cos^{4} x - \sin^{2} x + 2}} d x$

Answer:

$\int \frac{\sin (2 x) d x}{\sqrt{\cos^{4} x - \sin^{2} x + 2}} \Rightarrow \int \frac{2 \sin x \cos x d x}{\sqrt{\cos^{4} x - (1 - \cos^{2} x) + 2}} \Rightarrow \int \frac{2 \sin x \cos x}{\sqrt{\cos^{4} x + \cos^{2} x + 1}} Let \cos^{2} x = t \Rightarrow 2 \cos x \times - \sin x d x = d t \sin (2 x) d x = - d t Now, \int \frac{\sin (2 x) d x}{\sqrt{\cos^{4} x - \sin^{2} x + 2}} = \int \frac{- d t}{\sqrt{t^{2} + t + 1}} = \int \frac{- d t}{\sqrt{t^{2} + t + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1}} = - \int \frac{d t}{\sqrt{{(t + \frac{1}{2})}^{2} + \frac{3}{4}}} = - \int \frac{d t}{\sqrt{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}} = - \log |t + \frac{1}{2} + \sqrt{{(t + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}| + C = - \log |t + \frac{1}{2} + \sqrt{t^{2} + t + 1}| + C = - \log |\cos^{2} x + \frac{1}{2} + \sqrt{\cos^{4} x + \cos^{2} x + 1}| + C$

Page No 18.99:

Question 12:

$\int \frac{\cos x}{\sqrt{4 - \sin^{2} x}} d x$

Answer:

$\int \frac{\cos x d x}{\sqrt{4 - \sin^{2} x}} let \sin x = t \Rightarrow \cos x d x = d t Now, \int \frac{\cos x d x}{\sqrt{4 - \sin^{2} x}} = \int \frac{d t}{\sqrt{4 - t^{2}}} = \int \frac{d t}{\sqrt{2^{2} - t^{2}}} = \sin^{- 1} (\frac{t}{2}) + C = \sin^{- 1} (\frac{\sin x}{2}) + C$

Page No 18.99:

Question 13:

$\int \frac{1}{x^{2 / 3} \sqrt{x^{2 / 3} - 4}} d x$

Answer:

$\int \frac{d x}{x^{\frac{2}{3}} \sqrt{x^{\frac{2}{3}} - 2^{2}}} = \int \frac{d x}{x^{\frac{2}{3}} \sqrt{{(x^{\frac{1}{3}})}^{2} - 2^{2}}} Let x^{\frac{1}{3}} = t \Rightarrow \frac{1}{3} x^{\frac{- 2}{3}} d x = d t \Rightarrow \frac{1}{3 x^{\frac{2}{3}}} d x = d t \Rightarrow \frac{d x}{x^{\frac{2}{3}}} = 3 d t Now, \int \frac{d x}{x^{\frac{2}{3}} \sqrt{x^{\frac{2}{3}} - 2^{2}}} = 3 \int \frac{d t}{\sqrt{t^{2} - 2^{2}}} = 3 \log |t + \sqrt{t^{2} - 2^{2}}| + C = 3 \log |x^{\frac{1}{3}} + \sqrt{x^{\frac{2}{3}} - 4}| + C$

Page No 18.99:

Question 14:

$\int \frac{1}{\sqrt{(1 - x^{2}) \{9 + {(\sin^{- 1} x)}^{2}\}}} d x$

Answer:

$\int \frac{d x}{\sqrt{(1 - x^{2}) (9 + {(\sin^{- 1} x)}^{2})}} let \sin^{- 1} x = t \Rightarrow \frac{1}{\sqrt{1 - x^{2}}} d x = d t Now, \int \frac{d x}{\sqrt{(1 - x^{2}) (9 + {(\sin^{- 1} x)}^{2})}} = \int \frac{d t}{\sqrt{9 + t^{2}}} = \int \frac{d t}{\sqrt{3^{2} + t^{2}}} = \log |t + \sqrt{3^{2} + t^{2}}| + C = \log |\sin^{- 1} x + \sqrt{9 + {(\sin^{- 1} x)}^{2}}| + C$

Page No 18.99:

Question 15:

$\int \frac{\cos x}{\sqrt{\sin^{2} x - 2 \sin x - 3}} d x$

Answer:

$\int \frac{\cos x d x}{\sqrt{\sin^{2} x - 2 \sin x - 3}} let \sin x = t \Rightarrow \cos x d x = d t Now, \int \frac{\cos x d x}{\sqrt{\sin^{2} x - 2 \sin x - 3}} = \int \frac{d t}{\sqrt{t^{2} - 2 t - 3}} = \int \frac{d t}{\sqrt{t^{2} - 2 t + 1 - 1 - 3}} = \int \frac{d t}{\sqrt{{(t - 1)}^{2} - 2^{2}}} = \log |t - 1 + \sqrt{{(t - 1)}^{2} - 2^{2}}| + C = \log |t - 1 + \sqrt{t^{2} - 2 t - 3}| + C = \log |\sin x - 1 + \sqrt{\sin^{2} x - 2 \sin x - 3}| + C$

Page No 18.99:

Question 16:

$\int \sqrt{cosec x - 1} d x$

Answer:

$\int \sqrt{cosec x - 1} d x = \int \sqrt{\frac{1}{\sin x} - 1} d x = \int \frac{\sqrt{1 - \sin x}}{\sqrt{\sin x}} d x = \int \frac{\sqrt{(1 - \sin x) (1 + \sin x)}}{\sqrt{\sin x (1 + \sin x)}} d x = \int \frac{\sqrt{1 - \sin^{2} x}}{\sqrt{\sin^{2} x + \sin x}} d x = \int \frac{\cos x d x}{\sqrt{\sin^{2} x + \sin x}} Let \sin x = t \Rightarrow \cos x d x = d t Now, \int \frac{\cos x d x}{\sqrt{\sin^{2} x + \sin x}} = \int \frac{d t}{\sqrt{t^{2} + t}} \int \frac{d t}{\sqrt{t^{2} + t}} = \int \frac{d t}{\sqrt{t^{2} + t + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} = \int \frac{d t}{\sqrt{{(t + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} = \log |(t + \frac{1}{2}) + \sqrt{{(t + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}| + C = \log |t + \frac{1}{2} + \sqrt{t^{2} + t}| + C = \log |\sin x + \frac{1}{2} + \sqrt{\sin^{2} x + \sin x}| + C$

Page No 18.99:

Question 17:

$\int \frac{\sin x - \cos x}{\sqrt{\sin 2 x}} d x$

Answer:

$\int (\frac{\sin x - \cos x}{\sqrt{\sin 2 x}}) d x = \int (\frac{\sin x - \cos x}{\sqrt{1 + \sin 2 x - 1}}) d x = \int \frac{(\sin x - \cos x)}{\sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x \cos x - 1}} d x = \int \frac{(\sin x - \cos x)}{\sqrt{{(\sin x + \cos x)}^{2} - 1}} d x let \sin x + \cos x = t \Rightarrow (\cos x - \sin x) d x = d t \Rightarrow (\sin x - \cos x) d x = - d t Now, \int \frac{(\sin x - \cos x)}{\sqrt{{(\sin x + \cos x)}^{2} - 1}} d x = - \int \frac{d t}{\sqrt{t^{2} - 1^{2}}} = - \log |t + \sqrt{t^{2} - 1}| + C = - \log |\sin x + \cos x + \sqrt{{(\sin x + \cos x)}^{2} - 1}| + C = - \log |\sin x + \cos x + \sqrt{\sin^{2} x + \cos^{2} x + 2 \sin x . \cos x - 1}| + C = - \log |\sin x + \cos x + \sqrt{\sin 2 x}| + C$

Page No 18.99:

Question 18:

$\int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2 x}} d x$

Answer:

$Let I = \int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2 x}} d x = \int \frac{\cos x - \sin x}{\sqrt{9 - 1 - \sin 2 x}} d x = \int \frac{\cos x - \sin x}{\sqrt{9 - \sin^{2} x - \cos^{2} x - 2 \sin x \cos x}} d x = \int \frac{\cos x - \sin x}{\sqrt{9 - {(\sin x + \cos x)}^{2}}} d x Let (\sin x + \cos x) = t On differentiating both sides, we get (\cos x - \sin x) d x = d t ∴ I = \int \frac{1}{\sqrt{{(3)}^{2} - {(t)}^{2}}} d t = \sin^{- 1} (\frac{t}{3}) + c = \sin^{- 1} (\frac{\sin x + \cos x}{3}) + c Hence, \int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2 x}} d x = \sin^{- 1} (\frac{\sin x + \cos x}{3}) + c$

Page No 19.79:

Question 2:

Evaluate the following integrals:

$\int \frac{x^{7}}{{(a^{2} - x^{2})}^{5}} d x$

Answer:

$Let I = \int \frac{x^{7}}{{(a^{2} - x^{2})}^{5}} d x Let x = a \sin θ On differentiating both sides, we get d x = a \cos θ d θ ∴ I = \int \frac{a^{8} \sin^{7} θ \cos θ}{{(a^{2} - a^{2} \sin^{2} θ)}^{5}} d θ = \int \frac{a^{8} \sin^{7} θ \cos θ}{a^{10} {(1 - \sin^{2} θ)}^{5}} d θ = \int \frac{\sin^{7} θ}{a^{2} \cos^{9} θ} d θ = \frac{1}{a^{2}} \int \tan^{7} θ \sec^{2} θ d θ Let \tan θ = t On differentiating both sides, we get \sec^{2} θ d θ = d t ∴ I = \frac{1}{a^{2}} \int t^{7} d t = \frac{1}{a^{2}} \frac{t^{8}}{8} + c = \frac{1}{8 a^{2}} (\tan^{8} θ) + c = \frac{1}{8 a^{2}} {(\tan (\sin^{- 1} \frac{x}{a}))}^{8} + c = \frac{1}{8 a^{2}} {(\tan (\tan^{- 1} \frac{x}{\sqrt{a^{2} - x^{2}}}))}^{8} + c = \frac{1}{8 a^{2}} {(\frac{x}{\sqrt{a^{2} - x^{2}}})}^{8} + c = \frac{1}{8 a^{2}} \frac{x^{8}}{{(a^{2} - x^{2})}^{4}} + c Hence, \int \frac{x^{7}}{{(a^{2} - x^{2})}^{5}} d x = \frac{1}{8 a^{2}} \frac{x^{8}}{{(a^{2} - x^{2})}^{4}} + c$

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