Rd Sharma XII Vol 2 2020 for Class 12 Science Maths Chapter 1 - Definite Integrals

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Page No 19.110:

Question 1:

$\int_{0}^{3} (x + 4) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . + f (a + (n - 1) h)], where, h = \frac{b - a}{n}$

$Here, a = 0, b = 3, f (x) = x + 4, h = \frac{3 - 0}{n} = \frac{3}{n} Therefore, I = \int_{0}^{3} (x + 4) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [(0 + 4) + (h + 4) + . . . . . . . + ((n - 1) h + 4)] = \lim_{h \to 0} h [4 n + h (1 + 2 + . . . . . . . + (n - 1))] = \lim_{h \to 0} h [4 n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [4 n + \frac{3}{n} \times \frac{n (n - 1)}{2}] = \lim_{n \to \infty} [12 + \frac{9}{2} (1 - \frac{1}{n})] = 12 + \frac{9}{2} = \frac{33}{2}$

Page No 19.110:

Question 2:

$\int_{0}^{2} (x + 3) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x + 3, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x + 3) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [(0 + 3) + (0 + h + 3) + . . . . . . . . . . . . . . . + (0 + (n - 1) h + 3)] = \lim_{h \to 0} h [3 n + h \{1 + 2 + 3 . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [3 n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [3 n + n - 1] = \lim_{n \to \infty} 2 (4 - \frac{1}{n}) = 8$

Page No 19.110:

Question 3:

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

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$
$Here a = 1, b = 3, f (x) = 3 x - 2, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, I = \int_{1}^{3} (3 x - 2) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [(3 - 2) + (3 + 3 h - 2) + (3 + 6 h - 2) . . . . . . . . . . . . . . . + (3 (n - 1) h + 3 - 2)] = \lim_{h \to 0} h [n + 3 h (1 + 2 + 3 . . . . . . . . . + (n - 1))] = \lim_{h \to 0} h [n + 3 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [n + 3 n - 3] = \lim_{n \to \infty} 2 (4 - \frac{3}{n}) = 8$

Page No 19.110:

Question 4:

$\int_{- 1}^{1} (x + 3) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = - 1, b = 1, f (x) = x + 3, h = \frac{1 + 1}{n} = \frac{2}{n} Therefore, I = \int_{- 1}^{1} (x + 3) d x = \lim_{h \to 0} h [f (- 1) + f (- 1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{- 1 + (n - 1) h\}] = \lim_{h \to 0} h [(- 1 + 3) + (- 1 + h + 3) + . . . . . . . . . . . . . . . + \{- 1 + (n - 1) h + 3\}] = \lim_{h \to 0} h [2 n + h \{1 + 2 + 3 . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [2 n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [2 n + n - 1] = \lim_{n \to \infty} 2 (3 - \frac{1}{n}) = 6$

Page No 19.110:

Question 5:

$\int_{0}^{5} (x + 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here, a = 0, b = 5, f (x) = x + 1, h = \frac{5 - 0}{n} = \frac{5}{n} Therefore, I = \int_{0}^{5} (x + 1) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 1) + (h + 1) + . . . . . . . . . . . . . . . + \{(n - 1) h + 1\}] = \lim_{h \to 0} h [n + h \{1 + 2 + 3 . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{5}{n} [n + \frac{5 n - 5}{2}] = \lim_{n \to \infty} 5 (\frac{7}{2} - \frac{5}{n}) = \frac{35}{2}$

Page No 19.110:

Question 6:

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

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here, a = 1, b = 3, f (x) = 2 x + 3, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, I = \int_{1}^{3} (2 x + 3) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(2 + 3) + (2 + 2 h + 3) + . . . . . . . . . . . . . . . + \{2 + 2 (n - 1) h + 3\}] = \lim_{h \to 0} h [5 n + 2 h \{1 + 2 + 3 . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [5 n + 2 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [5 n + 2 n - 2] = \lim_{n \to \infty} 2 (7 - \frac{2}{n}) = 14$

Page No 19.110:

Question 7:

$\int_{3}^{5} (2 - x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 3, b = 5, f (x) = 2 - x, h = \frac{5 - 3}{n} = \frac{2}{n} Therefore, I = \int_{3}^{5} (2 - x) d x = \lim_{h \to 0} h [f (2) + f (2 + h) + . . . + f (2 + (n - 1) h)] = \lim_{h \to 0} h [(2 - 2) + (2 - h - 2) + . . . + (2 - (n - 1) h - 2)] = \lim_{h \to 0} h [- h (1 + 2 + 3 + . . . + (n - 1))] = \lim_{h \to 0} h [- 2 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [- 2 n + 2] = \lim_{n \to \infty} 2 (- 2 + \frac{2}{n}) = - 4$

Page No 19.110:

Question 8:

$\int_{0}^{2} (x^{2} + 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + 1, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + 1) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [(0 + 1) + (h^{2} + 1) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + 1\}] = \lim_{h \to 0} h [n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{2}{n} [n + \frac{2 (n - 1) (2 n - 1)}{3 n}] = \lim_{n \to \infty} 2 \{1 + \frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = 2 + \frac{8}{3} = \frac{14}{3}$

Page No 19.110:

Question 9:

$\int_{1}^{2} x^{2} d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 1, b = 2, f (x) = x^{2}, h = \frac{2 - 1}{n} = \frac{1}{n} Therefore, I = \int_{1}^{2} (x^{2}) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [(1) + {(h + 1)}^{2} + . . . . . . . . . . . . . . . + {((n - 1) h + 1)}^{2}] = \lim_{h \to 0} h [n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 2 h \{1 + 2 + 3 + . . . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 2 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{1}{n} [n + \frac{(n - 1) (2 n - 1)}{6 n} + n - 1] = \lim_{n \to \infty} \{2 + \frac{1}{6} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{1}{n}\} = 2 + \frac{1}{3} = \frac{7}{3}$

Page No 19.110:

Question 10:

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

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 2, b = 3, f (x) = 2 x^{2} + 1, h = \frac{3 - 2}{n} = \frac{1}{n} Therefore, I = \int_{2}^{3} (2 x^{2} + 1) d x = \lim_{h \to 0} h [f (2) + f (2 + h) + . . . . . . . . . . . . . . . . . . . . + f \{2 + (n - 1) h\}] = \lim_{h \to 0} h [2 (2 . 2^{2}) + 1 + \{2 {(2 + h)}^{2} + 1\} + . . . . . . . . . . . . . . . + \{2 {((2 + n - 1) h)}^{2} + 1\}] = \lim_{h \to 0} h [n + 2 \{2^{2} + {(2 + h)}^{2} + . . . . . . . . . . . . . {((2 + n - 1) h)}^{2}\}] = \lim_{h \to 0} h [n + 8 n + 2 h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 8 h \{1 + 2 + . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [9 n + h^{2} \frac{2 n (n - 1) (2 n - 1)}{6} + 8 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{1}{n} [9 n + \frac{(n - 1) (2 n - 1)}{3 n} + 4 n - 4] = \lim_{n \to \infty} \{13 + \frac{1}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{4}{n}\} = 13 + \frac{2}{3} = \frac{41}{3}$

Page No 19.110:

Question 11:

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

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 1, b = 2, f (x) = x^{2} - 1, h = \frac{2 - 1}{n} = \frac{1}{n} Therefore, I = \int_{1}^{2} (x^{2} - 1) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(1 - 1) + (h^{2} - 1) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} - 1\}] = \lim_{h \to 0} h [n - 1 + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [n - 1 + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{1}{n} [n - 1 + \frac{(n - 1) (2 n - 1)}{6 n}] = \lim_{n \to \infty} \{1 - \frac{1}{n} + \frac{1}{6} (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = 1 + \frac{1}{3} = \frac{4}{3}$

Page No 19.110:

Question 12:

$\int_{0}^{2} (x^{2} + 4) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + 4, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + 4) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 4) + (h^{2} + 4) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + 4\}] = \lim_{h \to 0} h [4 n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [4 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{2}{n} [4 n + \frac{2 (n - 1) (2 n - 1)}{3 n}] = \lim_{n \to \infty} 2 \{4 + \frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = 8 + \frac{8}{3} = \frac{32}{3}$

Page No 19.111:

Question 13:

$\int_{1}^{4} (x^{2} - x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 1, b = 4, f (x) = x^{2} - x, h = \frac{4 - 1}{n} = \frac{3}{n} Therefore, I = \int_{1}^{4} (x^{2} - x) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(1 - 1) + \{{(1 + h)}^{2} - (1 + h)\} + . . . . . . . . . . . . . . . + \{{(1 + (n - 1) h)}^{2} - (1 + (n - 1) h)\}] = \lim_{h \to 0} h [h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 1 + 2 h \{1 + 2 + . . . . . . + (n - 1)\} - n - h \{1 + 2 + . . . . . + (n - 1)\}] = \lim_{h \to 0} h [h^{2} \frac{n (n - 1) (2 n - 1)}{6} + h \frac{(n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [\frac{9 (n - 1) (2 n - 1)}{6 n} + \frac{3 (n - 1)}{2}] = \lim_{n \to \infty} 3 [\frac{3}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + \frac{3}{2} (1 - \frac{1}{n})] = 9 + \frac{9}{2} = \frac{27}{2}$

Page No 19.111:

Question 14:

$\int_{0}^{1} (3 x^{2} + 5 x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 1, f (x) = 3 x^{2} + 5 x, h = \frac{1 - 0}{n} = \frac{1}{n} Therefore, I = \int_{0}^{1} (3 x^{2} + 5 x) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 0) + (3 h^{2} + 5 h) + . . . . . . . . . . . . . . . + \{3 {(n - 1)}^{2} h^{2} + 5 (n - 1) h\}] = \lim_{h \to 0} h [5 h (1 + 2 + . . . . . . . . . + n) + 3 h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [5 h \frac{n (n - 1)}{2} + h^{2} \frac{3 n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{1}{n} [\frac{5 (n - 1)}{2} + \frac{(n - 1) (2 n - 1)}{2 n}] = \lim_{n \to \infty} [\frac{5}{2} (1 - \frac{1}{n}) + \frac{1}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n})] = \frac{5}{2} + 1 = \frac{7}{2}$

Page No 19.111:

Question 15:

$\int_{0}^{2} e^{x} d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = e^{x}, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} e^{x} d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [e^{0} + e^{h} + e^{2 h} + . . . . . . . + e^{(n - 1) h}] = \lim_{h \to 0} h [\frac{{(e^{h})}^{n} - 1}{e^{h} - 1}] = \lim_{h \to 0} [\frac{e^{2} - 1}{\frac{e^{h} - 1}{h}}] = \frac{e^{2} - 1}{1} = e^{2} - 1$

Page No 19.111:

Question 16:

$\int_{a}^{b} e^{x} d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = a, b = b, f (x) = e^{x}, h = \frac{b - a}{n} Therefore, I = \int_{a}^{b} e^{x} d x = \lim_{h \to 0} h [f (a) + f (a + h) + . . . . . . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] = \lim_{h \to 0} h [e^{a} + e^{a + h} + . . . . . . . . . . . . + e^{\{a + (n - 1) h\}}] = \lim_{h \to 0} h [e^{a} \{\frac{{(e^{h})}^{n} - 1}{e^{h} - 1}\}] = \lim_{h \to 0} h [e^{a} \frac{e^{b - a} - 1}{e^{h} - 1}] = \lim_{h \to 0} [\frac{e^{b} - e^{a}}{\frac{e^{h} - 1}{h}}] = \frac{e^{b} - e^{a}}{1} = e^{b} - e^{a}$

Page No 19.111:

Question 17:

$\int_{a}^{b} \cos x d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$
$Here a = a, b = b, f (x) = \cos x, h = \frac{b - a}{n} Therefore, I = \int_{a}^{b} \cos x d x = \lim_{h \to 0} h [f (a) + f (a + h) + . . . + f \{a + (n - 1) h\}] = \lim_{h \to 0} h [\cos (a) + \cos (a + h) + . . . + \cos \{a + (n - 1) h\}] = \lim_{h \to 0} h [\frac{\cos \{a + (n - 1) \frac{h}{2}\} \sin \frac{n h}{2}}{\sin \frac{h}{2}}] = \lim_{h \to 0} [\frac{\frac{h}{2}}{\sin \frac{h}{2}} 2 \cos (a + \frac{b - a}{2} - \frac{h}{2}) \sin (\frac{b - a}{2})] (Using n h = b - a) = \lim_{h \to 0} \frac{\frac{h}{2}}{\sin \frac{h}{2}} \times \lim_{h \to 0} 2 \cos (\frac{a + b}{2} - \frac{h}{2}) \sin (\frac{b - a}{2}) = 2 \cos (\frac{a + b}{2}) \sin (\frac{b - a}{2}) = \sin b - \sin a [Since, 2 \cos A \sin B = \sin (A + B) - \sin (A - B)]$

Page No 19.111:

Question 18:

$\int_{0}^{π / 2} \sin x d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$
$Here a = 0, b = \frac{π}{2}, f (x) = \sin x, h = \frac{\frac{π}{2} - 0}{n} = \frac{π}{2 n} Therefore, I = \int_{0}^{\frac{π}{2}} \sin x d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [\sin 0 + \sin h + \sin 2 h + . . . + \sin (n - 1) h] = \lim_{h \to 0} h [\frac{\sin ((n - 1) \frac{h}{2}) \sin \frac{n h}{2}}{\sin \frac{h}{2}}] = \lim_{h \to 0} [\frac{\frac{h}{2}}{\sin \frac{h}{2}} \times 2 \sin (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4}] (Using n h = \frac{π}{2}) = \lim_{h \to 0} \frac{\frac{h}{2}}{\sin \frac{h}{2}} \times \lim_{h \to 0} 2 \sin (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4} = 2 \sin \frac{π}{4} \sin \frac{π}{4} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1$

Page No 19.111:

Question 19:

$\int_{0}^{π / 2} \cos x d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$
$Here a = 0, b = \frac{π}{2}, f (x) = \cos x, h = \frac{\frac{π}{2} - 0}{n} = \frac{π}{2 n} Therefore, I = \int_{0}^{\frac{π}{2}} \cos x d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [\cos 0 + \cos h + \cos 2 h + . . . + \cos (n - 1) h] = \lim_{h \to 0} h [\frac{\cos ((n - 1) \frac{h}{2}) \sin \frac{n h}{2}}{\sin \frac{h}{2}}] = \lim_{h \to 0} h [\frac{\cos (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4}}{\sin \frac{h}{2}}] (Using, n h = \frac{π}{2}) = \lim_{h \to 0} [\frac{\frac{h}{2}}{\sin \frac{h}{2}} \times 2 \cos (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4}] = \lim_{h \to 0} \frac{\frac{h}{2}}{\sin \frac{h}{2}} \times \lim_{h \to 0} 2 \cos (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4} = 2 \cos \frac{π}{4} \sin \frac{π}{4} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1$

Page No 19.111:

Question 20:

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

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 1, b = 4, f (x) = 3 x^{2} + 2 x, h = \frac{4 - 1}{n} = \frac{3}{n} Therefore, I = \int_{1}^{4} (3 x^{2} + 2 x) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [(3 . 1^{2} + 2 \times 1) + (3 {(1 + h)}^{2} + 2 (1 + h)) + . . . . . . . . . . . . . . . + \{3 {(1 + (n - 1) h)}^{2} + 2 (1 + (n - 1) h)\}] = \lim_{h \to 0} h [3 \{1^{2} + {(1 + h)}^{2} + {(1 + 2 h)}^{2} + . . . . . . . . . . . + {(1 + (n - 1) h)}^{2}\} + 2 \{1 + (1 + h) + . . . . . . . . . . + (1 + (n - 1) h)\}] = \lim_{h \to 0} h [3 n + 3 h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 6 h \{1 + 2 + . . . . . . . . . (n - 1) h\} + 2 n + 2 h \{1 + 2 + . . . . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [5 n + 3 h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 8 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [5 n + \frac{9 (n - 1) (2 n - 1)}{2 n} + 12 n - 12] = \lim_{n \to \infty} 3 [17 - \frac{12}{n} + \frac{9}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n})] = 51 + 27 = 78$

Page No 19.111:

Question 21:

$\int_{0}^{2} (3 x^{2} - 2) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = 3 x^{2} - 2, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (3 x^{2} - 2) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 - 2) + (3 h^{2} - 2) + . . . . . . . . . . . . . . . + \{3 {(n - 1)}^{2} h^{2} - 2\}] = \lim_{h \to 0} h [- 2 n + 3 h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [- 2 n + 3 h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{2}{n} [- 2 n + \frac{2 (n - 1) (2 n - 1)}{n}] = \lim_{n \to \infty} 2 \{- 2 + 2 (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = - 4 + 8 = 4$

Page No 19.111:

Question 22:

$\int_{0}^{2} (x^{2} + 2) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + 2, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + 2) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 2) + (h^{2} + 2) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + 2\}] = \lim_{h \to 0} h [2 n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [2 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{2}{n} [2 n + \frac{2 (n - 1) (2 n - 1)}{3 n}] = \lim_{n \to \infty} 2 \{2 + \frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = 4 + \frac{8}{3} = \frac{20}{3}$

Page No 19.111:

Question 23:

$\int_{0}^{4} (x + e^{2 x}) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here, a = 0, b = 4, f (x) = x + e^{2 x}, h = \frac{4 - 0}{n} = \frac{4}{n} Therefore, I = \int_{0}^{4} (x + e^{2 x}) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + e^{0}) + (h + e^{2 h}) + . . . . . . . . . . . . . . . + \{(n - 1) h + e^{2 (n - 1) h}\}] = \lim_{h \to 0} h [h \{1 + 2 + . . . . . + (n - 1) h\} + e^{0} + e^{2 h} + e^{4 h} + . . . . . . . . . + e^{2 (n - 1) h}] = \lim_{h \to 0} h [h \frac{n (n - 1)}{2} + \frac{{(e^{2 h})}^{n} - 1}{e^{2 h} - 1}] = \lim_{n \to \infty} \frac{16}{n^{2}} \times \frac{n (n - 1)}{2} + \lim_{h \to 0} \frac{e^{8} - 1}{\frac{e^{2 h} - 1}{h}} = \lim_{n \to \infty} 8 (1 - \frac{1}{n}) + \lim_{h \to 0} \frac{e^{8} - 1}{\frac{2 (e^{2 h} - 1)}{2 h}} = 8 + \frac{e^{8} - 1}{2} = \frac{15 + e^{8}}{2}$

Page No 19.111:

Question 24:

$\int_{0}^{2} (x^{2} + x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + x, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + x) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 0) + (h^{2} + h) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + h\}] = \lim_{h \to 0} h [h^{2} (1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}) + h \{1 + 2 + 3 . . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [h^{2} \frac{n (n - 1) (2 n - 1)}{6} + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [\frac{2 (n - 1) (2 n - 1)}{3 n} + n - 1] = \lim_{n \to \infty} 2 [\frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + 1 - \frac{1}{n}] = \frac{8}{3} + 2 = \frac{14}{3}$

Page No 19.111:

Question 25:

$\int_{0}^{2} (x^{2} + 2 x + 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + 2 x + 1, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + 2 x + 1) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 0 + 1) + (h^{2} + 2 h + 1) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + 2 (n - 1) h + 1\}] = \lim_{h \to 0} h [n + h^{2} (1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}) + 2 h \{1 + 2 + . . . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 2 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [n + \frac{2 (n - 1) (2 n - 1)}{3 n} + 2 n - 2] = \lim_{n \to \infty} 2 \{3 + \frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{2}{n}\} = 6 + \frac{8}{3} = \frac{26}{3}$

Page No 19.111:

Question 26:

$\int_{0}^{3} (2 x^{2} + 3 x + 5) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 3, f (x) = 2 x^{2} + 3 x + 5, h = \frac{3 - 0}{n} = \frac{3}{n} Therefore, I = \int_{0}^{3} (2 x^{2} + 3 x + 5) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 0 + 5) + (2 h^{2} + 3 h + 5) + . . . . . . . . . . . . . . . + \{2 {(n - 1)}^{2} h^{2} + 3 (n - 1) h + 5\}] = \lim_{h \to 0} h [5 n + 2 h^{2} (1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}) + 3 h \{1 + 2 + . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [5 n + 2 h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 3 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [5 n + \frac{3 (n - 1) (2 n - 1)}{n} + \frac{9 (n - 1)}{2}] = \lim_{n \to \infty} 3 [5 + 3 (1 - \frac{1}{n}) (2 - \frac{1}{n}) + \frac{9}{2} (1 - \frac{1}{n})] = 15 + 18 + \frac{27}{2} = \frac{93}{3}$

Page No 19.112:

Question 27:

$\int_{a}^{b} x d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = a, b = b, f (x) = x, h = \frac{b - a}{n} Therefore, I = \int_{a}^{b} x d x = \lim_{h \to 0} h [f (a) + f (a + h) + . . . . . . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] = \lim_{h \to 0} h [a + (a + h) + (a + 2 h) + . . . . . . . . . . + \{a + (n - 1) h\}] = \lim_{h \to 0} h [n a + h \{1 + 2 + 3 + . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [n a + h \frac{n (n - 1)}{2}] = \lim_{h \to 0} \frac{b - a}{n} [n a + \frac{[b - a] (n - 1)}{2}] = \lim_{h \to 0} [(b - a) a + \frac{(b - a) (b - a - h)}{2}] = (b - a) a + \frac{{(b - a)}^{2}}{2} = \frac{2 a b - 2 a^{2} + b^{2} + a^{2} - 2 a b}{2} = \frac{b^{2} - a^{2}}{2}$

Page No 19.112:

Question 28:

$\int_{0}^{5} (x + 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 5, f (x) = x + 1, h = \frac{5 - 0}{n} = \frac{5}{n} Therefore, I = \int_{0}^{5} (x + 1) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 1) + (h + 1) + . . . . . . . . . . . . . . . + \{(n - 1) h + 1\}] = \lim_{h \to 0} h [n + h \{1 + 2 + 3 + . . . . . . . . . . . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{5}{n} [n + \frac{5 (n - 1)}{2}] = \lim_{n \to \infty} 5 [1 + \frac{5}{2} (1 - \frac{1}{n})] = 5 + \frac{25}{2} = \frac{35}{2}$

Page No 19.112:

Question 29:

$\int_{2}^{3} x^{2} d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 2, b = 3, f (x) = x^{2}, h = \frac{3 - 2}{n} = \frac{1}{n} Therefore, I = \int_{2}^{3} x^{2} d x = \lim_{h \to 0} h [f (2) + f (2 + h) + . . . . . . . . . . . . . . . . . . . . + f \{2 + (n - 1) h\}] = \lim_{h \to 0} h [2^{2} + {(2 + h)}^{2} + . . . . . . . . . . . + {\{2 + (n - 1) h\}}^{2}] = \lim_{h \to 0} h [4 n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 4 h \{1 + 2 + . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [4 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 4 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{1}{n} [4 n + \frac{(n - 1) (2 n - 1)}{6 n} + 2 n - 2] = \lim_{n \to \infty} [6 + \frac{1}{6} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{2}{n}] = 6 + \frac{1}{3} = \frac{19}{3}$

Page No 19.112:

Question 30:

$\int_{1}^{4} (x^{2} - x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 1, b = 4, f (x) = x^{2} - x, h = \frac{4 - 1}{n} = \frac{3}{n} Therefore, I = \int_{1}^{4} (x^{2} - x) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(1 - 1) + {(1 + h)}^{2} - (1 + h) + . . . . . . . . . . . . . . . + {\{(n - 1) h + 1\}}^{2} - \{(n - 1) h + 1\}] = \lim_{h \to 0} h [h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} - h \{1 + 2 + . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [h^{2} \frac{n (n - 1) (2 n - 1)}{6} - h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [\frac{3 (n - 1) (2 n - 1)}{2 n} + \frac{3 (n - 1)}{2}] = \lim_{n \to \infty} 3 [\frac{3}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + \frac{3}{2} (1 - \frac{1}{n})] = 9 + \frac{9}{3} = \frac{38}{3}$

Page No 19.112:

Question 31:

$\int_{0}^{2} (x^{2} - x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} - x, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} - x) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 - 0) + (h^{2} - h) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} - (n - 1) h\}] = \lim_{h \to 0} h [h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} - h \{1 + 2 + . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [h^{2} \frac{n (n - 1) (2 n - 1)}{6} - h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [\frac{2 (n - 1) (2 n - 1)}{3 n} - n + 1] = \lim_{n \to \infty} 2 [\frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - 1 + \frac{1}{n}] = \frac{8}{3} - 2 = \frac{2}{3}$

Page No 19.112:

Question 32:

$\int_{1}^{3} (2 x^{2} + 5 x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here, a = 1, b = 3, f (x) = 2 x^{2} + 5 x, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, I = \int_{1}^{3} (2 x^{2} + 5 x) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(2 + 5) + \{2 {(1 + h)}^{2} + 5 (1 + h)\} + . . . . . . . . . . . . . . . + \{2 {(1 + (n - 1) h)}^{2} + 5 (1 + (n - 1) h)\}] = \lim_{h \to 0} h [2 \{1^{2} + {(1 + h)}^{2} + . . . . . . . . . . . . + {\{1 + (n - 1) h\}}^{2}\} + 5 \{1 + (1 + h) + (1 + 2 h + . . . . . . . . + (1 + (n - 1) h))\}] = \lim_{h \to 0} h [2 n + 2 h^{2} (1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}) + 4 h \{1 + 2 + . . . . . . + (n - 1)\} + 5 n + 5 h \{1 + 2 + . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [7 n + 2 h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 9 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [7 n + \frac{4 (n - 1) (2 n - 1)}{3 n} + 9 n - 9] = \lim_{n \to \infty} 2 [16 + \frac{4}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{9}{n}] = 32 + \frac{16}{3} = \frac{112}{3}$

Page No 19.112:

Question 33:

Evaluate the following integrals as limit of sums:

$\int_{1}^{3} (3 x^{2} + 1) d x$ [CBSE 2014]

Answer:

We have,

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} \{f (a) + f (a + h) + f (a + 2 h) + . . . + f [(a + (n - 1) h)]\}$

Here, a = 1, b = 3, f(x) = 3x² + 1 and $h = \frac{3 - 1}{n} = \frac{2}{n} \Rightarrow n h = 2$

$∴ \int_{1}^{3} (3 x^{2} + 1) d x = \lim_{h \to 0} h \{f (1) + f (1 + h) + f (1 + 2 h) + . . . + f [1 + (n - 1) h]\} = \lim_{h \to 0} h \{[3 \times 1^{2} + 1] + [3 \times {(1 + h)}^{2} + 1] + [3 \times {(1 + 2 h)}^{2} + 1] + . . . + [3 \times {(1 + (n - 1) h)}^{2} + 1]\} = \lim_{h \to 0} h \{3 [1 + (1 + 2 h + h^{2}) + (1 + 4 h + 2^{2} h^{2}) + . . . + (1 + 2 (n - 1) h + {(n - 1)}^{2} h^{2})] + n\} = \lim_{h \to 0} h \{3 [n + 2 (1 + 2 + . . . + (n - 1)) h + (1^{2} + 2^{2} + . . . + {(n - 1)}^{2}) h^{2}] + n\} = \lim_{h \to 0} h [4 n + 6 \times \frac{n (n - 1)}{2} h + 3 \times \frac{(n - 1) n (2 n - 1)}{6} h^{2}]$
$= \lim_{h \to 0} [4 n h + 6 \times \frac{n h (n h - h)}{2} + 3 \times \frac{(n h - h) n h (2 n h - h)}{6}] = \lim_{h \to 0} [4 n h + 3 \times n h (n h - h) + 3 \times \frac{(n h - h) n h (2 n h - h)}{6}] = \lim_{h \to 0} [4 \times 2 + 3 \times 2 \times (2 - h) + 3 \times \frac{(2 - h) \times 2 \times (2 \times 2 - h)}{6}] = 8 + 6 \times (2 - 0) + \frac{(2 - 0) \times 2 \times (4 - 0)}{2} = 8 + 12 + 8 = 28$

Page No 19.112:

Question 34:

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

Answer:

$\int_{1}^{3} (x^{2} + 3 x + e^{x}) d x \Rightarrow a = 1, b = 3, f (x) = x^{2} + 3 x + e^{x}, h = \frac{2}{n} I = \int_{1}^{3} (x^{2} + 3 x + e^{x}) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + f (1 + 2 h) + . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [(4 + e) + [{(1 + h)}^{2} + 3 (1 + h) + e^{1 + h}] + [{(1 + 2 h)}^{2} + 3 (1 + 2 h) + e^{1 + 2 h}] + . . . + [{(1 + (n - 1) h)}^{2} + 3 (1 + (n - 1) h) + e^{1 + (n - 1) h}]] = \lim_{h \to 0} h [4 + e + [(n - 1) \cdot 1 + h^{2} (1 + 2^{2} + 3^{2} + . . . {(n - 1)}^{2}) + 2 h (1 + 2 + 3 + (n - 1))] + [3 (n - 1) + 3 h (1 + 2 + 3 . . . (n - 1))] + [e^{1 + h} + e^{1 + 2 h} + . . . . e^{1 + (n - 1) h}]] = \lim_{h \to 0} h [4 n + e + h^{2} \frac{(n - 1) (n - 2) (2 n - 3)}{6} + 2 h (\frac{n - 1 (n)}{2}) + 3 h (\frac{(n - 1) (n)}{2}) + e [\frac{e^{h} \cdot [e^{h (n - 1)} - 1]}{e^{h} - 1}]] = \lim_{n \to \infty} 4 n \cdot \frac{2}{n} + \frac{e \cdot 2}{n} + \frac{8}{n^{3}} \frac{(2 n^{3} . . .)}{6} + \frac{5.4}{n^{2}} (\frac{n^{2} - n}{2}) + e^{\frac{2}{n + 1}} [e^{\frac{2}{n} (n - 1)} - 1] = 8 + 0 + \frac{8}{3} + 10 + e (e^{2} - 1) = \frac{26}{3} + 12 + e^{3} - e = \frac{62}{3} + e^{3} - e$

Page No 19.115:

Question 1:

$\int_{0}^{4} x \sqrt{4 - x} d x$

Answer:

$Let, I = \int_{0}^{4} x \sqrt{4 - x} d x = \int_{0}^{4} (4 - x) \sqrt{4 - 4 + x} d x = \int_{0}^{4} (4 - x) \sqrt{x} d x = \int_{0}^{4} 4 \sqrt{x} - x^{\frac{3}{2}} d x = {[8 \frac{x^{\frac{3}{2}}}{3}]}_{0}^{4} - {[\frac{2 x^{\frac{5}{2}}}{5}]}_{0}^{4} = \frac{64}{3} - \frac{64}{5} = \frac{128}{15}$ →

Page No 19.115:

Question 2:

$\int_{1}^{2} x \sqrt{3 x - 2} d x$

Answer:

$\int_{1}^{2} x \sqrt{3 x - 2} d x Let, 3 x - 2 = t, then 3 d x = d t when, x = 1; t = 1 and x = 2; t = 4 Therefore the integral becomes \int_{1}^{4} \frac{t + 2}{3} \sqrt{t} \frac{d t}{3} = \frac{1}{9} \int_{1}^{4} t^{\frac{3}{2}} + 2 \sqrt{t} d t = \frac{1}{9} {[\frac{2 t^{\frac{5}{2}}}{5} + \frac{4 t^{\frac{3}{2}}}{3}]}_{1}^{4} = \frac{1}{9} [\frac{64}{5} + \frac{32}{3} - \frac{2}{5} - \frac{4}{3}] = \frac{46}{135}$

Page No 19.115:

Question 3:

$\int_{1}^{5} \frac{x}{\sqrt{2 x - 1}} d x$

Answer:

$Let I = \int_{1}^{5} \frac{x}{\sqrt{2 x - 1}} d x Let, 2 x - 1 = t, then 2 d x = d t, When, x \to 1; t \to 1 and x \to 5; t \to 9 x = \frac{t + 1}{2} I = \frac{1}{2} \int_{1}^{9} \frac{t + 1}{\sqrt{t}} \times \frac{d t}{2} = \frac{1}{4} {[\frac{2 t^{\frac{3}{2}}}{3} + 2 \sqrt{t}]}_{1}^{9} = \frac{1}{4} [18 + 6 - \frac{2}{3} - 2] = \frac{16}{3}$

Page No 19.115:

Question 4:

$\int_{0}^{1} \cos^{- 1} x d x$

Answer:

$\int_{0}^{1} \cos^{- 1} x d x = \int_{0}^{1} (\cos^{- 1} x \times 1) d x = {[\cos^{- 1} x x]}_{0}^{1} - \int_{0}^{1} \frac{- x}{\sqrt{1 - x^{2}}} d x = {[x \cos^{- 1} x]}_{0}^{1} - \frac{2}{2} {[\sqrt{1 - x^{2}}]}_{0}^{1} = 0 + 1 = 1$

Page No 19.115:

Question 5:

$\int_{0}^{1} \tan^{- 1} x d x$

Answer:

$\int_{0}^{1} \tan^{- 1} x d x = \int_{0}^{1} \tan^{- 1} x \times 1 d x = {[\tan^{- 1} x x]}_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x^{2}} d x = {[x \tan^{- 1} x]}_{0}^{1} - \frac{1}{2} {[\log (1 + x^{2})]}_{0}^{1} = \frac{π}{4} - 0 - \frac{1}{2} \log 2 + 0 = \frac{π}{4} - \frac{1}{2} \log 2$

Page No 19.115:

Question 6:

$\int_{0}^{1} \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x$

Answer:

$\int_{0}^{1} \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x Let, x = \tan θ, d x = s e c^{2} θ d θ When, x \to 0; θ \to 0 and x \to 1; θ \to \frac{π}{4} Therefore, the integral becomes \int_{0}^{\frac{π}{4}} \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) s e c^{2} θ d θ = \int_{0}^{\frac{π}{4}} \cos^{- 1} (\cos 2 θ) s e c^{2} θ d θ = 2 \int_{0}^{\frac{π}{4}} θ s e c^{2} θ d θ = 2 {[θ t an θ]}_{0}^{\frac{π}{4}} - 2 \int_{0}^{\frac{π}{4}} \tan θ d θ = 2 {[θ \tan θ]}_{0}^{\frac{π}{4}} + 2 {[\log (\cos θ)]}_{0}^{\frac{π}{4}} = 2 (\frac{π}{4} - 0) + 2 [\log \frac{1}{\sqrt{2}} - 0] = \frac{π}{2} - \log 2$

Page No 19.115:

Question 7:

$\int_{0}^{1} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x$

Answer:

$\int_{0}^{1} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x Let, x = \tan θ, then d x = s e c^{2} θ d θ When, x \to 0; θ \to 0 And x \to 1; θ \to \frac{π}{4} Therefore the integral becomes \int_{0}^{\frac{π}{4}} \tan^{- 1} (\frac{2 \tan θ}{1 - \tan^{2} θ}) s e c^{2} θ d θ = \int_{0}^{\frac{π}{4}} \tan^{- 1} (\tan 2 θ) s e c^{2} θ d θ = 2 \int_{0}^{\frac{π}{4}} θ s e c^{2} θ d θ = 2 {[θ \tan θ]}_{0}^{\frac{π}{4}} - 2 \int_{0}^{\frac{π}{4}} \tan θ d θ = 2 {[θ \tan θ]}_{0}^{\frac{π}{4}} - 2 {[- \log (\cos θ)]}_{0}^{\frac{π}{4}} = 2 (\frac{π}{4} - 0) + 2 [\log \frac{1}{\sqrt{2}} - 0] = \frac{π}{2} - \log 2$

Page No 19.115:

Question 8:

$\int_{0}^{1 / \sqrt{3}} \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) d x$

Answer:

$\int_{0}^{\frac{1}{\sqrt{3}}} \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) d x Let x = \tan θ, then d x = s e c^{2} θ d θ When, x \to 0; θ \to 0 And x \to \frac{1}{\sqrt{3}}; θ \to \frac{π}{6} Therefore the integral becomes \int_{0}^{\frac{π}{6}} \tan^{- 1} (\frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) s e c^{2} θ d θ = \int_{0}^{\frac{π}{6}} \tan^{- 1} (\tan 3 θ) s e c^{2} θ d θ = 3 \int_{0}^{\frac{π}{6}} θ s e c^{2} θ d θ = 3 {[θ \tan θ]}_{0}^{\frac{π}{6}} - 3 \int_{0}^{\frac{π}{6}} \tan θ d θ = 3 {[θ \tan θ]}_{0}^{\frac{π}{6}} - 3 {[- \log (\cos θ)]}_{0}^{\frac{π}{6}} = 3 (\frac{π}{6} \times \frac{1}{\sqrt{3}} - 0) + 3 [\log \frac{\sqrt{3}}{2}] = \frac{π}{2 \sqrt{3}} + 3 \log \frac{\sqrt{3}}{2} = \frac{π}{2 \sqrt{3}} - \frac{3}{2} \log \frac{4}{3}$

Page No 19.116:

Question 9:

$\int_{0}^{1} \frac{1 - x}{1 + x} d x$

Answer:

$\int_{0}^{1} \frac{1 - x}{1 + x} d x = \int_{0}^{1} \frac{1 - x - 1 + 1}{1 + x} d x = \int_{0}^{1} \frac{2 - (x + 1)}{1 + x} d x = \int_{0}^{1} \frac{2}{1 + x} - \int_{0}^{1} \frac{1 + x}{1 + x} d x = \int_{0}^{1} \frac{2}{1 + x} - \int_{0}^{1} d x = 2 {[\log (1 + x)]}_{0}^{1} - {[x]}_{0}^{1} = 2 \log 2 - 1$

Page No 19.116:

Question 10:

$\int_{0}^{π / 3} \frac{\cos x}{3 + 4 \sin x} d x$

Answer:

$\int_{0}^{\frac{π}{3}} \frac{\cos x}{3 + 4 \sin x} d x Let, \sin x = t \Rightarrow \cos x d x = d t When, \sin x \to 0; t \to 0 And \sin x \to \frac{π}{3}; t \to \frac{\sqrt{3}}{2} = \int_{0}^{\frac{\sqrt{3}}{2}} \frac{d t}{3 + 4 t} = \frac{1}{4} \log {[3 + 4 t]}_{0}^{\frac{\sqrt{3}}{2}} = \frac{1}{4} \log [\log (3 + 2 \sqrt{3}) - \log (3 + 0)] = \frac{1}{4} \log [\log (2 \sqrt{3} + 3) - \log (3)] = \frac{1}{4} (\log \frac{2 \sqrt{3} + 3}{3})$

Page No 19.116:

Question 11:

$\int_{0}^{π / 2} \frac{\sin^{2} x}{{(1 + \cos x)}^{2}} d x$

Answer:

$\int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{{(1 + \cos x)}^{2}} d x = \int_{0}^{\frac{π}{2}} \frac{1 - \cos^{2} x}{{(1 + \cos x)}^{2}} d x = \int_{0}^{\frac{π}{2}} \frac{(1 + \cos x) (1 - \cos x)}{{(1 + \cos x)}^{2}} d x = \int_{0}^{\frac{π}{2}} \frac{1 - \cos x}{1 + \cos x} d x = \int_{0}^{\frac{π}{2}} \frac{1 - \cos x - 1 + 1}{(1 + \cos x)} d x = \int_{0}^{\frac{π}{2}} \frac{2 - (1 + \cos x)}{(1 + \cos x)} d x = \int_{0}^{\frac{π}{2}} \frac{2}{1 + \cos x} d x - \int_{0}^{\frac{π}{2}} d x = \int_{0}^{\frac{π}{2}} \frac{2 (1 - \cos x)}{(1 + \cos x) (1 - \cos x)} d x - \int_{0}^{\frac{π}{2}} d x = 2 \int_{0}^{\frac{π}{2}} \frac{1 - \cos x}{\sin^{2} x} d x - {[x]}_{0}^{\frac{π}{2}} = 2 \int_{0}^{\frac{π}{2}} (\cos e c^{2} x - \cos e c x c o t x) d x - {[x]}_{0}^{\frac{π}{2}} = 2 {[- co t x + \cos e c x]}_{0}^{\frac{π}{2}} - {[x]}_{0}^{\frac{π}{2}} = 2 - \frac{π}{2}$

Page No 19.116:

Question 12:

$\int_{0}^{π / 2} \frac{\sin x}{\sqrt{1 + \cos x}} d x$

Answer:

$\int_{0}^{\frac{π}{2}} \frac{\sin x}{\sqrt{1 + \cos x}} d x Let 1 + \cos x = t, then - \sin x d x = d t When, x \to 0, t \to 2 and x \to \frac{π}{2}, t \to 1 Therefore, the integral becomes \int_{2}^{1} \frac{- 1}{\sqrt{t}} d t = \int_{1}^{2} \frac{1}{\sqrt{t}} d t = 2 {[\sqrt{t}]}_{1}^{2} = 2 (\sqrt{2} - 1)$

Page No 19.116:

Question 13:

$\int_{0}^{π / 2} \frac{\cos x}{1 + \sin^{2} x} d x$

Answer:

$\int_{0}^{\frac{π}{2}} \frac{\cos x}{1 + \sin^{2} x} d x Let \sin x = t, then \cos x d x = d t When x \to 0; t \to 0 And x \to \frac{π}{2}; t \to 1 Therefore the integral becomes \int_{0}^{1} \frac{d t}{1 + t^{2}} = {[\tan^{- 1} x]}_{0}^{1} = \frac{π}{4}$

Page No 19.116:

Question 14:

$\int_{0}^{π} \sin^{3} x (1 + 2 \cos x) {(1 + \cos x)}^{2} d x$

Answer:

$We have, I = \int_{0}^{π} \sin^{3} x (1 + 2 \cos x) {(1 + \cos x)}^{2} d x = \int_{0}^{π} \sin^{2} x (1 + 2 \cos x) {(1 + \cos x)}^{2} \sin x d x = \int_{0}^{π} (1 - \cos^{2} x) (1 + 2 \cos x) {(1 + \cos x)}^{2} \sin x d x Putting \cos x = t \Rightarrow - \sin x d x = d t When x \to 0; t \to 1 and x \to π; t \to - 1 ∴ I = - \int_{1}^{- 1} (1 - t^{2}) (1 + 2 t) {(1 + t)}^{2} d t = \int_{- 1}^{1} (1 - t^{2}) (1 + 2 t) {(1 + t)}^{2} d t = \int_{- 1}^{1} (1 + 2 t - t^{2} - 2 t^{3}) (1 + 2 t + t^{2}) d t = \int_{- 1}^{1} (1 + 2 t + t^{2} + 2 t + 4 t^{2} + 2 t^{3} - t^{2} - 2 t^{3} - t^{4} - 2 t^{3} - 4 t^{4} - 2 t^{5}) d t = \int_{- 1}^{1} (1 + 4 t + 4 t^{2} - 2 t^{3} - 5 t^{4} - 2 t^{5}) d t = {[t + 2 t^{2} + \frac{4 t^{3}}{3} - \frac{t^{4}}{2} - t^{5} - \frac{t^{6}}{3}]}_{- 1}^{1} = 1 + 2 + \frac{4}{3} - \frac{1}{2} - 1 - \frac{1}{3} - (- 1) - 2 {(- 1)}^{2} - \frac{4 {(- 1)}^{3}}{3} + \frac{{(- 1)}^{4}}{2} + {(- 1)}^{5} + \frac{{(- 1)}^{6}}{3} = 1 + 2 + \frac{4}{3} - \frac{1}{2} - 1 - \frac{1}{3} + 1 - 2 + \frac{4}{3} + \frac{1}{2} - 1 + \frac{1}{3} = \frac{8}{3}$

Page No 19.116:

Question 15:

$\int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x$

Answer:

I = $\int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x$

using partial fraction,
$\frac{x}{(1 + x) (1 + x^{2})} \frac{A}{1 + x} + \frac{B x + C}{1 + x^{2}} x = A (1 + x^{2}) + (B x + C) (1 + x) x = A + A x^{2} + B x + B x^{2} + C + C x B + C = 1 A + C = 0 A + B = 0 s o, A = \frac{- 1}{2}, B = \frac{1}{2}, C = \frac{1}{2}$

putting the values of A,B and C we get

$\frac{\frac{- 1}{2}}{1 + x} + \frac{\frac{1}{2} x + \frac{1}{2}}{1 + x^{2}} = \frac{- 1}{2} [\frac{1}{1 + x}] + \frac{1}{2} [\frac{x + 1}{1 + x^{2}}] T h e r e f o r e, I = \int_{0}^{\infty} \frac{- 1}{2} [\frac{1}{1 + x}] + \frac{1}{2} [\frac{x + 1}{1 + x^{2}}] I = \frac{- 1}{2} {[\log |1 + x|]}_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} [\frac{x}{1 + x^{2}} + \frac{1}{1 + x^{2}}] I = \frac{- 1}{2} {[l o g |1 + x|]}_{0}^{\infty} + \frac{1}{2 \times 2} \int_{0}^{\infty} [\frac{2 x}{1 + x^{2}}] + \frac{1}{2} \int_{0}^{\infty} \frac{1}{1 + x^{2}}$
$I = \frac{- 1}{2} {[\log |1 + x|]}_{0}^{\infty} + \frac{1}{4} {[\log |1 + x^{2}|]}_{0}^{\infty} + {[\frac{1}{2} t a n^{- 1} x]}_{0}^{\infty} I = \frac{- 1}{2} {[l o g |1 + x|]}_{0}^{\infty} + \frac{1}{2} \times \frac{1}{2} {[l o g |1 + x^{2}|]}_{0}^{\infty} + {[\frac{1}{2} t a n^{- 1} x]}_{0}^{\infty} I = \frac{1}{2} {[\log \frac{\sqrt{x^{2} + 1}}{x + 1}]}_{0}^{\infty} + {[\frac{1}{2} t a n^{- 1} x]}_{0}^{\infty} I = \frac{1}{2} {[l o g \frac{\sqrt{1 + \frac{1}{x^{2}}}}{1 + \frac{1}{x}}]}_{0}^{\infty} + {[\frac{1}{2} t a n^{- 1} x]}_{0}^{\infty} I = \frac{1}{2} [0] + \frac{1}{2} [t a n^{- 1} \infty - t a n^{- 1} 0]$
$I = \frac{π}{4}$

Page No 19.116:

Question 16:

$\int_{0}^{π / 4} \sin 2 x \sin 3 x d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{4}} \sin 2 x \sin 3 x d x . . . (i) \Rightarrow I = {[- \sin 2 x \frac{\cos 3 x}{3}]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} 2 \cos 2 x \frac{\cos 3 x}{3} d x \Rightarrow I = {[- \sin 2 x \frac{\cos 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{2}{3} {[\cos 2 x \frac{\sin 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{4}{9} \int_{0}^{\frac{π}{4}} \sin 2 x \sin 3 x d x \Rightarrow I = {[- \sin 2 x \frac{\cos 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{2}{3} {[\cos 2 x \frac{\sin 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{4}{9} I [From (i)] \Rightarrow \frac{5}{9} I = {[- \sin 2 x \frac{\cos 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{2}{3} {[\cos 2 x \frac{\sin 3 x}{3}]}_{0}^{\frac{π}{4}} \Rightarrow \frac{5}{9} I = \frac{1}{3 \sqrt{2}} + 0 \Rightarrow \frac{5}{9} I = \frac{1}{3 \sqrt{2}} ∴ I = \frac{3}{5 \sqrt{2}}$

Page No 19.116:

Question 17:

$\int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x$

Answer:

$\int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x = \int_{0}^{1} \sqrt{\frac{1 - x}{1 + x} \times \frac{1 - x}{1 - x}} d x = \int_{0}^{1} \frac{1 - x}{\sqrt{1 - x^{2}}} d x = \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} d x - \int_{0}^{1} \frac{x}{\sqrt{1 - x^{2}}} d x = {[\sin^{- 1} x]}_{0}^{1} + {[\sqrt{1 - x^{2}}]}_{0}^{1} = \frac{π}{2} - 1$

Page No 19.116:

Question 18:

$\int_{1}^{2} \frac{1}{x^{2}} e^{- 1 / x} d x$

Answer:

$\int_{1}^{2} \frac{1}{x^{2}} e^{\frac{- 1}{x}} d x Let \frac{- 1}{x} = t, then \frac{1}{x^{2}} d x = d t When, x \to 1; t \to - 1 And x \to 2; t \to \frac{- 1}{2} Therefore the integral becomes \int_{- 1}^{\frac{- 1}{2}} e^{t} d t = {[e^{t}]}_{- 1}^{\frac{- 1}{2}} = e^{\frac{- 1}{2}} - e^{- 1} = \frac{\sqrt{e} - 1}{e}$

Page No 19.116:

Question 19:

$\int_{0}^{π / 4} \cos^{4} x \sin^{3} x d x$

Answer:

$\int_{0}^{\frac{π}{4}} \cos^{4} x \sin^{3} x d x = \int_{0}^{\frac{π}{4}} \cos^{4} x \sin x (1 - \cos^{2} x) d x = \int_{0}^{\frac{π}{4}} \cos^{4} x \sin x d x - \int_{0}^{\frac{π}{4}} \cos^{6} x \sin x d x = - {[\frac{\cos^{5} x}{5}]}_{0}^{\frac{π}{4}} + {[\frac{\cos^{7} x}{7}]}_{0}^{\frac{π}{4}} = \frac{- 1}{20 \sqrt{2}} + \frac{1}{5} + \frac{1}{56 \sqrt{2}} - \frac{1}{7} = \frac{- \sqrt{2}}{40} + \frac{2}{35} + \frac{\sqrt{2}}{112} = \frac{2}{35} - \frac{9 \sqrt{2}}{560}$

Page No 19.116:

Question 20:

$\int_{π / 3}^{π / 2} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{5 / 2}} d x$

Answer:

$\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{\frac{5}{2}}} d x = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{\frac{5}{2}}} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} d x = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sin x}{{(1 - \cos x)}^{3}} d x = - \frac{1}{2} {[{(1 - \cos x)}^{- 2}]}_{\frac{π}{3}}^{\frac{π}{2}} = - \frac{1}{2} [1 - 4] = \frac{3}{2}$

Page No 19.116:

Question 21:

$\int_{0}^{π / 2} x^{2} \cos 2 x d x$

Answer:

$\int_{0}^{\frac{π}{2}} x^{2} \cos 2 x d x = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x \frac{\sin 2 x}{2} d x = {[x^{2} \frac{s i n 2 x}{2}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} x \sin 2 x d x = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} - {[- x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} + [- \int_{0}^{\frac{π}{2}} \frac{\cos 2 x}{2} d x] = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} + {[x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{\cos 2 x}{2} d x = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} + {[x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} - \frac{1}{2} {[\frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} = 0 - \frac{π}{4} - 0 = \frac{- π}{4}$

Page No 19.116:

Question 22:

$\int_{0}^{1} \log (1 + x) d x$

Answer:

$\int_{0}^{1} \log (1 + x) d x = \int_{0}^{1} \log (1 + x) \times 1 d x = {[\log (1 + x) x]}_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x} d x = {[\log (1 + x) x]}_{0}^{1} - \int_{0}^{1} (1 - \frac{1}{1 + x}) d x = {[x \log (1 + x)]}_{0}^{1} - {[x - \log (1 + x)]}_{0}^{1} = \log 2 - 1 + \log 2 = 2 \log 2 - 1 = \log 4 - \log e = \log \frac{4}{e}$

Page No 19.116:

Question 23:

Evaluate the following integrals:

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

Answer:

Let I = $\int_{2}^{4} \frac{x^{2} + x}{\sqrt{2 x + 1}} d x$

Put 2x + 1 = z^{2

$\Rightarrow 2 d x = 2 z d z \Rightarrow d x = z d z$}

When $x \to 2, z \to \sqrt{5}$

When $x \to 4, z \to 3$

$∴ I = \int_{\sqrt{5}}^{3} \frac{{(\frac{z^{2} - 1}{2})}^{2} + \frac{z^{2} - 1}{2}}{z} \times z d z \Rightarrow I = \int_{\sqrt{5}}^{3} \frac{z^{4} - 2 z^{2} + 1 + 2 z^{2} - 2}{4} d z \Rightarrow I = \frac{1}{4} \int_{\sqrt{5}}^{3} (z^{4} - 1) d z \Rightarrow I = \frac{1}{4} \times {(\frac{z^{5}}{5} - z)}_{\sqrt{5}}^{3}$
$\Rightarrow I = \frac{1}{4} [(\frac{243}{5} - 3) - (\frac{25 \sqrt{5}}{5} - \sqrt{5})] \Rightarrow I = \frac{1}{4} \times \frac{228}{5} - \frac{1}{4} \times 4 \sqrt{5} \Rightarrow I = \frac{57}{5} - \sqrt{5}$

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Question 24:

$\int_{0}^{1} x {(\tan^{- 1} x)}^{2} d x$

Answer:

$We have, I = \int_{0}^{1} x {(\tan^{- 1} x)}^{2} d x Putting \tan^{- 1} x = u \Rightarrow x = \tan u \Rightarrow d x = \sec^{2} u d u When x \to 0; u \to 0 and x \to 1; u \to \frac{π}{4} ∴ I = \int_{0}^{\frac{π}{4}} (\tan u) u^{2} \sec^{2} u d u = \int_{0}^{\frac{π}{4}} u^{2} \tan u \sec^{2} u d u = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} 2 u \frac{\tan^{2} u}{2} d u = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} u (\sec^{2} u - 1) d u = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} u \sec^{2} u d u + \int_{0}^{\frac{π}{4}} u d u = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - {[u \tan u]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} \tan u d u + {[\frac{u^{2}}{2}]}_{0}^{\frac{π}{4}} = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - {[u \tan u]}_{0}^{\frac{π}{4}} + {[\log |\sec u|]}_{0}^{\frac{π}{4}} + {[\frac{u^{2}}{2}]}_{0}^{\frac{π}{4}} = \frac{π^{2}}{16} \times \frac{1}{2} - \frac{π}{4} + \log \sqrt{2} + \frac{π^{2}}{32} = \frac{π^{2}}{16} - \frac{π}{4} + \log \sqrt{2} = \frac{π^{2}}{16} - \frac{π}{4} + \frac{1}{2} \log 2$

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Question 25:

$\int_{0}^{1} {(\cos^{- 1} x)}^{2} d x$

Answer:

$I = \int_{0}^{1} (\cos^{- 1} x)^{2} d x l e t c o s^{- 1} x = θ \Rightarrow x = \cos θ \Rightarrow d x = - \sin θ d θ w h e n x = 0, θ = \frac{π}{2} a n d w h e n x = 1, θ = 0 t h e r e f o r e, I = \int_{\frac{π}{2}}^{0} θ^{2} (- \sin θ) d θ I = - \int_{\frac{π}{2}}^{0} θ^{2} (s i n θ) d θ I = \int_{0}^{\frac{π}{2}} θ^{2} (s i n θ) d θ I = [θ^{2} (- c o s θ)]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 θ \int_{0}^{\frac{π}{2}} \sin θ d θ] I = [θ^{2} (- \cos θ)]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} [2 θ (- \cos θ) d θ]$
$= [- θ^{2} \cos θ]_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} 2 θ (\cos θ) d θ = [- θ^{2} c o s θ]_{0}^{\frac{π}{2}} + 2 [θ \sin θ - \int_{0}^{\frac{π}{2}} \sin θ d θ] = [- θ^{2} c o s θ]_{0}^{\frac{π}{2}} + 2 [θ s i n θ + \cos θ]_{0}^{\frac{π}{2}}$

$I = 2 [(\frac{π}{2} + 0) - 1] I = π - 2$

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Question 26:

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

Answer:

$\int_{1}^{2} \frac{x + 3}{x (x + 2)} d x = \int_{1}^{2} \frac{x + 2 + 1}{x (x + 2)} d x = \int_{1}^{2} \frac{1}{x} d x + \int_{1}^{2} \frac{1}{x (x + 2)} d x = \int_{1}^{2} \frac{1}{x} d x + \frac{1}{2} \int_{1}^{2} \frac{(x + 2) - x}{x (x + 2)} d x = \int_{1}^{2} \frac{1}{x} d x + \frac{1}{2} \int_{1}^{2} \frac{1}{x} d x - \frac{1}{2} \int_{1}^{2} \frac{1}{x + 2} d x = \frac{3}{2} \int_{1}^{2} \frac{1}{x} d x - \frac{1}{2} \int_{1}^{2} \frac{1}{x + 2} d x = \frac{3}{2} {[\log x]}_{1}^{2} - \frac{1}{2} {[\log (x + 2)]}_{1}^{2} = \frac{3}{2} \log 2 - \frac{1}{2} \log 4 + \frac{1}{2} \log 3 = \frac{3}{2} \log 2 - \log 2 + \frac{1}{2} \log 3 = \frac{1}{2} \log 2 + \frac{1}{2} \log 3 = \frac{1}{2} \log 6$

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Question 27:

$\int_{0}^{π / 4} e^{x} \sin x d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{4}} e^{x} \sin x d x . . . (i) = {[- e^{x} \cos x]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} e^{x} \cos x d x = {[- e^{x} \cos x]}_{0}^{\frac{π}{4}} + {[e^{x} \sin x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} e^{x} \sin x d x \Rightarrow I = {[- e^{x} \cos x]}_{0}^{\frac{π}{4}} + {[e^{x} \sin x]}_{0}^{\frac{π}{4}} - I [Using (i)] \Rightarrow 2 I = {[- e^{x} \cos x]}_{0}^{\frac{π}{4}} + {[e^{x} \sin x]}_{0}^{\frac{π}{4}} = - \frac{1}{\sqrt{2}} e^{\frac{π}{4}} + 1 + \frac{1}{\sqrt{2}} e^{\frac{π}{4}} - 0 = 1 Hence I = \frac{1}{2}$

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Question 28:

$\int_{0}^{π / 4} \tan^{4} x d x$

Answer:

$\int_{0}^{\frac{π}{4}} \tan^{4} x d x = \int_{0}^{\frac{π}{4}} \tan^{2} x (s e c^{2} x - 1) d x = \int_{0}^{\frac{π}{4}} \tan^{2} x s e c^{2} x d x - \int_{0}^{\frac{π}{4}} \tan^{2} x d x = {[\frac{\tan^{3} x}{3}]}_{0}^{\frac{π}{4}} - {[\tan x - x]}_{0}^{\frac{π}{4}} = \frac{1}{3} - 1 + \frac{π}{4} = \frac{π}{4} - \frac{2}{3}$

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Question 29:

$\int_{0}^{1} |2 x - 1| d x$

Answer:

$We have, |2 x - 1| = \{\begin{cases} - (2 x - 1), 0 \leq x \leq \frac{1}{2} \\ 2 x - 1, \frac{1}{2} \leq x \leq 1 \end{cases} ∴ \int_{0}^{1} |2 x - 1| d x = \int_{0}^{\frac{1}{2}} - (2 x - 1) d x + \int_{\frac{1}{2}}^{1} (2 x - 1) d x = {[- x^{2} + x]}_{0}^{\frac{1}{2}} + {[x^{2} - x]}_{\frac{1}{2}}^{1} = \frac{- 1}{4} + \frac{1}{2} + 1 - 1 - \frac{1}{4} + \frac{1}{2} = \frac{1}{2}$

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Question 30:

$\int_{1}^{3} |x^{2} - 2 x| d x$

Answer:

$We have, |x^{2} - 2 x| = \{\begin{cases} - (x^{2} - 2 x), 1 \leq x \leq 2 \\ x^{2} - 2 x, 2 \leq x \leq 3 \end{cases} ∴ \int_{1}^{3} |x^{2} - 2 x| d x = \int_{1}^{2} - (x^{2} - 2 x) d x + \int_{2}^{3} (x^{2} - 2 x) d x = {[- \frac{x^{3}}{3} + x^{2}]}_{1}^{2} + {[\frac{x^{3}}{3} - x^{2}]}_{2}^{3} = \frac{- 8}{3} + 4 + \frac{1}{3} - 1 + 9 - 9 - \frac{8}{3} + 4 = 2$

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Question 31:

$\int_{0}^{π / 2} |\sin x - \cos x| d x$

Answer:

$\int_{0}^{\frac{π}{2}} |\sin x - \cos x| d x = \sqrt{2} \int_{0}^{\frac{π}{2}} |\sin x \frac{1}{\sqrt{2}} - \cos x \frac{1}{\sqrt{2}}| d x = \sqrt{2} \int_{0}^{\frac{π}{2}} |\sin x \cos \frac{π}{4} - \cos x \sin \frac{π}{4}| d x = \sqrt{2} \int_{0}^{\frac{π}{2}} |\sin (x - \frac{π}{4})| d x We have, |\sin (x - \frac{π}{4})| = \{\begin{cases} - \sin (x - \frac{π}{4}), 0 \leq x \leq \frac{π}{4} \\ \sin (x - \frac{π}{4}), \frac{π}{4} \leq x \leq \frac{π}{2} \end{cases} ∴ \int_{0}^{\frac{π}{2}} |\sin x - \cos x| d x = \sqrt{2} \int_{0}^{\frac{π}{4}} - \sin (x - \frac{π}{4}) d x + \sqrt{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \sin (x - \frac{π}{4}) d x = \sqrt{2} {[\cos (x - \frac{π}{4})]}_{0}^{\frac{π}{4}} - \sqrt{2} {[\cos (x - \frac{π}{4})]}_{\frac{π}{4}}^{\frac{π}{2}} = \sqrt{2} [\cos (0) - \cos (- \frac{π}{4})] - \sqrt{2} [\cos (\frac{π}{4}) - \cos (0)] = \sqrt{2} (1 - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + 1) = \sqrt{2} (2 - \frac{2}{\sqrt{2}}) = 2 \sqrt{2} - 2 = 2 (\sqrt{2} - 1)$
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Question 32:

$\int_{0}^{1} |\sin 2 π x| d x$

Answer:

$We have, |\sin 2 πx| = \{\begin{cases} (\sin 2 πx), 0 \leq x \leq \frac{1}{2} \\ - (\sin 2 πx), \frac{1}{2} \leq x \leq 1 \end{cases} ∴ \int_{0}^{1} |\sin 2 πx| d x = \int_{0}^{\frac{1}{2}} \sin 2 πx dx + \int_{\frac{1}{2}}^{1} - \sin 2 πx dx = {[\frac{- \cos 2 πx}{2 π}]}_{0}^{\frac{1}{2}} + {[\frac{\cos 2 πx}{2 π}]}_{\frac{1}{2}}^{1} = \frac{1}{2 π} + \frac{1}{2 π} + \frac{1}{2 π} + \frac{1}{2 π} = \frac{2}{π}$

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Question 33:

$\int_{1}^{3} |x^{2} - 4| d x$

Answer:

$\int_{1}^{3} |x^{2} - 4| d x = \int_{1}^{2} - (x^{2} - 4) d x + \int_{2}^{3} (x^{2} - 4) d x = {[- \frac{x^{3}}{3} + 4 x]}_{1}^{2} + {[\frac{x^{3}}{3} - 4 x]}_{2}^{3} = \frac{- 8}{3} + 8 + \frac{1}{3} - 4 + 9 - 12 - \frac{8}{3} + 8 = 4$

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Question 34:

$\int_{- π / 2}^{π / 2} \sin^{9} x d x$

Answer:

$\int_{\frac{- π}{2}}^{\frac{π}{2}} \sin^{9} x d x Let f (x) = \sin^{9} x Consider, f (- x) = \sin^{9} (- x) = - \sin^{9} x = - f (x) Thus f (x) is an odd function Therefore, \int_{\frac{- π}{2}}^{\frac{π}{2}} \sin^{9} x d x = 0$

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Question 35:

$\int_{- 1 / 2}^{1 / 2} \cos x \log (\frac{1 + x}{1 - x}) d x$

Answer:

$\int_{\frac{- 1}{2}}^{\frac{1}{2}} \cos x \log (\frac{1 + x}{1 - x}) d x Let f (x) = \cos x \log (\frac{1 + x}{1 - x}) Consider f (- x) = \cos (- x) \log (\frac{1 - x}{1 + x}) = \cos x \{- \log (\frac{1 + x}{1 - x})\} = - \cos x \log (\frac{1 + x}{1 - x}) = - f (x) Thus f (x) is an odd function Therefore, \int_{\frac{- 1}{2}}^{\frac{1}{2}} \cos x \log (\frac{1 + x}{1 - x}) d x = 0$

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Question 36:

$\int_{- a}^{a} \frac{x e^{x^{2}}}{1 + x^{2}} d x$

Answer:

$\int_{- a}^{a} \frac{x e^{x^{2}}}{1 + x^{2}} d x Let f (x) = \frac{x e^{x^{2}}}{1 + x^{2}} Consider f (- x) = - \frac{x e^{x^{2}}}{1 + x^{2}} = - f (x) Thus f (x) i s an odd function Therefore, \int_{- a}^{a} \frac{x e^{x^{2}}}{1 + x^{2}} d x = 0$ ode is 4430. →

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Question 37:

$\int_{0}^{π / 2} \frac{1}{1 + \cot^{7} x} d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t^{7} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t^{7} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{7} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t^{7} x} + \frac{1}{1 + \tan^{7} x} d x = \int_{0}^{\frac{π}{2}} \frac{2 + c o t^{7} x + \tan^{7} x}{(1 + c o t^{7} x) (1 + \tan^{7} x)} d x = \int_{0}^{\frac{π}{2}} \frac{2 + c o t^{7} x + \tan^{7} x}{2 + c o t^{7} x + \tan^{7} x} d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence, I = \frac{π}{4}$

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Question 38:

$\int_{0}^{2 π} \cos^{7} x d x$

Answer:

$Let, I = \int_{0}^{2 π} \cos^{7} x d x . . . (i) = \int_{0}^{2 π} \cos^{7} (2 π - x) d x = \int_{0}^{2 π} - \cos^{7} x d x \Rightarrow I = - \int_{0}^{2 π} \cos^{7} x d x . . . (ii) Adding (i) and (ii) we get, 2 I = \int_{0}^{2 π} \cos^{7} x d x - \int_{0}^{2 π} \cos^{7} x d x \Rightarrow 2 I = 0 ∴ I = 0$

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Question 39:

$\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x$

Answer:

$Let, I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x . . . (i) = \int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{a - a + x}} d x = \int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{x}} d x \Rightarrow I = \int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{a} [\frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} + \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}}] d x = \int_{0}^{a} d x = {[x]}_{0}^{a} = a Hence, I = \frac{a}{2}$

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Question 40:

$\int_{0}^{π / 2} \frac{1}{1 + \tan^{3} x} d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{3} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{3} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t^{3} x} d x . . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} [\frac{1}{1 + \tan^{3} x} + \frac{1}{1 + c o t^{3} x}] d x = \int_{0}^{\frac{π}{2}} \frac{2 + \tan^{3} x + c o t^{3} x}{(1 + \tan^{3} x) (1 + c o t^{3} x)} d x = \int_{0}^{\frac{π}{2}} \frac{2 + \tan^{3} x + c o t^{3} x}{2 + \tan^{3} x + c o t^{3} x} d x = \int_{0}^{\frac{π}{2}} d x = {(x)}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence, I = \frac{π}{4}$

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Question 41:

$\int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x$

Answer:

$Let, I = \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x . . . (i) = \int_{0}^{π} \frac{(π - x) \sin (π - x)}{1 + \cos^{2} (π - x)} d x = \int_{0}^{π} \frac{(π - x) \sin x}{1 + \cos^{2} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{π} [\frac{x \sin x}{1 + \cos^{2} x} + \frac{(π - x) \sin x}{1 + \cos^{2} x}] d x = \int_{0}^{π} \frac{π \sin x}{1 + \cos^{2} x} d x = π {[- \tan^{- 1} (cosx)]}_{0}^{π} = - π [\tan^{- 1} (- 1) - \tan^{- 1} (1)] = - π (- \frac{π}{4} - \frac{π}{4}) = \frac{π^{2}}{2} Hence, I = \frac{π^{2}}{4}$

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Question 42:

$\int_{0}^{π} x \sin x \cos^{4} x d x$

Answer:

$Le t, I = \int_{0}^{π} x \sin x \cos^{4} x d x . . . (i) = \int_{0}^{π} (π - x) \sin (π - x) \cos^{4} (π - x) d x = \int_{0}^{π} (π - x) \sin x \cos^{4} x d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{π} [x \sin x \cos^{4} x + (π - x) \sin x \cos^{4} x] d x = \int_{0}^{π} (x + π - x) \sin x \cos^{4} x d x = π \int_{0}^{π} \sin x \cos^{4} x d x = π {[\frac{- \cos^{5} x}{5}]}_{0}^{π} = π [\frac{1}{5} + \frac{1}{5}] = \frac{2 π}{5} Hence, I = \frac{π}{5}$

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Question 43:

$\int_{0}^{π} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x$

Answer:

$We have, I = \int_{0}^{π} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x . . . . . (1) = \int_{0}^{π} \frac{(π - x)}{a^{2} \cos^{2} (π - x) + b^{2} \sin^{2} (π - x)} d x = \int_{0}^{π} \frac{π - x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x . . . . . (2) Adding (1) and (2) 2 I = \int_{0}^{π} \frac{x + π - x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x = π \int_{0}^{π} \frac{1}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x = π \int_{0}^{π} \frac{\sec^{2} x}{a^{2} + b^{2} \tan^{2} x} dx (Dividing numerator and denominator by \cos^{2} x) = 2 π \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{a^{2} + b^{2} \tan^{2} x} dx [Using \int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x] Putting \tan x = t \Rightarrow \sec^{2} x d x = d t When x \to 0; t \to 0 and x \to \frac{π}{2}; t \to \infty ∴ 2 I = 2 π \int_{0}^{\frac{π}{2}} \frac{d t}{a^{2} + b^{2} t^{2}} \Rightarrow I = \frac{π}{b^{2}} \int_{0}^{\frac{π}{2}} \frac{d t}{\frac{a^{2}}{b^{2}} + t^{2}} = \frac{π}{b^{2}} \times \frac{b}{a} {[\tan^{- 1} (\frac{b t}{a})]}_{0}^{\infty} = \frac{π}{a b} [\frac{π}{2} - 0] = \frac{π}{a b} \times \frac{π}{2} = \frac{π^{2}}{2 a b} Hence I = \frac{π^{2}}{2 a b}$

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Question 44:

$\int_{- π / 4}^{π / 4} |\tan x| d x$

Answer:

$\int_{\frac{- π}{4}}^{\frac{π}{4}} |\tan x| d x = \int_{\frac{- π}{4}}^{0} - \tan x d x + \int_{0}^{\frac{π}{4}} \tan x d x = {[\log (\cos x)]}_{\frac{- π}{4}}^{0} + {[- \log (\cos x)]}_{0}^{\frac{π}{4}} = - \log \frac{1}{\sqrt{2}} - \log \frac{1}{\sqrt{2}} = 2 \log \sqrt{2} = \log 2$

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Question 45:

$\int_{0}^{15} [x^{2}] d x$

Answer:

$We have, I = \int_{0}^{1.5} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x + \int_{\sqrt{2}}^{1.5} [x^{2}] d x = \int_{0}^{1} (0) d x + \int_{1}^{\sqrt{2}} (1) d x + \int_{\sqrt{2}}^{1.5} (2) d x (∵ [x^{2}] = \{\begin{cases} \begin{array}{l} 0 where, 0 < x < 1 \\ 1 where, 1 < x < \sqrt{2} \end{array} \\ 2 where, \sqrt{2} < x < 1.5 \end{cases}) = 0 + {[x]}_{1}^{\sqrt{2}} + {[2 x]}_{\sqrt{2}}^{1.5} = {[x]}_{1}^{\sqrt{2}} + 2 {[x]}_{\sqrt{2}}^{1.5} = (\sqrt{2} - 1) + 2 (1.5 - \sqrt{2}) = \sqrt{2} - 1 + 3 - 2 \sqrt{2} = 2 - \sqrt{2}$

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Question 46:

$\int_{0}^{π} \frac{x}{1 + \cos α \sin x} d x$

Answer:

$We have, I = \int_{0}^{π} \frac{x}{1 + \cos α \sin x} d x . . . . . (1) \Rightarrow I = \int_{0}^{π} \frac{π - x}{1 + \cos α \sin (π - x)} d x (∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x) \Rightarrow I = \int_{0}^{π} \frac{π - x}{1 + \cos α \sin x} d x . . . . . (2) Adding (1) and (2), we get$

$2 I = \int_{0}^{π} \frac{π}{1 + \cos α \sin x} d x \Rightarrow I = \frac{π}{2} \int_{0}^{π} \frac{1}{1 + \cos α \sin x} d x = \frac{π}{2} \int_{0}^{π} \frac{1}{1 + \cos α \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \frac{π}{2} \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} + \cos α 2 \tan \frac{x}{2}} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t When x \to 0; t \to 0 and x \to π; t \to \infty Now, integral becomes$

$I = π \int_{0}^{\infty} \frac{d t}{1 + t^{2} + 2 t \cos α} = π \int_{0}^{\infty} \frac{d t}{{(t + \cos α)}^{2} + 1 - \cos^{2} α} = π \int_{0}^{\infty} \frac{d t}{{(t + \cos α)}^{2} + \sin^{2} α} = π {[\frac{1}{\sin α} \tan^{- 1} \frac{t + \cos α}{\sin α}]}_{0}^{\infty} = \frac{π}{\sin α} {[\tan^{- 1} \frac{t + \cos α}{\sin α}]}_{0}^{\infty} = \frac{π}{\sin α} [\frac{π}{2} - \tan^{- 1} (\cot α)] = \frac{π}{\sin α} [\frac{π}{2} - \tan^{- 1} \{\tan (\frac{π}{2} - α)\}] = \frac{π}{\sin α} [\frac{π}{2} - (\frac{π}{2} - α)] = \frac{π α}{\sin α}$

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Question 47:

$\int_{0}^{π / 2} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \sin (\frac{π}{2} - x) \cos (\frac{π}{2} - x)}{\sin^{4} (\frac{π}{2} - x) + \cos^{4} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \cos x \sin x}{\cos^{4} x + \sin^{4} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} \frac{(x + \frac{π}{2} - x) \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x} d x = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 - 2 \sin^{2} x \cos^{2} x} d x = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 - 2 \sin^{2} x (1 - \sin^{2} x)} d x = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 - 2 \sin^{2} x + 2 \sin^{4} x} d x Let, \sin^{2} x = t, then 2 \sin x \cos x d x = d t When, x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1 2 I = \frac{π}{4} \int_{0}^{1} \frac{1}{1 - 2 t + 2 t^{2}} d t = \frac{π}{8} \int_{0}^{1} \frac{1}{{(t - \frac{1}{2})}^{2} + \frac{1}{4}} = \frac{π}{8} {[2 \tan^{- 1} (2 t - 1)]}_{0}^{1} = \frac{π}{4} [\tan^{- 1} (1) - \tan^{- 1} (- 1)] = \frac{π}{4} [\frac{π}{4} + \frac{π}{4}] = \frac{π^{2}}{8} Hence, I = \frac{π^{2}}{16}$

Page No 19.117:

Question 48:

$\int_{0}^{π / 2} \frac{\cos^{2} x}{\sin x + \cos x} d x$

Answer:

$We have, I = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\sin x + \cos x} d x . . . . . (1) = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{\cos x + \sin x} d x . . . . . (2)$

$Adding (1) and (2) 2 I = \int_{0}^{\frac{π}{2}} [\frac{\cos^{2} x}{\sin x + \cos x} + \frac{\sin^{2} x}{\cos x + \sin x}] d x = \int_{0}^{\frac{π}{2}} [\frac{1}{\sin x + \cos x}] d x = \int_{0}^{\frac{π}{2}} [\frac{1}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}}] d x$

$= - \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{\tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2} - 1} d x = - \int_{0}^{\frac{π}{2}} \frac{\sec^{2} \frac{x}{2}}{\tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2} - 1} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t When x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1$

$∴ 2 I = - 2 \int_{0}^{1} \frac{d t}{t^{2} - 2 t - 1} \Rightarrow I = - \int_{0}^{1} \frac{d t}{{(t - 1)}^{2} - {(\sqrt{2})}^{2}} = - \frac{1}{2 \sqrt{2}} {[\log |\frac{t - 1 - \sqrt{2}}{t - 1 + \sqrt{2}}|]}_{0}^{1} = - \frac{1}{2 \sqrt{2}} [\log |- 1| - \log |\frac{- 1 - \sqrt{2}}{- 1 + \sqrt{2}}|] = - \frac{1}{2 \sqrt{2}} [\log 1 - \log \frac{\sqrt{2} + 1}{\sqrt{2} - 1}]$

$= - \frac{1}{2 \sqrt{2}} [- \log \frac{\sqrt{2} + 1}{\sqrt{2} - 1}] = \frac{1}{2 \sqrt{2}} \log [\frac{(\sqrt{2} + 1) (\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)}] = \frac{1}{2 \sqrt{2}} \log [\frac{{(\sqrt{2} + 1)}^{2}}{(2 - 1)}] = \frac{1}{2 \sqrt{2}} \log {(\sqrt{2} + 1)}^{2} = \frac{1}{2 \sqrt{2}} \times 2 \log (\sqrt{2} + 1) = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$

Page No 19.117:

Question 49:

$\int_{0}^{π} \cos 2 x \log \sin x d x$

Answer:

$\int_{0}^{π} \cos 2 x \log \sin x d x = {[\log \sin x \frac{\sin 2 x}{2}]}_{0}^{π} - \int_{0}^{π} \frac{\cos x}{\sin x} \frac{\sin 2 x}{2} d x = {[\log \sin x \frac{\sin 2 x}{2}]}_{0}^{π} - \int_{0}^{π} \cos^{2} x d x = {[\log \sin x \frac{\sin 2 x}{2}]}_{0}^{π} - \int_{0}^{π} \frac{1 + \cos 2 x}{2} d x = {[\log \sin x \frac{\sin 2 x}{2}]}_{0}^{π} - \frac{1}{2} {[x + \frac{\sin 2 x}{2}]}_{0}^{π} = 0 - \frac{1}{2} (π + 0) = - \frac{π}{2}$

Page No 19.117:

Question 50:

$\int_{0}^{π} \frac{x}{a^{2} - \cos^{2} x} d x, a > 1$

Answer:

$Let I = \int_{0}^{π} \frac{x}{a^{2} - \cos^{2} x} d x . . . (i) = \int_{0}^{π} \frac{π - x}{a^{2} - \cos^{2} (π - x)} d x = \int_{0}^{π} \frac{π - x}{a^{2} - \cos^{2} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{π} \frac{π}{a^{2} - \cos^{2} x} d x = \frac{π}{2 a} \int_{0}^{π} [\frac{1}{a - cosx} + \frac{1}{a + cosx}] dx = \frac{π}{2 a} \int_{0}^{π} [\frac{\sec^{2} \frac{x}{2}}{(a - 1) + (a + 1) \tan^{2} \frac{x}{2}} + \frac{\sec^{2} \frac{x}{2}}{(a + 1) + (a - 1) \tan^{2} \frac{x}{2}}] dx Let, \tan \frac{x}{2} = t, then \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt 2 I = \frac{π}{a} \int_{0}^{\infty} [\frac{1}{(a - 1) + (a + 1) t^{2}} + \frac{1}{(a + 1) + (a - 1) t^{2}}] dt = \frac{π}{a \sqrt{(a^{2} - 1)}} {[\tan^{- 1} \sqrt{\frac{a + 1}{a - 1}} t + \tan^{- 1} \sqrt{\frac{a - 1}{a + 1}} t]}_{0}^{\infty} = \frac{π}{a \sqrt{(a^{2} - 1)}} [\frac{π}{2} + \frac{π}{2}] = \frac{π^{2}}{a \sqrt{(a^{2} - 1)}} ∴ I = \frac{π^{2}}{2 a \sqrt{(a^{2} - 1)}}$

Page No 19.117:

Question 51:

$\int_{0}^{π} \frac{x \tan x}{\sec x + \tan x} d x$

Answer:

$Let I = \int_{0}^{π} \frac{x \tan x}{s e c x + \tan x} d x . . . (i) = \int_{0}^{π} \frac{(π - x) \tan (π - x)}{s e c (π - x) + \tan (π - x)} d x = \int_{0}^{π} \frac{(π - x) \tan x}{s e c x + \tan x} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{π} \frac{π \tan x}{s e c x + \tan x} d x = π \int_{0}^{π} \frac{sinx}{1 + sinx} dx = π \int_{0}^{π} \frac{1 + sinx - 1}{1 + sinx} dx = π \int_{0}^{π} [1 - \frac{1}{1 + sinx}] dx = π {[x]}_{0}^{π} - π \int_{0}^{π} \frac{1}{1 + \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} dx = π^{2} - π \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2}} dx = π^{2} - π \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{{(1 + \tan \frac{x}{2})}^{2}} dx = π^{2} + π {[\frac{2}{1 + \tan \frac{x}{2}}]}_{0}^{π} = π^{2} + π (0 - 2) = π^{2} - 2 π = π (π - 2) Hence I = \frac{π}{2} (π - 2)$

Page No 19.117:

Question 52:

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

Answer:

$Let, I = \int_{2}^{3} \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} d x . . . (i) = \int_{2}^{3} \frac{\sqrt{5 - x}}{\sqrt{5 - 5 + x} + \sqrt{5 - x}} d x = \int_{2}^{3} \frac{\sqrt{5 - x}}{\sqrt{x} + \sqrt{5 - x}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{2}^{3} [\frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} + \frac{\sqrt{5 - x}}{\sqrt{x} + \sqrt{5 - x}}] d x = \int_{2}^{3} \frac{\sqrt{5 - x} + \sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} d x = \int_{2}^{3} d x = {[x]}_{2}^{3} = 3 - 1 = 1 Hence, I = \frac{1}{2}$

Page No 19.117:

Question 53:

$\int_{0}^{π / 2} \frac{\sin^{2} x}{\sin x + \cos x} d x$

Answer:

$We have, I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{\sin x + \cos x} d x . . . . . (1) = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\cos x + \sin x} d x . . . . . (2) Adding (1) and (2) 2 I = \int_{0}^{\frac{π}{2}} [\frac{\sin^{2} x}{\sin x + \cos x} + \frac{\cos^{2} x}{\cos x + \sin x}] d x = \int_{0}^{\frac{π}{2}} [\frac{1}{\sin x + \cos x}] d x = \int_{0}^{\frac{π}{2}} [\frac{1 + \tan^{2} \frac{x}{2}}{2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}}] d x = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} \frac{x}{2}}{2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t When x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1 ∴ 2 I = \int_{0}^{1} \frac{2 d t}{2 t + 1 - t^{2}} d x = 2 \int_{0}^{1} \frac{d t}{{(\sqrt{2})}^{2} - {(t - 1)}^{2}} = \frac{2}{2 \sqrt{2}} {[\log |\frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1}|]}_{0}^{1} = \frac{1}{\sqrt{2}} [\log (\frac{\sqrt{2}}{\sqrt{2}}) - l o g |\frac{\sqrt{2} - 1}{\sqrt{2} + 1}|] = \frac{1}{\sqrt{2}} [0 - \log |\frac{\sqrt{2} - 1}{\sqrt{2} + 1}|] = - \frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} - 1}{\sqrt{2} + 1}| = \frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} + 1}{\sqrt{2} - 1}| = \frac{1}{\sqrt{2}} \log [\frac{(\sqrt{2} + 1) (\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)}] 2 I = \frac{1}{\sqrt{2}} \log [\frac{{(\sqrt{2} + 1)}^{2}}{2 - 1}] 2 I = \frac{2}{\sqrt{2}} \log (\sqrt{2} + 1) Hence I = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$

Page No 19.117:

Question 54:

$\int_{0}^{π / 2} \frac{x}{\sin^{2} x + \cos^{2} x} d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \frac{x}{\sin^{2} x + \cos^{2} x} d x = \int_{0}^{\frac{π}{2}} \frac{x}{1} d x = \int_{0}^{\frac{π}{2}} x d x = {[\frac{x^{2}}{2}]}_{0}^{\frac{π}{2}} ∴ I = \frac{π^{2}}{8}$

Page No 19.117:

Question 55:

$\int_{- π}^{π} x^{10} \sin^{7} x d x$

Answer:

$\int_{- π}^{π} x^{10} \sin^{7} x d x Let f (x) = x^{10} \sin^{7} x Consider f (- x) = {(- x)}^{10} \sin^{7} (- x) = - x^{10} \sin^{7} x = - f (x) Hence f (x) is an odd function Therefore \int_{- π}^{π} x^{10} \sin^{7} x d x = 0$

Page No 19.117:

Question 56:

$\int_{0}^{1} \cot^{- 1} (1 - x + x^{2}) d x$

Answer:

$\int_{0}^{1} c o t^{- 1} (1 - x + x^{2}) d x = \int_{0}^{1} c o t^{- 1} [x (x - 1) + 1] d x = \int_{0}^{1} c o t^{- 1} [\frac{(x (x - 1) + 1)}{x - (x - 1)}] d x = \int_{0}^{1} c o t^{- 1} x - c o t^{- 1} (x - 1) d x = {[x c o t^{- 1} x]}_{0}^{1} + \int_{0}^{1} \frac{x}{1 + x^{2}} d x - {[(x - 1) c o t^{- 1} (x - 1)]}_{0}^{1} - \int_{0}^{1} \frac{(x - 1)}{1 + {(x - 1)}^{2}} d x = {[x c o t^{- 1} x]}_{0}^{1} + \frac{1}{2} {[\log (1 + x^{2})]}_{0}^{1} - {[(x - 1) c o t^{- 1} (x - 1)]}_{0}^{1} - \frac{1}{2} {[\log (1 + {(1 - x)}^{2})]}_{0}^{1} = \frac{π}{4} - \frac{1}{2} \log 2 + \frac{π}{4} - \frac{1}{2} \log 2 = \frac{π}{2} - \log 2$

Page No 19.117:

Question 57:

$\int_{0}^{π} \frac{d x}{6 - \cos x}$

Answer:

$\int_{0}^{π} \frac{1}{6 - \cos x} d x = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{6 + 6 \tan^{2} \frac{x}{2} - 1 + \tan^{2} \frac{x}{2}} d x = \int_{0}^{π} \frac{s e c^{2} \frac{x}{2}}{5 + 7 \tan^{2} \frac{x}{2}} d x Let, \tan \frac{x}{2} = t, then \frac{1}{2} s e c^{2} \frac{x}{2} d x = d t Therefore the integralbecomes \int_{0}^{\infty} \frac{2 d t}{5 + 7 t^{2}} = \frac{2}{7} \int_{0}^{\infty} \frac{d t}{\frac{5}{7} + t^{2}} = \frac{2}{\sqrt{35}} {[\tan^{- 1} \frac{\sqrt{7} t}{\sqrt{5}}]}_{0}^{\infty} = \frac{π}{\sqrt{35}}$

Page No 19.117:

Question 58:

$\int_{0}^{π / 2} \frac{1}{2 \cos x + 4 \sin x} d x$

Answer:

$We have, I = \int_{0}^{\frac{π}{2}} \frac{1}{2 \cos x + 4 \sin x} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{2 - 2 \tan^{2} \frac{x}{2} + 8 \tan \frac{x}{2}} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} s e c^{2} \frac{x}{2} d x = d t When x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1 ∴ I = 2 \int_{0}^{1} \frac{d t}{2 - 2 t^{2} + 8 t} = - \frac{2}{2} \int_{0}^{1} \frac{d t}{t^{2} - 4 t - 1} = - \int_{0}^{1} \frac{d t}{{(t - 2)}^{2} - 5} = \int_{0}^{1} \frac{d t}{{(\sqrt{5})}^{2} - {(t - 2)}^{2}} = \frac{1}{2 \sqrt{5}} {[\log |\frac{\sqrt{5} + t - 2}{\sqrt{5} - t + 2}|]}_{0}^{1} = \frac{1}{2 \sqrt{5}} [\log \frac{\sqrt{5} - 1}{\sqrt{5} + 1} - \log \frac{\sqrt{5} - 2}{\sqrt{5} + 2}] = \frac{1}{2 \sqrt{5}} \log [\frac{\sqrt{5} - 1}{\sqrt{5} + 1} \times \frac{\sqrt{5} + 2}{\sqrt{5} - 2}] = \frac{1}{2 \sqrt{5}} \log [\frac{5 + 2 \sqrt{5} - \sqrt{5} - 2}{5 - 2 \sqrt{5} + \sqrt{5} - 2}] = \frac{1}{2 \sqrt{5}} \log [\frac{\sqrt{5} + 3}{- \sqrt{5} + 3}]$
$I = \frac{1}{2 \sqrt{5}} \log (\frac{3 + \sqrt{5}}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}}) I = \frac{1}{2 \sqrt{5}} l o g {(\frac{3 + \sqrt{5}}{2})}^{2} I = \frac{2}{2 \sqrt{5}} l o g (\frac{3 + \sqrt{5}}{2}) I = \frac{1}{\sqrt{5}} l o g (\frac{3 + \sqrt{5}}{2})$

Page No 19.117:

Question 59:

$\int_{π / 6}^{π / 2} \frac{cosec x \cot x}{1 + {cosec}^{2} x} d x$

Answer:

$\int_{\frac{π}{6}}^{\frac{π}{2}} \frac{\cos e c x c o t x}{1 + \cos e c^{2} x} d x = \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{\cos x}{1 + \sin^{2} x} d x = {[\tan^{- 1} (\sin x)]}_{\frac{π}{6}}^{\frac{π}{2}} = \tan^{- 1} 1 - \tan^{- 1} \frac{1}{2} = \tan^{- 1} \frac{1 - \frac{1}{2}}{1 + 1 \times \frac{1}{2}} = \tan^{- 1} \frac{1}{3}$

Page No 19.117:

Question 60:

$\int_{0}^{π / 2} \frac{d x}{4 \cos x + 2 \sin x}$

Answer:

$\int_{0}^{\frac{π}{2}} \frac{1}{4 \cos x + 2 \sin x} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{4 - 4 \tan^{2} \frac{x}{2} + 4 \tan \frac{x}{2}} d x Let \tan \frac{x}{2} = t, then \frac{1}{2} s e c^{2} \frac{x}{2} d x = d t When x = 0, t = 0, x = \frac{π}{2}, t = 1 = \frac{- 1}{4} \int_{0}^{1} \frac{d t}{{(t - \frac{1}{2})}^{2} - \frac{5}{4}} = \frac{- 1}{4} \times \frac{- 4}{\sqrt{5}} {[\log \frac{2 t - 1 - \sqrt{5}}{2 t - 1 + \sqrt{5}}]}_{0}^{1} = \frac{1}{\sqrt{5}} \log \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$

Page No 19.117:

Question 61:

$\int_{0}^{4} x d x$

Answer:

$\int_{0}^{4} x d x = {[\frac{x^{2}}{2}]}_{0}^{4} = 8 - 0 = 8$

Page No 19.117:

Question 62:

$\int_{0}^{2} (2 x^{2} + 3) d x$

Answer:

$\int_{0}^{2} (2 x^{2} + 3) d x = {[\frac{2 x^{3}}{3} + 3 x]}_{0}^{2} = \frac{16}{3} + 6 = \frac{34}{3}$

Page No 19.117:

Question 63:

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

Answer:

$Here a = 1, b = 4, f (x) = x^{2} + x, h = \frac{4 - 1}{n} = \frac{3}{n} Therefore, \int_{1}^{4} (x^{2} + x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [1 + 1 + {(1 + h)}^{2} + (1 + h) + {(1 + 2 h)}^{2} + (1 + 2 h) + . . . . . . . . . + {(1 + (n - 1) h)}^{2} + (1 + (n - 1) h)] = \lim_{h \to 0} h [2 n + h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2}) + 2 h (1 + 2 + . . . . . . + (n - 1)) + h (1 + 2 + . . . . . . + (n - 1))] = \lim_{h \to 0} h [2 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 3 h \frac{n (n - 1)}{2}] = \lim_{n \to 0 \infty} [6 + \frac{9}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + \frac{9}{2} (1 - \frac{1}{n})] = 6 + 9 + \frac{9}{2} = \frac{27}{2}$

Page No 19.117:

Question 64:

$\int_{- 1}^{1} e^{2 x} d x$

Answer:

$Here a = - 1, b = 1, f (x) = e^{2 x}, h = \frac{1 + 1}{n} = \frac{2}{n} Therefore, \int_{- 1}^{1} e^{2 x} d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (- 1) + f (- 1 + h) + . . . . . . . . . . + f (- 1 + (n - 1) h)] = \lim_{h \to 0} h [e^{- 2} + e^{2 (- 1 + h)} + e^{2 (- 1 + 2 h)} + . . . . . . . + e^{2 (- 1 + (n - 1) h)}] = \lim_{h \to 0} h e^{- 2} [\frac{{(e^{2 h})}^{n} - 1}{e^{2 h} - 1}] = \lim_{h \to 0} e^{- 2} [\frac{e^{4} - 1}{\frac{e^{2 h} - 1}{2 h}}] \times \frac{1}{2} (Since, n h = 2) = \frac{1}{2} (e^{2} - e^{- 2})$

Page No 19.117:

Question 65:

$\int_{2}^{3} e^{- x} d x$

Answer:

$Here a = 2, b = 3, f (x) = e^{- x}, h = \frac{3 - 2}{n} = \frac{1}{n} Therefore, \int_{2}^{3} e^{- x} d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (2) + f (2 + h) + . . . . . . . . . . + f (2 + (n - 1) h)] = \lim_{h \to 0} h [e^{- 2} + e^{- (2 + h)} + e^{- (2 + 2 h)} + . . . . . . . + e^{- (2 + (n - 1) h)}] = \lim_{h \to 0} h e^{- 2} [\frac{{(e^{- h})}^{n} - 1}{e^{- h} - 1}] = \lim_{h \to 0} e^{- 2} [\frac{e^{- 1} - 1}{\frac{e^{- h} - 1}{- h}}] \times - 1 (Since n h = 1) = (e^{- 2} - e^{- 3})$

Page No 19.117:

Question 66:

$\int_{1}^{3} (2 x^{2} + 5 x) d x$

Answer:

$Here, a = 1, b = 3, f (x) = 2 x^{2} + 5 x, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, \int_{1}^{3} (2 x^{2} + 5 x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [2 + 5 + 2 {(1 + h)}^{2} + 5 (1 + h) + 2 {(1 + 2 h)}^{2} + 5 (1 + 2 h) + . . . . . . . . . + 2 {((n - 1) h)}^{2} + 5 ((n - 1) h)] = \lim_{h \to 0} h [2 n + 2 h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2}) + 4 h (1 + 2 + . . . . . . . . . . . . (n - 1)) + 5 n + 5 h (1 + 2 + . . . . . . . . . . . . (n - 1))] = \lim_{h \to 0} h [7 n + 2 h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 9 h \frac{n (n - 1)}{2}] = \lim_{n \to 0 \infty} [14 + \frac{8}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + 18 (1 - \frac{1}{n})] = 14 + \frac{16}{3} + 18 = \frac{112}{3}$

Page No 19.117:

Question 67:

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

Answer:

$Here a = 1, b = 3, f (x) = x^{2} + 3 x, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, \int_{1}^{3} (x^{2} + 3 x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [1 + 3 + {(1 + h)}^{2} + 3 (1 + h) + {(1 + 2 h)}^{2} + 3 (1 + 2 h) + . . . . . . . . . + {((n - 1) h)}^{2} + 3 ((n - 1) h)] = \lim_{h \to 0} h [n + h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2}) + 2 h (1 + 2 + . . . . . . . . . . . . (n - 1)) + 3 n + 3 h (1 + 2 + . . . . . . . . . . . . (n - 1))] = \lim_{h \to 0} h [4 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 5 h \frac{n (n - 1)}{2}] = \lim_{n \to 0 \infty} [8 + \frac{4}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + 10 (1 - \frac{1}{n})] = 8 + \frac{8}{3} + 10 = \frac{62}{3}$

Page No 19.117:

Question 68:

$\int_{0}^{2} (x^{2} + 2) d x$

Answer:

$Here a = 0, b = 2, f (x) = x^{2} + 2, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, \int_{0}^{2} (x^{2} + 2) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [0 + 2 + {(0 + h)}^{2} + 2 + {(0 + 2 h)}^{2} + 2 + . . . . . . . . . + {((n - 1) h)}^{2} + 2] = \lim_{h \to 0} h [2 n + h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2})] = \lim_{h \to 0} h [2 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to 0 \infty} [4 + \frac{4}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n})] = 4 + \frac{8}{3} = \frac{20}{3}$

Page No 19.117:

Question 69:

$\int_{0}^{3} (x^{2} + 1) d x$

Answer:

$Here, a = 0, b = 3, f (x) = x^{2} + 1, h = \frac{3 - 0}{n} = \frac{3}{n} Therefore, \int_{0}^{3} (x^{2} + 1) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [0 + 1 + h^{2} + 1 + {(2 h)}^{2} + 1 + . . . . . . . . . + {((n - 1) h)}^{2} + 1] = \lim_{h \to 0} h [n + h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2})] = \lim_{h \to 0} h [n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{h \to 0} [3 + \frac{9}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n})] = 3 + 9 = 12$

Page No 19.118:

Question 1:

$\int_{0}^{1} \sqrt{x (1 - x)} d x$ equals

(a) π/2
(b) π/4
(c) π/6
(d) π/8

Answer:

(d) $π$ /8

$Let, I = \int_{0}^{1} \sqrt{x (1 - x)} d x = \int_{0}^{1} \sqrt{x - x^{2}} d x = \int_{0}^{1} \sqrt{\frac{1}{4} - (x^{2} - x + \frac{1}{4})} d x = \int_{0}^{1} \sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{1}{2})}^{2}} d x = {[\frac{(x - \frac{1}{2})}{2} \sqrt{x - x^{2}} + \frac{1}{2} \times \frac{1}{4} \sin^{- 1} (2 x - 1)]}_{0}^{1} = \frac{1}{8} {[\sin^{- 1} (1) - \sin^{- 1} (- 1)]}_{0}^{1} = \frac{1}{8} [\frac{π}{2} + \frac{π}{2}] = \frac{π}{8}$

Page No 19.118:

Question 2:

$\int_{0}^{π} \frac{1}{1 + \sin x} d x$ equals

(a) 0
(b) 1/2
(c) 2
(d) 3/2

Answer:

(c) 2

$\int_{0}^{π} \frac{1}{1 + \sin x} d x = \int_{0}^{π} \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} d x = \int_{0}^{π} \frac{1 - \sin x}{1 - \sin^{2} x} d x = \int_{0}^{π} \frac{1 - \sin x}{\cos^{2} x} d x = \int_{0}^{π} (s e c^{2} x - s e c x \tan x) d x = {[\tan x - s e c x]}_{0}^{π} = 0 + 1 - 0 + 1 = 2$

Page No 19.118:

Question 3:

The value of $\int_{0}^{π} \frac{x \tan x}{\sec x + \cos x} d x$ is
(a) $\frac{π^{2}}{4}$

(b) $\frac{π^{2}}{2}$

(c) $\frac{3 π^{2}}{2}$

(d) $\frac{π^{2}}{3}$

Answer:

$(a) \frac{π^{2}}{4}$ π24

$Let I = \int_{0}^{π} \frac{x \tan x}{s e c x + \cos x} d x (i) = \int_{0}^{π} \frac{(π - x) \tan (π - x)}{s e c (π - x) + \cos (π - x)} d x = \int_{0}^{π} \frac{(π - x) tanx}{s e c x + \cos x} d x (i i) Adding (i) and (i i) 2 I = \int_{0}^{π} \frac{x \tan x}{s e c x + \cos x} + \frac{(π - x) tanx}{s e c x + \cos x} d x = \int_{0}^{π} \frac{π \tan x}{s e c x + \cos x} d x = π \int_{0}^{π} \frac{sinx}{1 + \cos^{2} x} dx = - π {[\tan^{- 1} cosx]}_{0}^{π} = - π (- \frac{π}{4} - \frac{π}{4}) = \frac{π^{2}}{2} Hence I = \frac{π^{2}}{4}$ $We have, I = \int_{0}^{π} \frac{x \tan x}{\sec x + \cos x} d x . . . . . (1) = \int_{0}^{π} \frac{(π - x) \tan (π - x)}{\sec (π - x) + \cos (π - x)} d x = \int_{0}^{π} \frac{(π - x) tanx}{\sec x + \cos x} d x . . . . . (2) Adding (1) and (2), we get 2 I = \int_{0}^{π} [\frac{x \tan x}{\sec x + \cos x} + \frac{(π - x) tanx}{\sec x + \cos x}] d x \Rightarrow I = \frac{1}{2} \int_{0}^{π} \frac{π \tan x}{\sec x + \cos x} d x = \frac{π}{2} \int_{0}^{π} \frac{sinx}{1 + \cos^{2} x} dx Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t When x \to 0; t \to 1 and x \to π; t \to - 1 \Rightarrow I = \frac{π}{2} \int_{1}^{- 1} \frac{- d t}{1 + t^{2}} = \frac{π}{2} \int_{- 1}^{1} \frac{d t}{1 + t^{2}} = \frac{π}{2} {[\tan^{- 1} t]}_{- 1}^{1} = \frac{π}{2} [\tan^{- 1} (1) - \tan^{- 1} (- 1)] = \frac{π}{2} [\frac{π}{4} - (- \frac{π}{4})] = \frac{π}{2} \times \frac{π}{2} = \frac{π^{2}}{4} Hence I = \frac{π^{2}}{4}$

Page No 19.119:

Question 4:

The value of $\int_{0}^{2 π} \sqrt{1 + \sin \frac{x}{2}} d x$ is
(a) 0
(b) 2
(c) 8
(d) 4

Answer:

(c) 8

$\int_{0}^{2 π} \sqrt{1 + \sin \frac{x}{2}} d x = \int_{0}^{2 π} \sqrt{\sin^{2} \frac{x}{4} + \cos^{2} \frac{x}{4} + 2 \sin \frac{x}{4} \cos \frac{x}{4}} d x = \int_{0}^{2 π} (\sin \frac{x}{4} + \cos \frac{x}{4}) d x = {[\frac{- \cos \frac{x}{4}}{\frac{1}{4}} + \frac{\sin \frac{x}{4}}{\frac{1}{4}}]}_{0}^{2 π} = 4 {[\sin \frac{x}{4} - \cos \frac{x}{4}]}_{0}^{2 π} = 4 [\sin \frac{2 π}{4} - \cos \frac{2 π}{4} - \sin 0 + \cos 0] = 4 [\sin \frac{π}{2} - \cos \frac{π}{2} - 0 + 1] = 4 [1 - 0 - 0 + 1] = 4 \times 2 = 8$

Page No 19.119:

Question 5:

The value of the integral $\int_{0}^{π / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$ is
(a) 0
(b) π/2
(c) π/4
(d) none of these

Answer:

(c) π/4

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cos (\frac{π}{2} - x)}}{\sqrt{\cos (\frac{π}{2} - x)} + \sqrt{\sin (\frac{π}{2} - x)}} d x = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} [\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} + \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}] d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 19.119:

Question 6:

$\int_{0}^{\infty} \frac{1}{1 + e^{x}} d x$ equals

(a) log 2 − 1
(b) log 2
(c) log 4 − 1
(d) − log 2

Answer:

(b) log 2

$We have, I = \int_{0}^{\infty} \frac{1}{1 + e^{x}} d x Putting e^{x} = t \Rightarrow e^{x} d x = d t \Rightarrow d x = \frac{d t}{t} When x \to 0; t \to 1 and x \to \infty; t \to \infty ∴ I = \int_{1}^{\infty} \frac{1}{t (1 + t)} d t = \int_{1}^{\infty} \frac{1}{t + t^{2}} d t = \int_{1}^{\infty} \frac{1}{{(t + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d t$

$= \frac{1}{2 \times \frac{1}{2}} {[\log |\frac{t + \frac{1}{2} - \frac{1}{2}}{t + \frac{1}{2} + \frac{1}{2}}|]}_{1}^{\infty} = {[\log |\frac{t}{t + 1}|]}_{1}^{\infty} = {[\log |\frac{\frac{t}{t}}{\frac{t}{t} + \frac{1}{t}}|]}_{1}^{\infty} = {[\log |\frac{1}{1 + \frac{1}{t}}|]}_{1}^{\infty} = \log \frac{1}{1 + 0} - \log \frac{1}{1 + 1} = \log (1) - \log (\frac{1}{2}) = 0 - (- \log 2) = \log 2$

Page No 19.119:

Question 7:

$\int_{0}^{π^{2} / 4} \frac{\sin \sqrt{x}}{\sqrt{x}} d x$ equals
(a) 2
(b) 1
(c) π/4
(d) π²/8

Answer:

(a) 2

$\int_{0}^{\frac{π^{2}}{4}} \frac{\sin \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t, then \frac{1}{2 \sqrt{x}} d x = d t When x = 0, t = 0, x = \frac{π^{2}}{4}, t = \frac{π}{2} Therefore the integral becomes \int_{0}^{\frac{π}{2}} 2 sint dt = - 2 {[cost]}_{0}^{\frac{π}{2}} = 2$

Page No 19.119:

Question 8:

$\int_{0}^{π / 2} \frac{\cos x}{(2 + \sin x) (1 + \sin x)} d x$ equals

(a) $\log (\frac{2}{3})$

(b) $\log (\frac{3}{2})$

(c) $\log (\frac{3}{4})$

(d) $\log (\frac{4}{3})$

Answer:

(d) $\log (\frac{4}{3})$

$Let, I = \int_{0}^{\frac{π}{2}} \frac{\cos x}{(2 + \sin x) (1 + \sin x)} d x Let \sin x, then \cos x d x = d t When x = 0, t = 0, x = \frac{π}{2}, t = 1 Therefore the integral becomes I = \int_{0}^{1} \frac{dt}{(2 + t) (1 + t)} = \int_{0}^{1} [\frac{- 1}{2 + t} + \frac{1}{1 + t}] dt = {[- \log (2 + t) + \log (1 + t)]}_{0}^{1} = {[\log (1 + t) - \log (2 + t)]}_{0}^{1} = \log 2 - \log 3 - \log 1 + \log 2 = \log \frac{4}{3}$

Page No 19.119:

Question 9:

$\int_{0}^{π / 2} \frac{1}{2 + \cos x} d x$ equals

(a) $\frac{1}{3} \tan^{- 1} (\frac{1}{\sqrt{3}})$

(b) $\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}})$

(c) $\sqrt{3} \tan^{- 1} (\sqrt{3})$

(d) $2 \sqrt{3} \tan^{- 1} \sqrt{3}$

Answer:

$(b) \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}})$

$We have, I = \int_{0}^{\frac{π}{2}} \frac{1}{2 + \cos x} d x = \int_{0}^{\frac{π}{2}} \frac{1}{2 + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{2 + 2 \tan^{2} \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} \frac{x}{2}}{3 + \tan^{2} \frac{x}{2}} d x$

$Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t When, x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1 ∴ I = \int_{0}^{1} \frac{2}{3 + t^{2}} d t = 2 \int_{0}^{1} \frac{1}{{(\sqrt{3})}^{2} + t^{2}} d t = \frac{2}{\sqrt{3}} {[\tan^{- 1} \frac{t}{\sqrt{3}}]}_{0}^{1} = \frac{2}{\sqrt{3}} [\tan^{- 1} \frac{1}{\sqrt{3}} - \tan^{- 1} \frac{0}{\sqrt{3}}] = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}})$
3√tan−1(13√)

Page No 19.119:

Question 10:

$\int_{0}^{π} \sqrt{\frac{1 - x}{1 + x}} d x =$

(a) $\frac{π}{2}$

(b) $\frac{π}{2} - 1$

(c) $\frac{π}{2} + 1$

(d) π + 1

Answer:

Disclaimer: None of the given option is correct.

$We have, I = \int_{0}^{π} \sqrt{\frac{1 - x}{1 + x}} d x = \int_{0}^{π} [\sqrt{\frac{1 - x}{1 + x}} \times \frac{\sqrt{1 - x}}{\sqrt{1 - x}}] d x = \int_{0}^{π} \frac{1 - x}{\sqrt{1 - x^{2}}} d x = \int_{0}^{π} \frac{1}{\sqrt{1 - x^{2}}} d x - \int_{0}^{π} \frac{x}{\sqrt{1 - x^{2}}} d x Putting 1 - x^{2} = t \Rightarrow - 2 x d x = d t \Rightarrow x d x = \frac{- d t}{2} When x \to 0; t \to 1 and x \to π; t \to 1 - π^{2} ∴ I = \int_{0}^{π} \frac{1}{\sqrt{1 - x^{2}}} d x - \int_{1}^{(1 - π^{2})} \frac{- d t}{2 \sqrt{t}} = {[\sin^{- 1} x]}_{0}^{π} + \frac{2}{2} {[\sqrt{t}]}_{1}^{(1 - π^{2})} = [0 - 0] + [\sqrt{1 - π^{2}} - \sqrt{1}] = \sqrt{1 - π^{2}} - 1$

Page No 19.119:

Question 11:

$\int_{0}^{π} \frac{1}{a + b \cos x} d x =$

(a) $\frac{π}{\sqrt{a^{2} - b^{2}}}$

(b) $\frac{π}{a b}$

(c) $\frac{π}{a^{2} + b^{2}}$

(d) (a + b) π

Answer:

$(a) \frac{π}{\sqrt{a^{2} - b^{2}}} We have, I = \int_{0}^{π} \frac{1}{a + b \cos x} d x = \int_{0}^{π} \frac{1}{a + b \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$

$= \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{a (1 + \tan^{2} \frac{x}{2}) + b (1 - \tan^{2} \frac{x}{2})} d x = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{(a + b) + (a - b) \tan^{2} \frac{x}{2}} d x = \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{(a + b) + (a - b) \tan^{2} \frac{x}{2}} d x$

$Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t When x \to 0; t \to 0 and x \to π; t \to \infty ∴ I = \int_{0}^{\infty} \frac{2 d t}{(a + b) + (a - b) t^{2}} = \frac{2}{a - b} \int_{0}^{\infty} \frac{1}{(\frac{a + b}{a - b}) + t^{2}} d t$

$= \frac{2}{(a - b)} \int_{0}^{\infty} \frac{1}{{(\sqrt{\frac{a + b}{a - b}})}^{2} + t^{2}} d t = \frac{2}{(a - b)} \times \sqrt{\frac{a - b}{a + b}} {[\tan^{- 1} \frac{t}{\sqrt{\frac{a + b}{a - b}}}]}_{0}^{\infty} = \frac{2}{\sqrt{a^{2} - b^{2}}} [\frac{π}{2} - 0] = \frac{2}{\sqrt{a^{2} - b^{2}}} [\frac{π}{2}] = \frac{π}{\sqrt{a^{2} - b^{2}}}$

Page No 19.119:

Question 12:

$\int_{π / 6}^{π / 3} \frac{1}{1 + \sqrt{\cot} x} d x$ is

(a) π/3
(b) π/6
(c) π/12
(d) π/2

Answer:

$(c) \frac{π}{12} Let, I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{c o t x}} d x . . . (i) = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{c o t (\frac{π}{3} + \frac{π}{6} - x)}} d x [Using \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{\tan x}} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} [\frac{1}{1 + \sqrt{c o t x}} + \frac{1}{1 + \sqrt{\tan x}}] d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{2 + \sqrt{c o t x} + \sqrt{\tan x}}{(1 + \sqrt{c o t x}) (1 + \sqrt{\tan x})} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} [\frac{2 + \sqrt{c o t x} + \sqrt{\tan x}}{2 + \sqrt{c o t x} + \sqrt{\tan x}}] d x = \int_{\frac{π}{6}}^{\frac{π}{3}} d x = {[x]}_{\frac{π}{6}}^{\frac{π}{3}} = \frac{π}{3} - \frac{π}{6} = \frac{π}{6} Hence, I = \frac{π}{12}$

Page No 19.119:

Question 13:

Given that $\int_{0}^{\infty} \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2}) (x^{2} + c^{2})} d x = \frac{π}{2 (a + b) (b + c) (c + a)},$ the value of $\int_{0}^{\infty} \frac{d x}{(x^{2} + 4) (x^{2} + 9)},$ is

(a) $\frac{π}{60}$

(b) $\frac{π}{20}$

(c) $\frac{π}{40}$

(d) $\frac{π}{80}$

Answer:

$(b) \frac{π}{60} \int_{0}^{\infty} \frac{1}{(x^{2} + 4) (x^{2} + 9)} d x = \frac{1}{5} \int_{0}^{\infty} \frac{1}{(x^{2} + 4)} - \frac{1}{(x^{2} + 9)} d x = \frac{1}{5} {[\frac{1}{2} \tan^{- 1} \frac{x}{2} - \frac{1}{3} \tan^{- 1} \frac{x}{3}]}_{0}^{\infty} = \frac{1}{5} [\frac{1}{2} \times \frac{π}{2} - \frac{1}{3} \times \frac{π}{2}] = \frac{1}{5} \times \frac{π}{12} = \frac{π}{60}$

Page No 19.120:

Question 14:

$\int_{1}^{e} \log x d x =$

(a) 1
(b) e − 1
(c) e + 1
(d) 0

Answer:

(a) 1

$\int_{1}^{e} \log x d x = \int_{1}^{e} \log x x^{0} d x = {[x \log x]}_{1}^{e} - \int_{1}^{e} \frac{1}{x} x d x = {[x \log x]}_{1}^{e} - {[x]}_{1}^{e} = (e - 0) - (e - 1) = e - e + 1 = 1$

Page No 19.120:

Question 15:

$\int_{1}^{\sqrt{3}} \frac{1}{1 + x^{2}} d x$ is equal to

(a) $\frac{π}{12}$

(b) $\frac{π}{6}$

(c) $\frac{π}{4}$

(d) $\frac{π}{3}$

Answer:

(a) $\frac{π}{12}$

$\int_{1}^{\sqrt{3}} \frac{1}{1 + x^{2}} d x = {[\tan^{- 1} x]}_{1}^{\sqrt{3}} = \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$
30. →

Page No 19.120:

Question 16:

$\int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} d x =$

(a) $\frac{π}{12} + \log (2 \sqrt{2})$

(b) $\frac{π}{2} + \log (2 \sqrt{2})$

(c) $\frac{π}{6} + \log (2 \sqrt{2})$

(d) $\frac{π}{3} + \log (2 \sqrt{2})$

Answer:

$(a) \frac{π}{12} + \log (2 \sqrt{2})$

$We have, I = \int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} d x I = \int_{0}^{3} \frac{3 x}{x^{2} + 9} d x + \int_{0}^{3} \frac{1}{x^{2} + 9} d x I_{1} = \int_{0}^{3} \frac{3 x}{x^{2} + 9} d x and I_{2} = \int_{0}^{3} \frac{1}{x^{2} + 9} d x Putting x^{2} + 9 = t in I_{1} \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} When x \to 0; t \to 9 and x \to 3; t \to 18 ∴ I = \int_{9}^{18} \frac{3 d t}{2 t} + \int_{0}^{3} \frac{1}{x^{2} + 9} d x = \frac{3}{2} \int_{9}^{18} \frac{d t}{t} + \int_{0}^{3} \frac{1}{x^{2} + 3^{2}} d x = \frac{3}{2} {[\log (t)]}_{9}^{18} + \frac{1}{3} {[\tan^{- 1} (\frac{x}{3})]}_{0}^{3} = \frac{3}{2} [\log 18 - \log 9] + \frac{1}{3} (\frac{π}{4} - 0) = \frac{3}{2} [\log \frac{18}{9}] + \frac{π}{12} = \frac{3}{2} [\log 2] + \frac{π}{12} = \log (\sqrt{8}) + \frac{π}{12} = \log (2 \sqrt{2}) + \frac{π}{12} = \frac{π}{12} + \log (2 \sqrt{2})$

Page No 19.120:

Question 17:

The value of the integral $\int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x$ is

(a) $\frac{π}{2}$

(b) $\frac{π}{4}$

(c) $\frac{π}{6}$

(d) $\frac{π}{3}$

Answer:

$(b) \frac{π}{4}$

$We have, I = \int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x Putting x = \tan θ \Rightarrow d x = \sec^{2} θ d θ When x \to 0; θ \to 0 and x \to \infty; θ \to \frac{π}{2} Now, integral becomes$

$I = \int_{0}^{\frac{π}{2}} \frac{\tan θ}{(1 + \tan θ) \sec^{2} θ} \sec^{2} θ d θ = \int_{0}^{\frac{π}{2}} \frac{\tan θ}{1 + \tan θ} d θ = \int_{0}^{\frac{π}{2}} \frac{\frac{\sin θ}{co s θ}}{1 + \frac{\sin θ}{\cos θ}} d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sin θ + \cos θ} d θ . . . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - θ)}{\sin (\frac{π}{2} - θ) + \cos (\frac{π}{2} - θ)} d θ [∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\cos θ + \sin θ} d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ + \cos θ} d θ . . . . . (2)$

$Adding (1) and (2), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin θ + \cos θ}{\sin θ + \cos θ} d θ \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} d θ \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4} ∴ \int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x = \frac{π}{4}$

Page No 19.120:

Question 18:

$\int_{- π / 2}^{π / 2} \sin |x| d x$ is equal to

(a) 1
(b) 2
(c) − 1
(d) − 2

Answer:

(b) 2

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \sin |x| d x = - \int_{- \frac{π}{2}}^{0} \sin x d x + \int_{0}^{\frac{π}{2}} \sin x d x = - {[- \cos x]}_{- \frac{π}{2}}^{0} + {[- \cos x]}_{0}^{\frac{π}{2}} = 1 - 0 - 0 + 1 = 2$

Page No 19.120:

Question 19:

$\int_{0}^{π / 2} \frac{1}{1 + \tan x} d x$ is equal to

(a) $\frac{π}{4}$

(b) $\frac{π}{3}$

(c) $\frac{π}{2}$

(d) π

Answer:

(a) $\frac{π}{4}$

$Let, I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t x} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{\frac{π}{2}} [\frac{1}{1 + \tan x} + \frac{1}{1 + c o t x}] d x = \int_{0}^{\frac{π}{2}} [\frac{(1 + c o t x) + (1 + \tan x)}{(1 + \tan x) (1 + c o t x)}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan x + \cot x}{1 + \tan x + c o t x + \tan x \cot x}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan x + \cot x}{2 + \tan x + \cot x}] d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence, I = \frac{π}{4}$

Page No 19.120:

Question 20:

The value of $\int_{0}^{π / 2} \cos x e^{\sin x} d x$ is
(a) 1
(b) e − 1
(c) 0
(d) − 1

Answer:

(b) e − 1

$Let, I = \int_{0}^{\frac{π}{2}} \cos x e^{\sin x} d x Let \sin x = t, then \cos x d x = d t When x = 0, t = 0 and x = \frac{π}{2}, t = 1 Therefore the integral becomes I = \int_{0}^{1} e^{t} dt = {[e^{t}]}_{0}^{1} = e - 1$

Page No 19.120:

Question 21:

If $\int_{0}^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8},$ then a equals

(a) $\frac{π}{2}$

(b) $\frac{1}{2}$

(c) $\frac{π}{4}$

(d) 1

Answer:

(b) $\frac{1}{2}$

$\int_{0}^{α} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8} \Rightarrow \int_{0}^{α} \frac{1}{1 + {(2 x)}^{2}} d x = \frac{π}{8} \Rightarrow \frac{1}{2} {[\tan^{- 1} 2 x]}_{0}^{α} = \frac{π}{8} \Rightarrow \frac{1}{2} \tan^{- 1} 2 α = \frac{π}{8} \Rightarrow 2 α = \tan \frac{π}{4} \Rightarrow 2 α = 1 ∴ α = \frac{1}{2}$ 4430. →

Page No 19.120:

Question 22:

If $\int_{0}^{1} f (x) d x = 1, \int_{0}^{1} x f (x) d x = a, \int_{0}^{1} x^{2} f (x) d x = a^{2}, then \int_{0}^{1} {(a - x)}^{2} f (x) d x$ equals

(a) 4a²
(b) 0
(c) 2a²
(d) none of these

Answer:

(b) 0

$\int_{0}^{1} {(a - x)}^{2} f (x) d x = a^{2} \int_{0}^{1} f (x) d x + \int_{0}^{1} x^{2} f (x) d x - 2 a \int_{0}^{1} x f (x) d x = a^{2} \times 1 + a^{2} - 2 a a (As per given values) = 2 a^{2} - 2 a^{2} = 0$

Page No 19.120:

Question 23:

The value of $\int_{- π}^{π} \sin^{3} x \cos^{2} x d x$ is

(a) $\frac{π^{4}}{2}$

(b) $\frac{π^{4}}{4}$

(c) 0

(d) none of these

Answer:

(c) 0

$\int_{- π}^{π} \sin^{3} x \cos^{2} x d x = \int_{- π}^{π} \sin x (1 - \cos^{2} x) \cos^{2} x d x Let \cos x = t, then - \sin x d x = d t, When, x = - π, t = - 1, x = π, t = - 1 Therefore the integral becomes \int_{- 1}^{- 1} - (1 - t^{2}) t^{2} dt = 0$

Page No 19.121:

Question 24:

$\int_{π / 6}^{π / 3} \frac{1}{\sin 2 x} d x$ is equal to

(a) log_e 3

(b) $\log_{e} \sqrt{3}$

(c) $\frac{1}{2} \log (- 1)$

(d) log (−1)

Answer:

(b) $\log_{e} \sqrt{3}$

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{\sin 2 x} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \cos e c 2 x d x = \frac{1}{2} \int_{\frac{π}{6}}^{\frac{π}{3}} 2 \cos e c 2 x d x = \frac{- 1}{2} {[\log (\cos e c 2 x + c o t 2 x)]}_{\frac{π}{6}}^{\frac{π}{3}} = \frac{- 1}{2} [- 2 \log \sqrt{3}] = \log \sqrt{3}$

Page No 19.121:

Question 25:

$\int_{- 1}^{1} |1 - x| d x$ is equal to

(a) −2
(b) 2
(c) 0
(d) 4

Answer:

(b) 2

$\int_{- 1}^{1} |1 - x| d x = \int_{- 1}^{0} (1 - x) d x + \int_{0}^{1} (1 - x) d x = {[x - \frac{x^{2}}{2}]}_{- 1}^{0} + {[x - \frac{x^{2}}{2}]}_{0}^{1} = 0 + 1 + \frac{1}{2} + 1 - \frac{1}{2} - 0 = 2$

Page No 19.121:

Question 26:

The derivative of $f (x) = \int_{x^{2}}^{x^{3}} \frac{1}{\log_{e} t} d t, (x > 0),$ is

(a) $\frac{1}{3 \ln x}$

(b) $\frac{1}{3 \ln x} - \frac{1}{2 \ln x}$

(c) (ln x)⁻¹ x (x − 1)

(d) $\frac{3 x^{2}}{\ln x}$

Answer:

(c) (ln x)⁻¹ x (x − 1)

Using Newton Leibnitz formula

$\begin{matrix} f^{'} (x) = \frac{1}{\log_{e} x^{3}} (3 x^{2}) - \frac{1}{\log_{e} x^{2}} (2 x) \\ = \frac{3 x^{2}}{3 \ln x} - \frac{2 x}{2 \ln x} \\ = \frac{x^{2}}{\ln x} - \frac{x}{\ln x} \\ = \frac{1}{\ln x} x (x - 1) \\ = {(\ln x)}^{- 1} x (x - 1) \end{matrix}$

Page No 19.121:

Question 27:

If $I_{10} = \int_{0}^{π / 2} x^{10} \sin x d x,$ then the value of I₁₀ + 90I₈ is

(a) $9 {(\frac{π}{2})}^{9}$

(b) $10 {(\frac{π}{2})}^{9}$

(c) ${(\frac{π}{2})}^{9}$

(d) $9 {(\frac{π}{2})}^{8}$

Answer:

$(b) 10 {(\frac{π}{2})}^{9} We have, I_{10} = \int_{0}^{π / 2} x^{10} \sin x d x = {[x^{10} (- \cos x)]}_{0}^{\frac{π}{2}} - \int_{0}^{π / 2} [10 x^{9} \int \sin x d x] d x = {[- x^{10} \cos x]}_{0}^{\frac{π}{2}} - 10 \int_{0}^{π / 2} x^{9} (- \cos x) d x = - {[x^{10} \cos x]}_{0}^{\frac{π}{2}} + 10 \int_{0}^{π / 2} x^{9} \cos x d x = - {[x^{10} \cos x]}_{0}^{\frac{π}{2}} + 10 {[x^{9} \sin x]}_{0}^{\frac{π}{2}} - 10 \int_{0}^{π / 2} 9 x^{8} \sin x d x = - [{(\frac{π}{2})}^{10} \times 0 - 0^{10} \cos 0] + 10 [{(\frac{π}{2})}^{9} \times 1 - 0^{9} \times 0] - 90 \int_{0}^{π / 2} x^{8} \sin x d x = 10 [{(\frac{π}{2})}^{9} \times 1] - 90 I_{8} = 10 {(\frac{π}{2})}^{9} - 90 I_{8} ∴ I_{10} + 90 I_{8} = 10 {(\frac{π}{2})}^{9}$

Page No 19.121:

Question 28:

$\int_{0}^{1} \frac{x}{{(1 - x)}^{54}} d x =$

(a) $\frac{15}{16}$

(b) $\frac{3}{16}$

(c) $- \frac{3}{16}$

(d) $- \frac{16}{3}$

Answer:

Disclaimer: The question given is not correct because the function provided does not converge in the given domain.

Page No 19.121:

Question 29:

$\int_{0}^{\frac{π}{2}} \sqrt{1 - \sin 2 x} d x$ is equal to

(a) $2 \sqrt{2}$

(b) $2 (\sqrt{2} + 1)$

(c) 2

(b) $2 (\sqrt{2} - 1)$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \sqrt{1 - \sin 2 x} d x = \int_{0}^{\frac{π}{2}} \sqrt{\cos^{2} x + \sin^{2} x - 2 \sin x \cos x} d x (using the identity : \cos^{2} x + \sin^{2} x = 1 and 2 \sin x \cos x = \sin 2 x) = \int_{0}^{\frac{π}{2}} \sqrt{{(\cos x - \sin x)}^{2}} d x = \int_{0}^{\frac{π}{2}} |\cos x - \sin x| d x = \int_{0}^{\frac{π}{4}} |\cos x - \sin x| d x + \int_{\frac{π}{4}}^{\frac{π}{2}} |\cos x - \sin x| d x = \int_{0}^{\frac{π}{4}} (\cos x - \sin x) d x + \int_{\frac{π}{4}}^{\frac{π}{2}} - (\cos x - \sin x) d x = \int_{0}^{\frac{π}{4}} (\cos x - \sin x) d x + \int_{\frac{π}{4}}^{\frac{π}{2}} (\sin x - \cos x) d x = {[\sin x + \cos x]}_{0}^{\frac{π}{4}} + {[- \cos x - \sin x]}_{\frac{π}{4}}^{\frac{π}{2}} = [\sin \frac{π}{4} + \cos \frac{π}{4} - \sin 0 - \cos 0] + [- \cos \frac{π}{2} - \sin \frac{π}{2} - (- \cos \frac{π}{4} - \sin \frac{π}{4})] = [\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 0 - 1] + [- 0 - 1 - (- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})] = [\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1] + [- 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}] = \frac{2}{\sqrt{2}} - 1 + - 1 + \frac{2}{\sqrt{2}} = \sqrt{2} - 1 + - 1 + \sqrt{2} = 2 \sqrt{2} - 2 = 2 (\sqrt{2} - 1)$

Hence, the correct option is (d).

Page No 19.121:

Question 30:

The value of the integral $\int_{- 2}^{2} |1 - x^{2}| d x$ is
(a) 4
(b) 2
(c) −2
(d) 0

Answer:

(a) 4

$We have, I = \int_{- 2}^{2} |1 - x^{^{2}}| d x |1 - x^{^{2}}| = \{\begin{cases} \begin{array}{l} - (1 - x^{^{2}}), - 2 < x < - 1 \\ (1 - x^{^{2}}), - 1 < x < 1 \end{array} \\ - (1 - x^{^{2}}), 1 < x < 2 \end{cases} ∴ I = \int_{- 2}^{- 1} |1 - x^{^{2}}| d x + \int_{- 1}^{1} |1 - x^{^{2}}| d x + \int_{1}^{2} |1 - x^{^{2}}| d x = \int_{- 2}^{- 1} - (1 - x^{^{2}}) d x + \int_{- 1}^{1} (1 - x^{^{2}}) d x + \int_{1}^{2} - (1 - x^{^{2}}) d x = - \int_{- 2}^{- 1} (1 - x^{^{2}}) d x + \int_{- 1}^{1} (1 - x^{^{2}}) d x - \int_{1}^{2} (1 - x^{^{2}}) d x = - {[x - \frac{x^{3}}{3}]}_{- 2}^{- 1} + {[x - \frac{x^{3}}{3}]}_{- 1}^{1} - {[x - \frac{x^{3}}{3}]}_{1}^{2} = - [- 1 + \frac{1}{3} + 2 - \frac{8}{3}] + [1 - \frac{1}{3} + 1 - \frac{1}{3}] - [2 - \frac{8}{3} - 1 + \frac{1}{3}] = - [1 - \frac{7}{3}] + [2 - \frac{2}{3}] - [1 - \frac{7}{3}] = - 1 + \frac{7}{3} + 2 - \frac{2}{3} - 1 + \frac{7}{3} = 4$

Page No 19.121:

Question 31:

$\int_{0}^{π / 2} \frac{1}{1 + \cot^{3} x} d x$ is equal to

(a) 0
(b) 1
(c) π/2
(d) π/4

Answer:

(d) π/4

$We have, I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \cot^{3} x} d x . . . . . (1) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \cot^{3} (\frac{π}{2} - x)} d x ∴ I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{3} x} d x . . . . . (2) Adding (1) and (2) we get 2 I = \int_{0}^{\frac{π}{2}} [\frac{1}{1 + c o t^{3} x} + \frac{1}{1 + \tan^{3} x}] d x$

$= \int_{0}^{\frac{π}{2}} [\frac{1 + \tan^{3} x + 1 + co t^{3} x}{(1 + c o t^{3} x) (1 + \tan^{3} x)}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan^{3} x + co t^{3} x}{1 + \tan^{3} x + co t^{3} x + c o t^{3} x \tan^{3} x}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan^{3} x + co t^{3} x}{1 + \tan^{3} x + co t^{3} x + 1}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan^{3} x + co t^{3} x}{2 + \tan^{3} x + co t^{3} x}] d x = \int_{0}^{\frac{π}{2}} [1] d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 19.121:

Question 32:

$\int_{0}^{π / 2} \frac{\sin x}{\sin x + \cos x} d x$ equals to

(a) π
(b) π/2
(c) π/3
(d) π/4

Answer:

(d) π/4

$We have, I = \int_{0}^{\frac{π}{2}} \frac{\sin x}{\sin x + \cos x} d x . . . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\cos x + \sin x} d x ∴ I = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\sin x + \cos x} d x . . . . . (2) Adding (1) and (2), we get 2 I = \int_{0}^{\frac{π}{2}} [\frac{\sin x}{\sin x + \cos x} + \frac{\cos x}{\cos x + \sin x}] d x = \int_{0}^{\frac{π}{2}} [\frac{\sin x + \cos x}{\sin x + \cos x}] d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 19.121:

Question 33:

$\int_{0}^{1} \frac{d}{d x} \{\sin^{- 1} (\frac{2 x}{1 + x^{2}})\} d x$ is equal to

(a) 0
(b) π
(c) π/2
(d) π/4

Answer:

(c) π/2

$We have, I = \int_{0}^{1} \frac{d}{d x} \{\sin^{- 1} (\frac{2 x}{1 + x^{2}})\} d x We know since \int f' (x) = f (x) f (x) = s i n^{- 1} (\frac{2 x}{1 + x^{2}}) a n d f' (x) = \frac{d}{d x} \{s i n^{- 1} (\frac{2 x}{1 + x^{2}})\} Therefore, I = {[\sin^{- 1} (\frac{2 x}{1 + x^{2}})]}_{0}^{1} = \sin^{- 1} (1) - \sin^{- 1} (0) = \frac{π}{2}$

Page No 19.122:

Question 34:

$\int_{0}^{π / 2} x \sin x d x$ is equal to

(a) π/4
(b) π/2
(c) π
(d) 1

Answer:

(d) 1

$We have, I = \int_{0}^{\frac{π}{2}} x \sin x d x = {[- x \cos x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 1 (- \cos x) d x = {[- x \cos x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} \cos x d x = - {[x \cos x]}_{0}^{\frac{π}{2}} + {[\sin x]}_{0}^{\frac{π}{2}} = - [0 - 0] + [1 - 0] = 1$

Page No 19.122:

Question 35:

$\int_{0}^{π / 2} \sin 2 x \log \tan x d x$ is equal to

(a) π
(b) π/2
(c) 0
(d) 2π

Answer:

(c) 0

$I = \int_{0}^{\frac{π}{2}} \sin 2 x \log \tan x d x . . . . . (1) I = \int_{0}^{\frac{π}{2}} \sin (π - 2 x) \log \tan (\frac{π}{2} - x) d x I = \int_{0}^{\frac{π}{2}} \sin 2 x \log \cot x d x . . . . . (2) Adding (1) and (2), we get, 2 I = \int_{0}^{\frac{π}{2}} \sin 2 x (\log \tan x + \log \cot x) d x 2 I = \int_{0}^{\frac{π}{2}} \sin 2 x (\log \tan x \cot x) d x 2 I = \int_{0}^{\frac{π}{2}} \sin 2 x (\log 1) d x I = 0$

Page No 19.122:

Question 36:

The value of $\int_{0}^{π} \frac{1}{5 + 3 \cos x} d x$ is

(a) π/4
(b) π/8
(c) π/2
(d) 0

Answer:

(a) π/4

$\int_{0}^{π} \frac{1}{5 + 3 \cos x} d x = \int_{0}^{π} \frac{1}{5 + 3 \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{5 + 5 \tan^{2} \frac{x}{2} + 3 - 3 \tan^{2} \frac{x}{2}} d x = \int_{0}^{π} \frac{s e c^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x Let \tan \frac{x}{2} = t, then s e c^{2} \frac{x}{2} d x = 2 d t When x = 0, t = 0, x = π, t = \infty Therefore the integral becomes \frac{1}{2} \int_{0}^{\infty} \frac{dt}{4 + t^{2}} = \frac{1}{2} {[\tan^{- 1} \frac{t}{2}]}_{0}^{\infty} = \frac{1}{2} (\frac{π}{2} - 0) = \frac{π}{4}$

Page No 19.122:

Question 37:

$\int_{0}^{\infty} \log (x + \frac{1}{x}) \frac{1}{1 + x^{2}} d x =$

(a) π ln 2
(b) −π ln 2
(c) 0
(d) $- \frac{π}{2} \ln 2$

Answer:

(a) π ln 2

$\int_{0}^{\infty} \log (x + \frac{1}{x}) \frac{1}{1 + x^{2}} d x$
Substitute x = tan θ
⇒ dx = sec² θ dθ.
when,
x = 0 ⇒ θ = 0
$x = \infty \Rightarrow θ = \frac{π}{2} \int_{0}^{\frac{π}{2}} (\tan θ + \frac{1}{\tan θ}) \frac{1}{1 + \tan^{2} θ} \times \sec^{2} θ d θ \int_{0}^{\frac{π}{2}} \log (\frac{\tan^{2} θ + 1}{\tan θ}) \frac{1}{1 + \tan^{2} θ} \times \sec^{2} θ d θ \Rightarrow \int_{0}^{\frac{π}{2}} \log (\frac{\sec^{2} θ}{\tan θ}) \frac{1}{\sec^{2} θ} \times \sec^{2} θ d θ [∵ 1 + \tan^{2} θ = \sec^{2} θ] \Rightarrow \int_{0}^{\frac{π}{2}} \log (\frac{\sec^{2} θ}{\tan θ}) d θ \Rightarrow \int_{0}^{\frac{π}{2}} \log (\frac{1}{\sin θ . \cos θ}) d θ \Rightarrow - \int_{0}^{\frac{π}{2}} \log (\sin θ . \cos θ) d θ \Rightarrow - \int_{0}^{\frac{π}{2}} [\log \sin θ + \log \cos θ] d θ \Rightarrow - \int_{0}^{\frac{π}{2}} \log \sin θ d θ - \int_{0}^{\frac{π}{2}} \log \cos θ d θ$
Let us consider,
$\int_{0}^{\frac{π}{2}} \log \sin θ d θ = I . . . . . (i) \Rightarrow I = \int_{0}^{\frac{π}{2}} \log (\sin (\frac{π}{2} - θ)) d θ = \int_{0}^{\frac{π}{2}} \log \cos θ d θ . . . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} \log \sin θ d θ + \int_{0}^{\frac{π}{2}} \log \cos θ d θ = \int_{0}^{\frac{π}{2}} \log (\sin θ . \cos θ) d θ = \int_{0}^{\frac{π}{2}} \log (\sin 2 θ) d θ - \int_{0}^{\frac{π}{2}} \log 2 d θ Let us consider 2 θ = t 2 d θ = d t 2 I = \frac{1}{2} \int_{0}^{π} \log (\sin t) d t - \frac{π}{2} \log 2 2 I = \frac{2}{2} \int_{0}^{\frac{π}{2}} \log (\sin t) d t - \frac{π}{2} \log 2 [∵ \sin θ is positive in both 1^{st} and 2^{nd} quadrants] 2 I = I - \frac{π}{2} \log 2 2 I - I = - \frac{π}{2} \log 2 I = - \frac{π}{2} \log 2, where I = \int_{0}^{\frac{π}{2}} \log \sin θ d θ Now, - \int_{0}^{\frac{π}{2}} \log (\sin θ) d θ - \int_{0}^{\frac{π}{2}} \log \cos θ d θ - 2 \int_{0}^{\frac{π}{2}} \log \sin θ d θ = - 2 \times I = - 2 \times - \frac{π}{2} \log 2 [∵ where I = - \frac{π}{2} \log 2] = π \log 2$

Page No 19.122:

Question 38:

$\int_{0}^{2 a} f (x) d x$ is equal to

(a) $2 \int_{0}^{a} f (x) d x$

(b) 0

(c) $\int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x$

(d) $\int_{0}^{a} f (x) d x + \int_{0}^{2 a} f (2 a - x) d x$

Answer:

$(c) \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x$

$A c c o r d i n g t o t h e a d d i t i v i t y p r o p e r t y o f i n t e g r a l s, \int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x, w h e r e a < c < b u \sin g t h i s p r o p e r t y, \int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{2 a} f (x) d x . . . . . . (1) N o w, c o n s i d e r t h e i n t e g r a l, \int_{0}^{2 a} f (x) d x L e t x = 2 a - t . T h e n d x = d (2 a - t) \Rightarrow d x = - d t A l s o, x = a \Rightarrow t = a a n d x = 2 a \Rightarrow t = 0 T h e r e f o r e, \int_{a}^{2 a} f (x) d x = - \int_{a}^{0} f (2 a - t) d t \Rightarrow \int_{a}^{2 a} f (x) d x = \int_{0}^{a} f (2 a - t) d t \Rightarrow \int_{a}^{2 a} f (x) d x = \int_{0}^{a} f (2 a - x) d x S u b s t i t u t i n g t h i s i n e q u a t i o n (1) w e g e t, \int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x$ →

Page No 19.122:

Question 39:

If f (a + b − x) = f (x), then $\int_{a}^{b}$ x f (x) dx is equal to

(a) $\frac{a + b}{2} \int_{a}^{b} f (b - x) d x$

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

(c) $\frac{b - a}{2} \int_{a}^{b} f (x) d x$

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

Answer:

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

$Let, I = \int_{a}^{b} x f (x) d x . . . (i) = \int_{a}^{b} (a + b - x) f (a + b - x) d x = \int_{a}^{b} (a + b - x) f (x) d x . . . (ii) Adding (i) and (ii) 2 I = \int_{a}^{b} (x + a + b - x) f (x) d x = (a + b) \int_{a}^{b} f (x) d x Hence I = \frac{a + b}{2} \int_{a}^{b} f (x) d x$

Page No 19.122:

Question 40:

The value of $\int_{0}^{1} \tan^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x,$ is

(a) 1
(b) 0
(c) −1
(d) π/4

Answer:

(b) 0

$Let, I = \int_{0}^{1} \tan^{- 1} \frac{2 x - 1}{1 + x - x^{2}} d x . . . (i) = \int_{0}^{1} \tan^{- 1} \frac{2 (1 - x) - 1}{1 + (1 - x) - {(1 - x)}^{2}} d x = \int_{0}^{1} \tan^{- 1} \frac{1 - 2 x}{2 - x - 1 - x^{2} + 2 x} d x = \int_{0}^{1} \tan^{- 1} \frac{1 - 2 x}{1 + x - x^{2}} d x = - \int_{0}^{1} \tan^{- 1} \frac{2 x - 1}{1 + x - x^{2}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{1} \tan^{- 1} \frac{2 x - 1}{1 + x - x^{2}} d x - \int_{0}^{1} \tan^{- 1} \frac{2 x - 1}{1 + x - x^{2}} d x = 0 Hence, I = 0$

Page No 19.122:

Question 41:

The value of $\int_{0}^{π / 2} \log (\frac{4 + 3 \sin x}{4 + 3 \cos x}) d x$ is

(a) 2

(b) $\frac{3}{4}$

(c) 0

(d) −2

Answer:

(c) 0

$Let I = \int_{0}^{\frac{π}{2}} \log (\frac{4 + 3 \sin x}{4 + 3 \cos x}) d x . . . (i) = \int_{0}^{\frac{π}{2}} \log [\frac{4 + 3 \sin (\frac{π}{2} - x)}{4 + 3 \cos (\frac{π}{2} - x)}] d x = \int_{0}^{\frac{π}{2}} \log (\frac{4 + 3 \cos x}{4 + 3 \sin x}) d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} [\log (\frac{4 + 3 \sin x}{4 + 3 \cos x}) + l og (\frac{4 + 3 \cos x}{4 + 3 \sin x})] d x = \int_{0}^{\frac{π}{2}} \log (\frac{4 + 3 \sin x}{4 + 3 \cos x} \times \frac{4 + 3 \cos x}{4 + 3 \sin x}) d x = \int_{0}^{\frac{π}{2}} \log 1 d x = 0 Hence I = 0$

Page No 19.122:

Question 42:

The value of $\int_{- π / 2}^{π / 2} (x^{3} + x \cos x + \tan^{5} x + 1) d x,$ is

(a) 0
(b) 2
(c) π
(d) 1

Answer:

(c) $π$

$\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{3} + x \cos x + \tan^{5} x + 1) d x = {[\frac{x^{4}}{4}]}_{- \frac{π}{2}}^{\frac{π}{2}} + {[x \sin x]}_{- \frac{π}{2}}^{\frac{π}{2}} - \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin x d x + \int_{- \frac{π}{2}}^{\frac{π}{2}} \tan^{3} x (s e c^{2} x - 1) d x + {[x]}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{π^{4}}{64} - \frac{π^{4}}{64} + \frac{π}{2} - \frac{π}{2} - {[- \cos x]}_{- \frac{π}{2}}^{\frac{π}{2}} + \int_{- \frac{π}{2}}^{\frac{π}{2}} \tan^{3} x s e c^{2} x d x - \int_{- \frac{π}{2}}^{\frac{π}{2}} \tan^{3} x d x + \frac{π}{2} + \frac{π}{2} = π + 0 + {[\frac{\tan^{4} x}{4}]}_{- \frac{π}{2}}^{\frac{π}{2}} - \int_{- \frac{π}{2}}^{\frac{π}{2}} tanx \sec^{2} x dx - \int_{- \frac{π}{2}}^{\frac{π}{2}} tanx dx = π - {[\frac{\tan^{2} x}{2}]}_{- \frac{π}{2}}^{\frac{π}{2}} - {[- \log (\cos x)]}_{- \frac{π}{2}}^{\frac{π}{2}} = π$

Page No 19.123:

Question 43:

$\int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{1}{1 + \cos 2 x} d x$ is equal to

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

$Let I = \int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{1}{1 + \cos 2 x} d x Let f (x) = \frac{1}{1 + \cos 2 x} f (- x) = \frac{1}{1 + \cos (- 2 x)} = \frac{1}{1 + \cos 2 x} = f (x) We know, \int_{- a}^{a} f (x) d x = \{\begin{cases} 0, if f (- x) = - f (x) \\ {2 \int}_{0}^{a} f (x) d x, if f (- x) = f (x) \end{cases} Thus, I = 2 \int_{0}^{\frac{π}{4}} \frac{1}{1 + \cos 2 x} d x = 2 \int_{0}^{\frac{π}{4}} \frac{1}{2 \cos^{2} x} d x (using the identity : \cos 2 x = 2 \cos^{2} x - 1) = \int_{0}^{\frac{π}{4}} \frac{1}{\cos^{2} x} d x = \int_{0}^{\frac{π}{4}} \sec^{2} x d x = {(\tan x)}_{0}^{\frac{π}{4}} = (\tan \frac{π}{4} - \tan 0) = 1 - 0 = 1$

Hence, the correct option is (a).

Page No 19.123:

Question 44:

$\int_{a + c}^{b + c} f (x) d x$ is equal to

(a) $\int_{a}^{b} f (x - c) d x$

(b) $\int_{a}^{b} f (x + c) d x$

(c) $\int_{a}^{b} f (x) d x$

(d) $\int_{a - c}^{b - c} f (x) d x$

Answer:

$Let I = \int_{a + c}^{b + c} f (x) d x Let x = t + c \Rightarrow d x = d t Also, if x = a + c, t = a if x = b + c, t = b Thus, I = \int_{a}^{b} f (t + c) d t = \int_{a}^{b} f (x + c) d x (∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t)$

Hence, the correct option is (b).

Page No 19.123:

Question 45:

If f and g are continuous functions in [0, 1] satisfying f(x) = f(a – x) and g(x) = g(a – x) = a, then $\int_{0}^{a} f (x) g (x) d x$ is equal to

(a) $\frac{a}{2}$

(b) $\frac{a}{2} \int_{0}^{a} f (x) d x$

(c) $\int_{0}^{a} f (x) d x$

(d) $a \int_{0}^{b} f (x) d x$

Answer:

Given: f(x) = f(a – x) and g(x) + g(a – x) = a

$Let I = \int_{0}^{a} f (x) g (x) d x . . . (1) \Rightarrow I = \int_{0}^{a} f (a - x) g (a - x) d x (∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x) \Rightarrow I = \int_{0}^{a} f (x) (a - g (x)) d x (∵ f (x) = f (a - x) and g (a - x) = a - g (x)) \Rightarrow I = a \int_{0}^{a} f (x) d x - \int_{0}^{a} f (x) g (x) d x \Rightarrow I = a \int_{0}^{a} f (x) d x - I (From (1)) \Rightarrow 2 I = a \int_{0}^{a} f (x) d x \Rightarrow I = \frac{a}{2} \int_{0}^{a} f (x) d x$

Hence, the correct option is (b).

Page No 19.123:

Question 1:

If $\int_{0}^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8},$ then a = _______________.

Answer:

$Given : \int_{0}^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8} . . . (1) Let I = \int_{0}^{a} \frac{1}{1 + 4 x^{2}} d x \Rightarrow I = \frac{1}{4} \int_{0}^{a} \frac{1}{\frac{1}{4} + x^{2}} d x \Rightarrow I = \frac{1}{4} \int_{0}^{a} \frac{1}{{(\frac{1}{2})}^{2} + x^{2}} d x \Rightarrow I = \frac{1}{4} {\{\frac{1}{\frac{1}{2}} \tan^{- 1} \frac{x}{\frac{1}{2}}\}}_{0}^{a} (∵ \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} + c) \Rightarrow I = \frac{1}{4} {\{2 \tan^{- 1} 2 x\}}_{0}^{a} \Rightarrow I = \frac{1}{4} {\{2 \tan^{- 1} 2 a - 2 \tan^{- 1} 0\}}_{0}^{a} \Rightarrow I = \frac{1}{2} \tan^{- 1} 2 a Since, it is given that \int_{0}^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8} Therefore, \frac{1}{2} \tan^{- 1} 2 a = \frac{π}{8} \Rightarrow \tan^{- 1} 2 a = \frac{π}{4} \Rightarrow 2 a = \tan \frac{π}{4} \Rightarrow 2 a = 1 \Rightarrow a = \frac{1}{2}$

Hence, a = $\frac{1}{2}$ .

Page No 19.123:

Question 2:

The value of $\int_{- π}^{π} \sin^{3} x \cos^{2} x d x$ is _______________.

Answer:

$Let I = \int_{- π}^{π} \sin^{3} x \cos^{2} x d x Let, f (x) = \sin^{3} x \cos^{2} x f (- x) = \sin^{3} (- x) \cos^{2} (- x) = - \sin^{3} x \cos^{2} x = - f (x) We know, \int_{- a}^{a} f (x) d x = \{\begin{cases} 0, if f (- x) = - f (x) \\ 2 \int_{0}^{a} f (x) d x, if f (- x) = f (x) \end{cases} Thus, \int_{- π}^{π} \sin^{3} x \cos^{2} x d x = 0$

Hence, the value of $\int_{- π}^{π} \sin^{3} x \cos^{2} x d x$ is 0.

Page No 19.123:

Question 3:

The value of $\int_{0}^{\frac{π}{2}} e^{\sin x} \cos x d x$ is _______________.

Answer:

$Let I = \int_{0}^{\frac{π}{2}} e^{\sin x} \cos x d x Let, \sin x = t \Rightarrow d (\sin x) = d t \Rightarrow \cos x d x = d t Also, if x = 0, t = 0 if x = \frac{π}{2}, t = 1 Thus, I = \int_{0}^{1} e^{t} d t = {[e^{t}]}_{0}^{1} = e^{1} - e^{0} = e - 1$

Hence, the value of $\int_{0}^{\frac{π}{2}} e^{\sin x} \cos x d x is e - 1 .$

Page No 19.123:

Question 4:

$\int_{0}^{a} \frac{x}{\sqrt{a^{2} + x^{2}}} d x =$ _______________.

Answer:

$Let I = \int_{0}^{a} \frac{x}{\sqrt{a^{2} + x^{2}}} d x Let, a^{2} + x^{2} = t \Rightarrow d (a^{2} + x^{2}) = d t \Rightarrow 2 x d x = d t Also, if x = 0, t = a^{2} if x = a, t = 2 a^{2} Thus, I = \frac{1}{2} \int_{a^{2}}^{2 a^{2}} \frac{1}{\sqrt{t}} d t = \frac{1}{2} {[\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}]}_{a^{2}}^{2 a^{2}} = \frac{1}{2} {[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}]}_{a^{2}}^{2 a^{2}} = \frac{1}{2} {[2 \sqrt{t}]}_{a^{2}}^{2 a^{2}} = \frac{1}{2} [2 \sqrt{2 a^{2}} - 2 \sqrt{a^{2}}] = \frac{1}{2} [2 a \sqrt{2} - 2 a] = [a \sqrt{2} - a] = a (\sqrt{2} - 1)$

Hence, $\int_{0}^{a} \frac{x}{\sqrt{a^{2} + x^{2}}} d x = a (\sqrt{2} - 1) .$

Page No 19.123:

Question 5:

The value of the integral $\int_{\frac{1}{π}}^{\frac{2}{π}} \frac{\sin (\frac{1}{π})}{x^{2}} d x$ is _______________.

Answer:

$Let I = \int_{\frac{1}{π}}^{\frac{2}{π}} \frac{\sin (\frac{1}{x})}{x^{2}} d x Let, \frac{1}{x} = t \Rightarrow d (\frac{1}{x}) = d t \Rightarrow - \frac{1}{x^{2}} d x = d t Also, if x = \frac{1}{π}, t = π if x = \frac{2}{π}, t = \frac{π}{2} Thus, I = \int_{π}^{\frac{π}{2}} - \sin (t) d t = {[\cos t]}_{π}^{\frac{π}{2}} = [\cos \frac{π}{2} - cosπ] = [0 - (- 1)] = 1$

Hence, the value of the integral $\int_{\frac{1}{π}}^{\frac{2}{π}} \frac{\sin (\frac{1}{x})}{x^{2}} d x$ is 1.

Page No 19.123:

Question 6:

The value of the integral $\int_{0}^{1} \frac{\tan^{- 1} x}{1 + x^{2}} d x$ is _______________.

Answer:

$Let I = \int_{0}^{1} \frac{\tan^{- 1} x}{1 + x^{2}} d x Let, \tan^{- 1} x = t \Rightarrow d (\tan^{- 1} x) = d t \Rightarrow \frac{1}{1 + x^{2}} d x = d t Also, if x = 0, t = 0 if x = 1, t = \frac{π}{4} Thus, I = \int_{0}^{\frac{π}{4}} t d t = {[\frac{t^{2}}{2}]}_{0}^{\frac{π}{4}} = \frac{1}{2} [{(\frac{π}{4})}^{2} - {(0)}^{2}] = \frac{1}{2} [\frac{π^{2}}{16}] = \frac{π^{2}}{32}$

Hence, the value of the integral $\int_{0}^{1} \frac{\tan^{- 1} x}{1 + x^{2}} d x is \frac{π^{2}}{32} .$

Page No 19.123:

Question 7:

The value of the integral $\int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x$ is _______________.

Answer:

$Let I = \int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x We know, \int e^{x} (f (x) + f' (x)) d x = e^{x} f (x) + c Here, f (x) = \frac{1}{x} \Rightarrow f' (x) = - \frac{1}{x^{2}} Thus, I = {[e^{x} (\frac{1}{x})]}_{1}^{2} = [e^{2} (\frac{1}{2}) - e^{1} (\frac{1}{1})] = \frac{e^{2}}{2} - e = \frac{e^{2} - 2 e}{2} = \frac{e (e - 2)}{2}$

Hence, the value of the integral $\int_{1}^{2} e^{x} (\frac{1}{x} - \frac{1}{x^{2}}) d x$ is $\frac{e (e - 2)}{2} .$

Page No 19.124:

Question 8:

$\int_{1}^{e} \frac{e^{x}}{x} (1 + \log x) d x =$ ________________.

Answer:

$Let I = \int_{1}^{e} \frac{e^{x}}{x} (1 + x \log x) d x = \int_{1}^{e} e^{x} (\frac{1}{x} + \log x) d x = \int_{1}^{e} e^{x} (\log x + \frac{1}{x}) d x We know, \int e^{x} (f (x) + f' (x)) d x = e^{x} f (x) + c Here, f (x) = \log x \Rightarrow f' (x) = \frac{1}{x} Thus, I = {[e^{x} \log x]}_{1}^{e} = [e^{e} (\log e) - e^{1} (\log 1)] = e^{e} (1) - e (0) = e^{e}$

Hence, $\int_{1}^{e} \frac{e^{x}}{x} (1 + x \log x) d x = e^{e}$ .

Page No 19.124:

Question 9:

The value of the integral $\int_{- 2}^{2} (a x^{5} + b x^{3} + c x + d) d x,$ where a, b, c, d are constants, depends only on ________________.

Answer:

$Let I = \int_{- 2}^{2} (a x^{5} + b x^{3} + c x + d) d x = \int_{- 2}^{2} (a x^{5} + b x^{3} + c x) d x + \int_{- 2}^{2} (d) d x Let, f (x) = a x^{5} + b x^{3} + c x f (- x) = a {(- x)}^{5} + b {(- x)}^{3} + c (- x) = - a x^{5} - b x^{3} - c x = - (a x^{5} + b x^{3} + c x) = - f (x) We know, \int_{- a}^{a} f (x) d x = \{\begin{cases} 0, if f (- x) = - f (x) \\ 2 \int_{0}^{a} f (x) d x, if f (- x) = f (x) \end{cases} Thus, I = \int_{- 2}^{2} (a x^{5} + b x^{3} + c x) d x + \int_{- 2}^{2} (d) d x = 0 + \int_{- 2}^{2} (d) d x = {(d x)}_{- 2}^{2} = (2 d - (- 2 d)) = 2 d + 2 d = 4 d$

Hence, the value of the integral $\int_{- 2}^{2} (a x^{5} + b x^{3} + c x + d) d x,$ where a, b, c, d are constants, depends only on d.

Page No 19.124:

Question 10:

$\int_{0}^{π} \frac{1}{1 + \sin x} d x =$ ________________.

Answer:

$Let I = \int_{0}^{π} \frac{1}{1 + \sin x} d x = \int_{0}^{π} \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} d x = \int_{0}^{π} \frac{1 - \sin x}{1 - \sin^{2} x} d x = \int_{0}^{π} \frac{1 - \sin x}{\cos^{2} x} d x = \int_{0}^{π} (\frac{1}{\cos^{2} x} - \frac{\sin x}{\cos^{2} x}) d x = \int_{0}^{π} (\sec^{2} x - \sec x \tan x) d x = {[\tan x - \sec x]}_{0}^{π} = [tanπ - secπ] - [\tan 0 - \sec 0] = [0 - (- 1)] - [0 - 1] = [1] - [- 1] = 1 + 1 = 2$

Hence, $\int_{0}^{π} \frac{1}{1 + \sin x} d x =$ 2.

Page No 19.124:

Question 11:

$\int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 + \sin^{4} x} d x =$ ________________.

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 + \sin^{4} x} d x Let \sin^{2} x = t \Rightarrow d (\sin^{2} x) = d t \Rightarrow 2 \sin x \cos x d x = d t Also, if x = 0, t = 0 if x = \frac{π}{2}, t = 1 Thus, I = \int_{0}^{1} \frac{1}{1 + t^{2}} \frac{d t}{2} = \frac{1}{2} \int_{0}^{1} \frac{1}{1 + t^{2}} d t = \frac{1}{2} {[\tan^{- 1} t]}_{0}^{1} = \frac{1}{2} [\tan^{- 1} 1 - \tan^{- 1} 0] = \frac{1}{2} [\frac{π}{4} - 0] = \frac{π}{8}$

Hence, $\int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 + \sin^{4} x} d x = \frac{π}{8} .$

Page No 19.124:

Question 12:

$\int_{0}^{\frac{π}{4}} \tan^{6} x \sec^{2} x d x =$ ________________.

Answer:

$Let I = \int_{0}^{\frac{π}{4}} \tan^{6} x \sec^{2} x d x Let \tan x = t \Rightarrow d (\tan x) = d t \Rightarrow \sec^{2} x d x = d t Also, if x = 0, t = 0 if x = \frac{π}{4}, t = 1 Thus, I = \int_{0}^{1} t^{6} d t = {[\frac{t^{7}}{7}]}_{0}^{1} = [\frac{1^{7}}{7} - \frac{0}{7}] = \frac{1}{7}$

Hence, $\int_{0}^{\frac{π}{4}} \tan^{6} x \sec^{2} x d x = \frac{1}{7} .$

Page No 19.124:

Question 13:

The value of $\int_{0}^{\frac{π}{4}} \frac{1 + \tan x}{1 - \tan x} d x$ is ________________.

Answer:

$Let I = \int_{0}^{\frac{π}{4}} \frac{1 + \tan x}{1 - \tan x} d x = \int_{0}^{\frac{π}{4}} \frac{\tan \frac{π}{4} + \tan x}{1 - \tan \frac{π}{4} \tan x} d x = \int_{0}^{\frac{π}{4}} \tan (\frac{π}{4} + x) d x = {[- \log (\cos (\frac{π}{4} + x))]}_{0}^{\frac{π}{4}} = - \log (\cos (\frac{π}{4} + \frac{π}{4})) + \log (\cos (\frac{π}{4} + 0)) = - \log (\cos (\frac{2 π}{4})) + \log (\cos (\frac{π}{4})) = - \log (\cos (\frac{π}{2})) + \log (\frac{1}{\sqrt{2}}) = - \log (0) + \log {(2)}^{- \frac{1}{2}} = - \frac{1}{2} \log (2)$

Hence, the value of $\int_{0}^{\frac{π}{4}} \frac{1 + \tan x}{1 - \tan x} d x$ is $- \frac{1}{2} \log (2) .$

Page No 19.124:

Question 14:

The value of $\int_{0}^{2} \frac{3^{\sqrt{x}}}{\sqrt{x}} d x$ is ________________.

Answer:

$Let I = \int_{0}^{2} \frac{3^{\sqrt{x}}}{\sqrt{x}} d x Let \sqrt{x} = t \Rightarrow d (\sqrt{x}) = d t \Rightarrow \frac{1}{2 \sqrt{x}} d x = d t Also, if x = 0, t = 0 if x = 2, t = \sqrt{2} Thus, I = \int_{0}^{\sqrt{2}} 3^{t} (2 d t) = 2 \int_{0}^{\sqrt{2}} 3^{t} d t = 2 {[\frac{3^{t}}{\log 3}]}_{0}^{\sqrt{2}} = 2 [\frac{3^{\sqrt{2}}}{\log 3} - \frac{3^{0}}{\log 3}] = 2 [\frac{3^{\sqrt{2}} - 1}{\log 3}]$

Hence, the value of $\int_{0}^{2} \frac{3^{\sqrt{x}}}{\sqrt{x}} d x$ is $2 [\frac{3^{\sqrt{2}} - 1}{\log 3}]$ .

Page No 19.124:

Question 15:

The value of $\int_{0}^{\frac{π}{2}} \frac{\sin x}{1 + \cos^{2} x} d x$ is ________________.

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin x}{1 + \cos^{2} x} d x Let \cos x = t \Rightarrow d (\cos x) = d t \Rightarrow - \sin x d x = d t Also, if x = 0, t = 1 if x = \frac{π}{2}, t = 0 Thus, I = \int_{1}^{0} \frac{- 1}{1 + t^{2}} d t = - {[\tan^{- 1} t]}_{1}^{0} = - [\tan^{- 1} 0 - \tan^{- 1} 1] = - [0 - \frac{π}{4}] = \frac{π}{4}$

Hence, the value of $\int_{0}^{\frac{π}{2}} \frac{\sin x}{1 + \cos^{2} x} d x$ is $\frac{π}{4}$ .

Page No 19.124:

Question 16:

If f(a – x) = x and $\int_{0}^{a} f (x) d x = k \int_{0}^{\frac{a}{2}} f (x) d x,$ then k = _____________.

Answer:

$\int_{0}^{a} f (x) d x = k \int_{0}^{\frac{a}{2}} f (x) d x$

Given:
f(x) = f(a – x) ...(1)
$\int_{0}^{a} f (x) d x = k \int_{0}^{\frac{a}{2}} f (x) d x$ ...(2)

$Let I = \int_{0}^{a} f (x) d x . . . (3) We know, \int_{0}^{2 a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, if f (2 a - x) = f (x) \\ 0, if f (2 a - x) = - f (x) \end{cases} Here, f (a - x) = f (x) (From (1)) Thus, I = 2 \int_{0}^{\frac{a}{2}} f (x) d x . . . (4) From (2) and (4), k \int_{0}^{\frac{a}{2}} f (x) d x = 2 \int_{0}^{\frac{a}{2}} f (x) d x \Rightarrow k = 2$

Hence, k = 2.

Page No 19.124:

Question 17:

The value of the integral $\int_{0}^{2 π} \cos^{7} x \sin^{4} x d x$ is ________________.

Answer:

$Let I = \int_{0}^{2 π} \cos^{7} x \sin^{4} x d x Let f (x) = \cos^{7} x \sin^{4} x f (2 π - x) = {(\cos (2 π - x))}^{7} {(\sin (2 π - x))}^{4} = {(\cos (x))}^{7} {(- \sin (x))}^{4} = \cos^{7} x \sin^{4} x = f (x) We know, \int_{0}^{2 a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, if f (2 a - x) = f (x) \\ 0, if f (2 a - x) = - f (x) \end{cases} Thus, I = 2 \int_{0}^{π} \cos^{7} x \sin^{4} x d x Again, Let f (x) = \cos^{7} x \sin^{4} x f (π - x) = {(\cos (π - x))}^{7} {(\sin (π - x))}^{4} = {(- \cos (x))}^{7} {(\sin (x))}^{4} = - \cos^{7} x \sin^{4} x = - f (x) Thus, I = 0$

Hence, the value of the integral $\int_{0}^{2 π} \cos^{7} x \sin^{4} x d x$ is 0.

Page No 19.124:

Question 18:

The value of the integral $\int_{- 1}^{1} x |x| d x$ is ________________.

Answer:

$Let I = \int_{- 1}^{1} x |x| d x Let f (x) = x |x| f (- x) = (- x) |- x| = - x |x| = - f (x) We know, \int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, if f (- x) = f (x) \\ 0, if f (- x) = - f (x) \end{cases} Thus, I = 0$

Hence, the value of the integral $\int_{- 1}^{1} x |x| d x$ is 0.

Page No 19.124:

Question 19:

The value of the integral $\int_{- 1}^{1} \log (\frac{2 - x}{2 + x}) d x$ is ________________.

Answer:

$Let I = \int_{- 1}^{1} \log (\frac{2 - x}{2 + x}) d x Let f (x) = \log (\frac{2 - x}{2 + x}) f (- x) = \log (\frac{2 - (- x)}{2 + (- x)}) = \log (\frac{2 + x}{2 - x}) = \log {(\frac{2 - x}{2 + x})}^{- 1} = - \log (\frac{2 - x}{2 + x}) = - f (x) We know, \int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, if f (- x) = f (x) \\ 0, if f (- x) = - f (x) \end{cases} Thus, I = 0$

Hence, the value of the integral $\int_{- 1}^{1} \log (\frac{2 - x}{2 + x}) d x$ is 0.

Page No 19.124:

Question 20:

The value of the integral $\int_{- 1}^{1} |1 - x| d x$ is ________________.

Answer:

$Let I = \int_{- 1}^{1} |1 - x| d x We know, |1 - x| = \{\begin{cases} (1 - x), if (1 - x) \geq 0 \\ - (1 - x), if (1 - x) < 0 \end{cases} = \{\begin{cases} (1 - x), if x \leq 1 \\ - (1 - x), if x > 1 \end{cases} Thus, I = \int_{- 1}^{1} (1 - x) d x = {[x - \frac{x^{2}}{2}]}_{- 1}^{1} = [1 - \frac{{(1)}^{2}}{2}] - [- 1 - \frac{{(- 1)}^{2}}{2}] = [1 - \frac{1}{2}] - [- 1 - \frac{1}{2}] = [\frac{2 - 1}{2}] - [\frac{- 2 - 1}{2}] = \frac{1}{2} - [- \frac{3}{2}] = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$

Hence, the value of the integral $\int_{- 1}^{1} |1 - x| d x$ is 2.

Page No 19.124:

Question 21:

The value of the integral $\int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{3} x} d x$ is ________________.

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{3} x} d x = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \frac{\sin^{3} x}{\cos^{3} x}} d x = \int_{0}^{\frac{π}{2}} \frac{1}{\frac{\cos^{3} x + \sin^{3} x}{\cos^{3} x}} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{3} x}{\cos^{3} x + \sin^{3} x} d x . . . (1) We know, \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x Thus, I = \int_{0}^{\frac{π}{2}} \frac{\cos^{3} (\frac{π}{2} - x)}{\cos^{3} (\frac{π}{2} - x) + \sin^{3} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{\sin^{3} x}{\sin^{3} x + \cos^{3} x} d x . . . (2) Adding (1) and (2), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\cos^{3} x}{\cos^{3} x + \sin^{3} x} d x + \int_{0}^{\frac{π}{2}} \frac{\sin^{3} x}{\sin^{3} x + \cos^{3} x} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{3} x + \sin^{3} x}{\sin^{3} x + \cos^{3} x} d x = \int_{0}^{\frac{π}{2}} 1 d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} \Rightarrow I = \frac{π}{4}$

Hence, the value of the integral $\int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{3} x} d x$ is $\frac{π}{4}$ .

Page No 19.124:

Question 22:

If $\int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} d x = k π,$ then k = ________________.

Answer:

$Given : \int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} d x = k π . . . (1) Let I = \int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} d x We know, \int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x Thus, I = \int_{0}^{a} [\sqrt{\frac{a - x}{a + x}} + \sqrt{\frac{a - (- x)}{a + (- x)}}] d x = \int_{0}^{a} [\sqrt{\frac{a - x}{a + x}} + \sqrt{\frac{a + x}{a - x}}] d x = \int_{0}^{a} [\frac{\sqrt{a - x}}{\sqrt{a + x}} + \frac{\sqrt{a + x}}{\sqrt{a - x}}] d x = \int_{0}^{a} [\frac{{(\sqrt{a - x})}^{2} + {(\sqrt{a + x})}^{2}}{\sqrt{a + x} \sqrt{a - x}}] d x = \int_{0}^{a} [\frac{a - x + a + x}{\sqrt{a^{2} - x^{2}}}] d x = \int_{0}^{a} [\frac{2 a}{\sqrt{a^{2} - x^{2}}}] d x = 2 a \int_{0}^{a} \frac{1}{\sqrt{a^{2} - x^{2}}} d x = 2 a {[\sin^{- 1} \frac{x}{a}]}_{0}^{a} = 2 a [\sin^{- 1} \frac{a}{a} - \sin^{- 1} 0] = 2 a [\sin^{- 1} 1 - 0] = 2 a [\frac{π}{2}] = a π . . . (2) From (1) and (2), a π = k π \Rightarrow k = a$

Hence, k = a.

Page No 19.124:

Question 23:

If f(x) = f(a – x) and $\int_{0}^{a} x f (x) d x = k \int_{0}^{a} f (x) d x,$ then k = ________________.

Answer:

Given:
f(x) = f(a – x) ...(1)
$\int_{0}^{a} x f (x) d x = k \int_{0}^{a} f (x) d x$ ...(2)

$Let I = \int_{0}^{a} x f (x) d x . . . (3) We know, \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x Thus, I = \int_{0}^{a} (a - x) f (a - x) d x \Rightarrow I = \int_{0}^{a} (a - x) f (x) d x (From (1)) \Rightarrow I = \int_{0}^{a} a f (x) d x - \int_{0}^{a} x f (x) d x \Rightarrow I = \int_{0}^{a} a f (x) d x - I (From (3)) \Rightarrow 2 I = \int_{0}^{a} a f (x) d x \Rightarrow I = \frac{1}{2} \int_{0}^{a} a f (x) d x \Rightarrow I = \frac{a}{2} \int_{0}^{a} f (x) d x . . . (4) From (2) and (4), k \int_{0}^{a} f (x) d x = \frac{a}{2} \int_{0}^{a} f (x) d x \Rightarrow k = \frac{a}{2}$

Hence, k = $\frac{a}{2}$ .

Page No 19.125:

Question 24:

The value of the integral $\int_{0}^{10} \frac{x^{10}}{{(10 - x)}^{10} + x^{10}} d x$ is ________________.

Answer:

$Let I = \int_{0}^{10} \frac{x^{10}}{{(10 - x)}^{10} + x^{10}} d x . . . (1) We know, \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x Thus, I = \int_{0}^{10} \frac{{(10 - x)}^{10}}{{(10 - (10 - x))}^{10} + {(10 - x)}^{10}} d x = \int_{0}^{10} \frac{{(10 - x)}^{10}}{{(10 - 10 + x)}^{10} + {(10 - x)}^{10}} d x = \int_{0}^{10} \frac{{(10 - x)}^{10}}{{(x)}^{10} + {(10 - x)}^{10}} d x . . . (2) Adding (1) and (2), we get 2 I = \int_{0}^{10} \frac{x^{10}}{{(10 - x)}^{10} + x^{10}} d x + \int_{0}^{10} \frac{{(10 - x)}^{10}}{{(x)}^{10} + {(10 - x)}^{10}} d x = \int_{0}^{10} \frac{x^{10} + {(10 - x)}^{10}}{{(x)}^{10} + {(10 - x)}^{10}} d x = \int_{0}^{10} 1 d x = {[x]}_{0}^{10} = 10 \Rightarrow I = 5$

Hence, the value of the integral $\int_{0}^{10} \frac{x^{10}}{{(10 - x)}^{10} + x^{10}} d x$ is 5.

Page No 19.125:

Question 25:

$\int_{- 1}^{1} e^{|x|} d x =$ ________________.

Answer:

$Let I = \int_{- 1}^{1} e^{|x|} d x Let f (x) = e^{|x|} f (- x) = e^{|- x|} = e^{|x|} = f (x) We know, \int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, if f (- x) = f (x) \\ 0, if f (- x) = - f (x) \end{cases} Thus, I = 2 \int_{0}^{1} e^{|x|} d x = 2 \int_{0}^{1} e^{x} d x = 2 {(e^{x})}_{0}^{1} = 2 (e^{1} - e^{0}) = 2 (e - 1)$

Hence, $\int_{- 1}^{1} e^{|x|} d x = 2 (e - 1) .$

Page No 19.125:

Question 1:

$\int_{0}^{π / 2} \sin^{2} x d x .$

Answer:

$\int_{0}^{\frac{π}{2}} \sin^{2} x d x = \int_{0}^{\frac{π}{2}} \frac{1 - \cos 2 x}{2} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 - \cos 2 x) d x = \frac{1}{2} {[x - \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} = \frac{1}{2} (\frac{π}{2} - 0) = \frac{π}{4}$

Page No 19.125:

Question 2:

$\int_{0}^{π / 2} \cos^{2} x d x .$

Answer:

$\int_{0}^{\frac{π}{2}} \cos^{2} x d x = \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 x}{2} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + \cos 2 x) d x = \frac{1}{2} {[x + \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} = \frac{1}{2} [\frac{π}{2} + 0] = \frac{π}{4}$

Page No 19.125:

Question 3:

$\int_{- π / 2}^{π / 2} \sin^{2} x d x .$

Answer:

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{2} x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 - \cos 2 x}{2} d x = \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} (1 - \cos 2 x) d x = \frac{1}{2} {[x - \frac{\sin 2 x}{2}]}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{1}{2} (\frac{π}{2} - 0 + \frac{π}{2} - 0) = \frac{π}{2}$

Page No 19.125:

Question 4:

$\int_{- π / 2}^{π / 2} \cos^{2} x d x .$

Answer:

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + \cos 2 x}{2} d x = \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} (1 + \cos 2 x) d x = \frac{1}{2} {[x + \frac{\sin 2 x}{2}]}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{1}{2} (\frac{π}{2} + 0 + \frac{π}{2} - 0) = \frac{π}{2}$

Page No 19.125:

Question 5:

$\int_{- π / 2}^{π / 2} \sin^{3} x d x .$

Answer:

$Let I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{3} x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin x \sin^{2} x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin x (1 - \cos^{2} x) d x Let \cos x = t, then - \sin x d x = d t, When, x \to - \frac{π}{2}; t \to 0 and x \to \frac{π}{2}; t \to 0 I = \int_{0}^{0} (- 1 + t^{2}) d t = 0$

Page No 19.125:

Question 6:

$\int_{- π / 2}^{π / 2} x \cos^{2} x d x .$

Answer:

$We have, I = \int_{- \frac{π}{2}}^{\frac{π}{2}} x \cos^{2} x d x Let f (x) = x \cos^{2} x \Rightarrow f (- x) = (- x) \cos^{2} (- x) = - x \cos^{2} x ∴ f (- x) = - f (x) i . e ., f (x) is odd function We know that \int_{- a}^{a} f (x) d x = 0, if f (x) is odd function . ∴ I = \int_{- \frac{π}{2}}^{\frac{π}{2}} x \cos^{2} x d x = 0$

Page No 19.125:

Question 7:

$\int_{0}^{π / 4} \tan^{2} x d x .$

Answer:

$\int_{0}^{\frac{π}{4}} \tan^{2} x d x = \int_{0}^{\frac{π}{4}} (s e c^{2} x - 1) d x = {[\tan x - x]}_{0}^{\frac{π}{4}} = 1 - \frac{π}{4} - 0 = 1 - \frac{π}{4}$ 4430. →

Page No 19.125:

Question 8:

$\int_{0}^{1} \frac{1}{x^{2} + 1} d x$ .

Answer:

$\int_{0}^{1} \frac{1}{1 + x^{2}} d x = {[\tan^{- 1} x]}_{0}^{1} = \frac{π}{4} - 0 = \frac{π}{4}$ ode is 4430. →

Page No 19.125:

Question 9:

$\int_{- 2}^{1} \frac{|x|}{x} d x .$

Answer:

$Let, I = \int_{- 2}^{1} \frac{|x|}{x} d x We have, |x| = \{\begin{cases} x 0 \leq x \leq 1 \\ - x - 2 \leq x < 0 \end{cases} ∴ \frac{|x|}{x} = \{\begin{cases} 1 0 \leq x \leq 1 \\ - 1 - 2 \leq x < 0 \end{cases} Therefore, I = \int_{- 2}^{0} - 1 d x + \int_{0}^{1} 1 d x = - {[x]}_{- 2}^{0} + {[x]}_{0}^{1} = 0 - 2 + 1 - 0 = - 1$

Page No 19.125:

Question 10:

$\int_{0}^{\infty} e^{- x} d x .$

Answer:

$\int_{0}^{\infty} e^{- x} d x = - {[e^{- x}]}_{0}^{\infty} = - (0 - 1) = 0 + 1 = 1$ 4430. →

Page No 19.125:

Question 11:

$\int_{0}^{4} \frac{1}{\sqrt{16 - x^{2}}} d x .$

Answer:

$\int_{0}^{4} \frac{1}{\sqrt{16 - x^{2}}} d x = \int_{0}^{4} \frac{1}{\sqrt{4^{2} - x^{2}}} d x = {[\sin^{- 1} \frac{x}{4}]}_{0}^{4} = (\frac{π}{2} - 0) = \frac{π}{2}$ s 4430. →

Page No 19.125:

Question 12:

$\int_{0}^{3} \frac{1}{x^{2} + 9} d x .$

Answer:

$\int_{0}^{3} \frac{1}{x^{2} + 9} d x = \int_{0}^{3} \frac{1}{x^{2} + 3^{2}} d x = \frac{1}{3} {[\tan^{- 1} \frac{x}{3}]}_{0}^{3} = \frac{1}{3} (\tan^{- 1} 1 - \tan^{- 1} 0) = \frac{1}{3} (\frac{π}{4} - 0) = \frac{π}{12}$

Page No 19.125:

Question 13:

$\int_{0}^{π / 2} \sqrt{1 - \cos 2 x} d x .$

Answer:

$\int_{0}^{\frac{π}{2}} \sqrt{1 - \cos 2 x} d x = \int_{0}^{\frac{π}{2}} \sqrt{2 \sin^{2} x} d x = \int_{0}^{\frac{π}{2}} \sqrt{2} \sin x d x = - \sqrt{2} {[\cos x]}_{0}^{\frac{π}{2}} = - (0 - \sqrt{2}) = \sqrt{2}$

Page No 19.125:

Question 14:

$\int_{0}^{π / 2} \log \tan x d x .$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \log \tan x d x . . . (i) = \int_{0}^{\frac{π}{2}} \log \tan (\frac{π}{2} - x) d x [Using, \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{\frac{π}{2}} \log c o t x d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{\frac{π}{2}} \log \tan x d x + \int_{0}^{\frac{π}{2}} \log c o t x d x = \int_{0}^{\frac{π}{2}} \log (\tan x \times c o t x) d x = \int_{0}^{\frac{π}{2}} \log 1 d x = 0 Hence, I = 0$

Page No 19.125:

Question 15:

$\int_{0}^{π / 2} \log (\frac{3 + 5 \cos x}{3 + 5 \sin x}) d x .$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \log (\frac{3 + 5 \cos x}{3 + 5 \sin x}) d x . . . (i) = \int_{0}^{\frac{π}{2}} l og [\frac{3 + 5 \cos (\frac{π}{2} - x)}{3 + 5 \sin (\frac{π}{2} - x)}] d x = \int_{0}^{\frac{π}{2}} l og (\frac{3 + 5 \sin x}{3 + 5 \cos x}) d x . . . (ii) Adding (i) and (i i) 2 I = \int_{0}^{\frac{π}{2}} [\log (\frac{3 + 5 \cos x}{3 + 5 \sin x}) + l og (\frac{3 + 5 \sin x}{3 + 5 \cos x})] d x = \int_{0}^{\frac{π}{2}} \log (\frac{3 + 5 \cos x}{3 + 5 \sin x} \times \frac{3 + 5 \sin x}{3 + 5 \cos x}) d x = \int_{0}^{\frac{π}{2}} \log 1 d x = 0 Hence I = 0$

Page No 19.125:

Question 16:

$\int_{0}^{π / 2} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x, n \in N .$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (\frac{π}{2} - x)}{\sin^{n} (\frac{π}{2} - x) + \cos^{n} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\cos^{n} x + \sin^{n} x} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} [\frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} + \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x}] d x = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x + \cos^{n} x}{\sin^{n} x + \cos^{n} x} d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 19.126:

Question 17:

$\int_{0}^{π} \cos^{5} x d x .$

Answer:

$Let I = \int_{0}^{π} \cos^{5} x d x = \int_{0}^{π} \cos x {(\cos^{2} x)}^{2} d x = \int_{0}^{π} \cos x {(1 - \sin^{2} x)}^{2} d x Let \sin x = t, then \cos x d x = d t When, x \to 0; t \to 0 and x \to π; t \to 0 Therefore, I = \int_{0}^{0} {(1 - t^{2})}^{2} d t = 0$

Page No 19.126:

Question 18:

$\int_{- π / 2}^{π / 2} \log (\frac{a - \sin θ}{a + \sin θ}) d θ$

Answer:

$Let, I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \log (\frac{a - \sin θ}{a + \sin θ}) d θ Here, f (θ) = \log (\frac{a - \sin θ}{a + \sin θ}) Consider, f (- θ) = \log [\frac{a - \sin (- θ)}{a + \sin (- θ)}] = - \log (\frac{a - \sin θ}{a + \sin θ}) = - f (θ) i . e ., f (θ) is odd function . Therefore, I = 0$

Page No 19.126:

Question 19:

$\int_{- 1}^{1} x |x| d x .$

Answer:

$|x| = \{\begin{cases} - x, - 1 < x < 0 \\ x, 0 < x < 1 \end{cases} ∴ x |x| = \{\begin{cases} - x^{2}, - 1 < x < 0 \\ x^{2}, 0 < x < 1 \end{cases} Now, \int_{- 1}^{1} x |x| d x = \int_{- 1}^{0} - x^{2} d x + \int_{0}^{1} x^{2} d x = - \int_{- 1}^{0} x^{2} d x + \int_{0}^{1} x^{2} d x = - {[\frac{x^{3}}{3}]}_{- 1}^{0} + {[\frac{x^{3}}{3}]}_{0}^{1} = - (0 + \frac{1}{3}) + (\frac{1}{3} - 0) = 0 - \frac{1}{3} + \frac{1}{3} - 0 = 0$

Page No 19.126:

Question 20:

$\int_{a}^{b} \frac{f (x)}{f (x) + f (a + b - x)} d x .$

Answer:

$Let I = \int_{a}^{b} \frac{f (x)}{f (x) + f (a + b - x)} d x . . . (i) = \int_{a}^{b} \frac{f (a + b - x)}{f (a + b - x) + f (a + b - a - b + x)} d x = \int_{a}^{b} \frac{f (a + b - x)}{f (a + b - x) + f (x)} d x ∴ I = \int_{a}^{b} \frac{f (a + b - x)}{f (x) + f (a + b - x)} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{a}^{b} [\frac{f (x)}{f (x) + f (a + b - x)} + \frac{f (a + b - x)}{f (x) + f (a + b - x)}] d x = \int_{a}^{b} \frac{f (x) + f (a + b - x)}{f (x) + f (a + b - x)} dx = {[x]}_{a}^{b} = b - a Hence, I = \frac{b - a}{2}$

Page No 19.126:

Question 21:

$\int_{0}^{1} \frac{1}{1 + x^{2}} d x$

Answer:

$\int_{0}^{1} \frac{1}{1 + x^{2}} d x = {[\tan^{- 1} x]}_{0}^{1} = \tan^{- 1} 1 - \tan^{- 1} 0 = \frac{π}{4} - 0 = \frac{π}{4}$

Page No 19.126:

Question 22:

Evaluate each of the following integrals:

$\int_{0}^{\frac{π}{4}} \tan x d x$

Answer:

$\int_{0}^{\frac{π}{4}} \tan x d x = {logsec x}_{0}^{\frac{π}{4}} = logsec \frac{π}{4} - logsec 0 = \log \sqrt{2} - \log 1 = \log 2^{\frac{1}{2}} - 0 = \frac{1}{2} \log 2$

Page No 19.126:

Question 23:

$\int_{2}^{3} \frac{1}{x} d x$

Answer:

$\int_{2}^{3} \frac{1}{x} d x = {[\log_{e} x]}_{2}^{3} = \log_{e} 3 - \log_{e} 2 = \log_{e} (\frac{3}{2})$ →

Page No 19.126:

Question 24:

$\int_{0}^{2} \sqrt{4 - x^{2}} d x$

Answer:

$\int_{0}^{2} \sqrt{4 - x^{2}} d x = \int_{0}^{2} \sqrt{2^{2} - x^{2}} d x = {[\frac{x}{2} \sqrt{4 - x^{2}} + \frac{1}{2} \times 2^{2} \sin^{- 1} \frac{x}{2}]}_{0}^{2} = {[\frac{x}{2} \sqrt{4 - x^{2}}]}_{0}^{2} + 2 {[\sin^{- 1} \frac{x}{2}]}_{0}^{2} = 0 + 2 (\frac{π}{2} - 0) = π$

Page No 19.126:

Question 25:

$\int_{0}^{1} \frac{2 x}{1 + x^{2}} d x$

Answer:

$We have, I = \int_{0}^{1} \frac{2 x}{1 + x^{2}} d x Putting 1 + x^{2} = t \Rightarrow 2 x d x = d t When x \to 0; t \to 1 And x \to 1; t \to 2 ∴ I = \int_{1}^{2} \frac{d t}{t} = {[\log_{e} |t|]}_{1}^{2} = \log_{e} 2 - \log_{e} 1 = \log_{e} 2 - 0 = \log_{e} 2$ 4430. →

Page No 19.126:

Question 26:

Evaluate each of the following integrals:

$\int_{0}^{1} x e^{x^{2}} d x$ [CBSE 2014]

Answer:

$I = \int_{0}^{1} x e^{x^{2}} d x = \frac{1}{2} \int_{0}^{1} e^{x^{2}} 2 x d x$

Put $x^{2} = z$

$\Rightarrow 2 x d x = d z$

When $x \to 0, z \to 0$

When $x \to 1, z \to 1$

$∴ I = \frac{1}{2} \int_{0}^{1} e^{z} d z = \frac{1}{2} \times {e^{z}}_{0}^{1} = \frac{1}{2} (e - e^{0}) = \frac{1}{2} (e - 1)$

Page No 19.126:

Question 27:

Evaluate each of the following integrals:

$\int_{0}^{\frac{π}{4}} \sin 2 x d x$ [CBSE 2014]

Answer:

$\int_{0}^{\frac{π}{4}} \sin 2 x d x = {\frac{- \cos 2 x}{2}}_{0}^{\frac{π}{4}} = - \frac{1}{2} (\cos \frac{π}{2} - \cos 0) = - \frac{1}{2} \times (0 - 1) = \frac{1}{2}$

Page No 19.126:

Question 28:

Evaluate each of the following integrals:

$\int_{e}^{e^{2}} \frac{1}{x \log x} d x$ [CBSE 2014]

Answer:

$\int_{e}^{e^{2}} \frac{1}{x \log x} d x = \int_{e}^{e^{2}} \frac{\frac{1}{x}}{\log x} d x = {\log (\log x)}_{e}^{e^{2}} [\int \frac{f' (x)}{f (x)} d x = \log f (x) + C] = \log (\log e^{2}) - \log (\log e) = \log (2 \log e) - \log (\log e) = \log 2 - \log 1 (\log e = 1) = \log 2 - 0 = \log 2$

Page No 19.126:

Question 29:

Evaluate each of the following integrals:

$\int_{0}^{\frac{π}{2}} e^{x} (\sin x - \cos x) d x$ [CBSE 2014]

Answer:

Disclaimer: The solution has been provided by taking the lower limit of integral as 0.

$\int_{0}^{\frac{π}{2}} e^{x} (\sin x - \cos x) d x = - \int_{0}^{\frac{π}{2}} e^{x} [\cos x + (- \sin x)] d x = - {e^{x} \cos x}_{0}^{\frac{π}{2}} \{\int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + C\} = - (e^{\frac{π}{2}} \cos \frac{π}{2} - e^{0} \cos 0) = - (e^{\frac{π}{2}} \times 0 - 1 \times 1) = - (0 - 1) = 1$

Page No 19.126:

Question 30:

Solve each of the following integrals:

$\int_{2}^{4} \frac{x}{x^{2} + 1} d x$ [CBSE 2014]

Answer:

$\int_{2}^{4} \frac{x}{x^{2} + 1} d x = \frac{1}{2} \int_{2}^{4} \frac{2 x}{x^{2} + 1} d x = \frac{1}{2} \times {\log (x^{2} + 1)}_{2}^{4} [\int \frac{f' (x)}{f (x)} d x = \log f (x) + C] = \frac{1}{2} (\log 17 - \log 5) = \frac{1}{2} \log (\frac{17}{5}) (\log a - \log b = \log \frac{a}{b})$

Page No 19.126:

Question 31:

If $\int_{0}^{1} (3 x^{2} + 2 x + k) d x = 0,$ find the value of k.

Answer:

$We have, \int_{0}^{1} (3 x^{2} + 2 x + k) d x = 0 \Rightarrow {[x^{3} + x^{2} + k x]}_{0}^{1} = 0 \Rightarrow 1 + 1 + k - 0 = 0 \Rightarrow k = - 2$ . →

Page No 19.126:

Question 32:

If $\int_{0}^{a} 3 x^{2} d x = 8,$ write the value of a.

Answer:

$We have, \int_{0}^{a} 3 x^{2} d x = 8 \Rightarrow {[3 \frac{x^{3}}{3}]}_{0}^{a} = 8 \Rightarrow {[x^{3}]}_{0}^{a} = 8 \Rightarrow a^{3} - 0 = 8 \Rightarrow a = \sqrt[3]{8} = 2$ 0. →

Page No 19.126:

Question 33:

If $f (x) = \int_{0}^{x} t \sin t d t$ , the write the value of $f' (x)$ . [CBSE 2014]

Answer:

$f (x) = \int_{0}^{x} t \sin t d t \Rightarrow f (x) = {t (- \cos t)}_{0}^{x} - \int_{0}^{x} \frac{d}{d t} (t) \times (- \cos t) d t \Rightarrow f (x) = - (x \cos x - 0) + \int_{0}^{x} \cos t d t \Rightarrow f (x) = - x \cos x + {\sin t}_{0}^{x}$
$\Rightarrow f (x) = - x \cos x + (\sin x - 0) \Rightarrow f (x) = - x \cos x + \sin x$

Differentiating both sides with respect to x, we get

$f' (x) = - [x \times (- \sin x) + \cos x \times 1] + \cos x \Rightarrow f' (x) = - (- x \sin x) - \cos x + \cos x \Rightarrow f' (x) = x \sin x$

Thus, the value of $f' (x)$ is x sinx.

Page No 19.126:

Question 34:

If $\int_{0}^{a} \frac{1}{4 + x^{2}} d x = \frac{π}{8}$ , find the value of a. [CBSE 2014]

Answer:

$\int_{0}^{a} \frac{1}{4 + x^{2}} d x = \frac{π}{8} \Rightarrow {\frac{1}{2} \tan^{- 1} \frac{x}{2}}_{0}^{a} = \frac{π}{8} [\int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C] \Rightarrow \frac{1}{2} (\tan^{- 1} \frac{a}{2} - \tan^{- 1} 0) = \frac{π}{8} \Rightarrow \tan^{- 1} \frac{a}{2} - 0 = \frac{π}{4}$
$\Rightarrow \tan^{- 1} \frac{a}{2} = \frac{π}{4} \Rightarrow \frac{a}{2} = \tan \frac{π}{4} = 1 \Rightarrow a = 2$

Thus, the value of a is 2.

Page No 19.126:

Question 35:

Write the coefficient a, b, c of which the value of the integral $\int_{- 3}^{3} (a x^{2} + b x + c) d x$ is independent.

Answer:

$\int_{- 3}^{3} (a x^{2} + b x + c) d x = {[a \frac{x^{3}}{3} + b \frac{x^{2}}{2} + c x]}_{- 3}^{3} = 9 a + \frac{9}{2} b + 3 c + 9 a - \frac{9}{2} b + 3 c = 18 a + 6 c$

Hence, the given integral is independent of b

Page No 19.126:

Question 36:

Evaluate : $\int_{2}^{3} 3^{x} d x .$

Answer:

$I = \int_{2}^{3} 3^{x} = {\frac{3^{x}}{\log 3}}_{2}^{3} + C (Use : \int a^{x} = \frac{a^{x}}{\log a} + C) = \frac{3^{3}}{\log 3} - \frac{3^{2}}{\log 3} + C$

$= \frac{1}{\log 3} (3^{3} - 3^{2}) + C = \frac{1}{\log 3} (27 - 9) + C = \frac{1}{\log 3} (18) + C$

Page No 19.126:

Question 37:

$\int_{0}^{2} [x] d x .$

Answer:

$We have, I = \int_{0}^{2} [x] d x We know that, [x] = \{\begin{cases} 0, 0 < x < 1 \\ 1, 1 < x < 2 \end{cases} ∴ I = \int_{0}^{2} [x] d x = \int_{0}^{1} [x] d x + \int_{1}^{2} [x] d x = \int_{0}^{1} (0) d x + \int_{1}^{2} (1) d x = 0 + {[x]}_{1}^{2} = 2 - 1 = 1$

Page No 19.126:

Question 38:

$\int_{0}^{15} [x] d x .$

Answer:

$We have, I = \int_{0}^{1.5} [x] d x = \int_{0}^{1} [x] d x + \int_{1}^{1.5} [x] d x = \int_{0}^{1} (0) d x + \int_{1}^{1.5} (1) d x [∵ [x] = \{\begin{cases} 0 0 \leq x < 1 \\ 1 1 \leq x < 1.5 \end{cases}] = 0 + {[x]}_{1}^{1.5} = 1.5 - 1 = 0.5 = \frac{1}{2}$

Page No 19.126:

Question 39:

$\int_{0}^{1} \{x\} d x,$ where {x} denotes the fractional part of x.

Answer:

$We have, I = \int_{0}^{1} \{x\} d x We know \{x\} = x, 0 < x < 1 ∴ I = \int_{0}^{1} x d x = {[\frac{x^{2}}{2}]}_{0}^{1} = \frac{1}{2} - \frac{0}{2} = \frac{1}{2}$
s 4430. →

Page No 19.126:

Question 40:

$\int_{0}^{1} e^{\{x\}} d x .$

Answer:

$We have, I = \int_{0}^{1} e^{\{x\}} d x We know that, \{x\} = x, when 0 < x < 1 ∴ I = \int_{0}^{1} e^{x} d x = {[e^{x}]}_{0}^{1} = e^{1} - e^{0} = e - 1$

Page No 19.127:

Question 41:

$\int_{0}^{2} x [x] d x .$

Answer:

$We have, I = \int_{0}^{2} x [x] d x We know that, x [x] = \{\begin{cases} x \times 0, 0 < x < 1 \\ x \times 1, 1 < x < 2 \end{cases} i . e ., x [x] = \{\begin{cases} 0, 0 < x < 1 \\ x, 1 < x < 2 \end{cases} ∴ I = \int_{0}^{2} x [x] d x = \int_{0}^{1} x [x] d x + \int_{1}^{2} x [x] d x = \int_{0}^{1} (0) d x + \int_{1}^{2} (x) d x = 0 + {[\frac{x^{2}}{2}]}_{1}^{2} = \frac{2^{2}}{2} - \frac{1^{2}}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$

Page No 19.127:

Question 42:

$\int_{0}^{1} 2^{x - [x]} d x$

Answer:

$We have, I = \int_{0}^{1} 2^{x - [x]} d x = \int_{0}^{1} 2^{x - 0} d x (∵ [x] = 0 where, 0 < x < 1) = \int_{0}^{1} 2^{x} d x = {[\frac{2^{x}}{\log_{e} 2}]}_{0}^{1} = \frac{2^{1}}{\log_{e} 2} - \frac{2^{0}}{\log_{e} 2} = \frac{2}{\log_{e} 2} - \frac{1}{\log_{e} 2} = \frac{1}{\log_{e} 2}$

Page No 19.127:

Question 43:

$\int_{1}^{2} \log_{e} [x] d x .$

Answer:

$We have, I = \int_{1}^{2} \log_{e} [x] d x We know that, [x] = 1, when 1 < x < 2 ∴ I = \int_{1}^{2} \log_{e} 1 d x I = \int_{1}^{2} (0) d x = 0$

Page No 19.127:

Question 44:

$\int_{0}^{\sqrt{2}} [x^{2}] d x .$

Answer:

$We have, I = \int_{0}^{\sqrt{2}} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x = \int_{0}^{1} (0) d x + \int_{1}^{\sqrt{2}} (1) d x (∵ [x^{2}] = \{\begin{cases} 0 0 < x < 1 \\ 1 1 < x < \sqrt{2} \end{cases}) = 0 + {[x]}_{1}^{\sqrt{2}} = \sqrt{2} - 1$

Page No 19.127:

Question 45:

If $[\cdot] and \{\cdot\}$ denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
$\int_{0}^{π / 4} \sin \{x\} d x$

Answer:

$We have, I = \int_{0}^{π / 4} \sin \{x\} d x We know that, \{x\} = x, when 0 < x < \frac{π}{4} (As π = 3.14 \Rightarrow \frac{π}{4} = 0.785 < 1) ∴ I = \int_{0}^{π / 4} \sin x d x = {[- \cos x]}_{0}^{\frac{π}{4}} = - (\cos \frac{π}{4} - \cos 0) = \cos 0 - \cos \frac{π}{4} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$

Page No 19.16:

Question 1:

$\int_{4}^{9} \frac{1}{\sqrt{x}} d x$

Answer:

$Let I = \int_{4}^{9} \frac{1}{\sqrt{x}} d x . Then, I = 2 \int_{4}^{9} \frac{1}{2 \sqrt{x}} d x \Rightarrow I = 2 {[\sqrt{x}]}_{4}^{9} \Rightarrow I = 2 (3 - 2) \Rightarrow I = 2$

Page No 19.16:

Question 2:

$\int_{- 2}^{3} \frac{1}{x + 7} d x$

Answer:

$Let I = \int_{- 2}^{3} \frac{1}{x + 7} d x . Then, I = {[\log (x + 7)]}_{- 2}^{3} \Rightarrow I = \log 10 - \log 5 \Rightarrow I = \log \frac{10}{5} [∵ \log a - \log b = \log \frac{a}{b}] \Rightarrow I = \log 2$

Page No 19.16:

Question 3:

$\int_{0}^{1 / 2} \frac{1}{\sqrt{1 - x^{2}}} d x$

Answer:

$Let I = \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{2}}} d x . Then, I = {[\sin^{- 1} x]}_{0}^{\frac{1}{2}} \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} 0 \Rightarrow I = \frac{π}{6} - 0 \Rightarrow I = \frac{π}{6}$

Page No 19.16:

Question 4:

$\int_{0}^{1} \frac{1}{1 + x^{2}} d x$

Answer:

$Let I = \int_{0}^{1} \frac{1}{1 + x^{2}} d x . Then, I = {[\tan^{- 1} x]}_{0}^{1} \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} 0 \Rightarrow I = \frac{π}{4} - 0 \Rightarrow I = \frac{π}{4}$

Page No 19.16:

Question 5:

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

Answer:

$Let I = \int_{2}^{3} \frac{x}{x^{2} + 1} d x . Then, I = \frac{1}{2} \int_{2}^{3} \frac{2 x}{x^{2} + 1} \Rightarrow I = \frac{1}{2} {[\log (x^{2} + 1)]}_{2}^{3} \Rightarrow I = \frac{1}{2} (\log 10 - \log 5) \Rightarrow I = \frac{1}{2} \log \frac{10}{5} [∵ \log a - \log b = \log \frac{a}{b}] \Rightarrow I = \frac{1}{2} \log 2$

Page No 19.16:

Question 6:

$\int_{0}^{\infty} \frac{1}{a^{2} + b^{2} x^{2}} d x$

Answer:

$Let I = \int_{0}^{\infty} \frac{1}{a^{2} + b^{2} x^{2}} d x . Then, I = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{1 + \frac{b^{2} x^{2}}{a^{2}}} d x \Rightarrow I = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{1 + {(\frac{b x}{a})}^{2}} d x \Rightarrow I = \frac{a}{b a^{2}} {[\tan^{- 1} (\frac{b x}{a})]}_{0}^{\infty} \Rightarrow I = \frac{1}{a b} (\tan^{- 1} \infty - \tan^{- 1} 0) \Rightarrow I = \frac{π}{2 a b}$

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Question 7:

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

Answer:

$Let I = \int_{- 1}^{1} \frac{1}{1 + x^{2}} d x . Then, I = {[\tan^{- 1} x]}_{- 1}^{1} \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} (- 1) \Rightarrow I = \frac{π}{4} - (- \frac{π}{4}) \Rightarrow I = \frac{π}{2}$

Page No 19.16:

Question 8:

$\int_{0}^{\infty} e^{- x} d x$

Answer:

$Let I = \int_{0}^{\infty} e^{- x} d x . Then, I = {[- e^{- x}]}_{0}^{\infty} \Rightarrow I = - e^{- \infty} + e^{0} \Rightarrow I = 0 + 1 \Rightarrow I = 1$

Page No 19.16:

Question 9:

$\int_{0}^{1} \frac{x}{x + 1} d x$

Answer:

$Let I = \int_{0}^{1} \frac{x}{x + 1} d x . Then, I = \int_{0}^{1} 1 - \frac{1}{x + 1} d x \Rightarrow I = {[x - \log (x + 1)]}_{0}^{1} \Rightarrow I = 1 - \log 2 - (0 - \log 1) \Rightarrow I = \log e - \log 2 \Rightarrow I = \log \frac{e}{2}$

Page No 19.16:

Question 10:

$\int_{0}^{π / 2} (\sin x + \cos x) d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} (\sin x + \cos x) d x . Then, I = {[- \cos x + \sin x]}_{0}^{\frac{π}{2}} \Rightarrow I = 0 + 1 - (- 1 + 0) \Rightarrow I = 2$

Page No 19.16:

Question 11:

$\int_{π / 4}^{π / 2} \cot x d x$

Answer:

$Let I = \int_{\frac{π}{4}}^{\frac{π}{2}} \cot x d x . Then, I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \cot x \frac{- (cosec x + \cot x)}{cosec x + \cot x} d x \Rightarrow I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- cosec x \cot x - \cot^{2} x}{cosec x + \cot x} d x \Rightarrow I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- cosec x \cot x - {cosec}^{2} x + 1}{cosec x + \cot x} d x [∵ {cosec}^{2} x = 1 + \cot^{2} x] \Rightarrow I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- cosec x \cot x - {cosec}^{2} x}{cosec x + \cot x} d x - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{1}{cosec x + \cot x} d x \Rightarrow I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- cosec x \cot x - {cosec}^{2} x}{cosec x + \cot x} d x - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\sin x}{1 + \cos x} d x \Rightarrow I = - {[\log (cosec x + \cot x)]}_{\frac{π}{4}}^{\frac{π}{2}} + {[\log (1 + \cos x)]}_{\frac{π}{4}}^{\frac{π}{2}} \Rightarrow I = - \log (1 + \infty) + \log (\sqrt{2} + 1) + \log (1 + 0) - \log (1 + \frac{1}{\sqrt{2}}) \Rightarrow I = \log (\sqrt{2} + 1) - \log (\frac{\sqrt{2} + 1}{\sqrt{2}}) \Rightarrow I = \log (\frac{\sqrt{2} (\sqrt{2} + 1)}{(\sqrt{2} + 1)}) \Rightarrow I = \log \sqrt{2} \Rightarrow I = \frac{1}{2} \log 2$

Page No 19.16:

Question 12:

$\int_{0}^{π / 4} \sec x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{4}} \sec x d x . Then, I = \int_{0}^{\frac{π}{4}} \sec x \frac{\sec x + \tan x}{\sec x + \tan x} d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{\sec^{2} x + \sec x \tan x}{\sec x + \tan x} d x Put u = \sec x + \tan x \Rightarrow d u = \sec^{2} x + \sec x \tan x d x ∴ \int_{0}^{\frac{π}{4}} \frac{\sec^{2} x + \sec x \tan x}{\sec x + \tan x} d x = \int \frac{d u}{u} \Rightarrow I = [\log u] \Rightarrow I = {[\log (\sec x + \tan x)]}_{0}^{\frac{π}{4}} \Rightarrow I = \log (\sec \frac{π}{4} + \tan \frac{π}{4}) - \log (\sec 0 + \tan 0) \Rightarrow I = \log (\sqrt{2} + 1) - \log 1 \Rightarrow I = \log (\sqrt{2} + 1)$

Page No 19.16:

Question 13:

$\int_{π / 6}^{π / 4} cosec x d x$

Answer:

$Let I = \int_{\frac{π}{6}}^{\frac{π}{4}} cosec x d x . Then, I = \int_{\frac{π}{6}}^{\frac{π}{4}} cosec x \frac{cosec x - \cot x}{cosec x - \cot x} d x \Rightarrow I = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{{cosec}^{2} x - cosec x \cot x}{cosec x - \cot x} d x \Rightarrow I = {[\log (cosec x - \cot x)]}_{\frac{π}{6}}^{\frac{π}{4}} \Rightarrow I = \log (\sqrt{2} - 1) - \log (2 - \sqrt{3})$

Page No 19.16:

Question 14:

$\int_{0}^{1} \frac{1 - x}{1 + x} d x$

Answer:

$Let I = \int_{0}^{1} \frac{1 - x}{1 + x} d x . Then, I = \int_{0}^{1} (\frac{1}{1 + x} - \frac{1 + x - 1}{1 + x}) d x I = \int_{0}^{1} (\frac{1}{1 + x} - 1 + \frac{x}{1 + x}) d x \Rightarrow I = {[\log (1 + x) - x + \log (1 + x)]}_{0}^{1} \Rightarrow I = (\log 2 - 1 + \log 2) - (\log 1 - 0 + \log 1) = 2 \log 2 - 1$

Page No 19.16:

Question 15:

$\int_{0}^{π} \frac{1}{1 + \sin x} d x$

Answer:

$Let I = \int_{0}^{π} \frac{1}{1 + \sin x} d x . Then, I = \int_{0}^{π} \frac{1 - \sin x}{(1 + \sin x) (1 - \sin x)} d x \Rightarrow I = \int_{0}^{π} \frac{1 - \sin x}{1 - \sin^{2} x} d x \Rightarrow I = \int_{0}^{π} \frac{1 - \sin x}{\cos^{2} x} d x [∵ \sin^{2} x + \cos^{2} x = 1] \Rightarrow I = \int_{0}^{π} \sec^{2} x - \sec x \tan x d x \Rightarrow I = {[\tan x - \sec x]}_{0}^{π} \Rightarrow I = (\tan π - \sec π) - (\tan 0 - \sec 0) \Rightarrow I = 0 + 1 - (0 - 1) \Rightarrow I = 1 + 1 \Rightarrow I = 2$

Page No 19.16:

Question 16:

$\int_{- π / 4}^{π / 4} \frac{1}{1 + \sin x} d x$

Answer:

$Let I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{1 + \sin x} d x . Then, I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1 - \sin x}{1 - \sin^{2} x} d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1 - \sin x}{\cos^{2} x} d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\frac{1}{\cos^{2} x} - \frac{\sin x}{\cos^{2} x}) d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\sec^{2} x - \sec x \tan x) d x \Rightarrow I = {[\tan x - \sec x]}_{- \frac{π}{4}}^{\frac{π}{4}} \Rightarrow I = (1 - \sqrt{2}) - (- 1 - \sqrt{2}) \Rightarrow I = 2$

Page No 19.16:

Question 17:

$\int_{0}^{π / 2} \cos^{2} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \cos^{2} x d x . Then, I = \int_{0}^{\frac{π}{2}} \cos^{2} x d x \Rightarrow I = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + \cos 2 x) d x [∵ \cos 2 x = 2 \cos^{2} x - 1] \Rightarrow I = {[\frac{x}{2} + \frac{\sin 2 x}{4}]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{π}{4} + 0 - 0 \Rightarrow I = \frac{π}{4}$

Page No 19.16:

Question 18:

$\int_{0}^{π / 2} \cos^{3} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \cos^{3} x d x . Then, I = \int_{0}^{\frac{π}{2}} \cos^{2} x \cos x d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (1 - \sin^{2} x) \cos x d x Let u = \sin x, d u = \cos x d x \Rightarrow I = \int (1 - u^{2}) d u \Rightarrow I = [u - \frac{u^{3}}{3}] \Rightarrow I = {[\sin x - \frac{\sin^{3} x}{3}]}_{0}^{\frac{π}{2}} \Rightarrow I = 1 - \frac{1}{3} - 0 \Rightarrow I = \frac{2}{3}$

Page No 19.16:

Question 19:

$\int_{0}^{π / 6} \cos x \cos 2 x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{6}} \cos x \cos 2 x d x . Then, I = \int_{0}^{\frac{π}{6}} \cos x (\cos^{2} x - \sin^{2} x) d x \Rightarrow I = \int_{0}^{\frac{π}{6}} (2 \cos^{3} x - \cos x) d x \Rightarrow I = \int_{0}^{\frac{π}{6}} (2 \cos x (1 - \sin^{2} x) - \cos x) d x \Rightarrow I = {[2 (\sin x - \frac{\sin^{3} x}{3}) - \sin x]}_{0}^{\frac{π}{6}} \Rightarrow I = [2 (\frac{1}{2} - \frac{1}{24}) - \frac{1}{2}] - 0 \Rightarrow I = \frac{5}{12}$

Page No 19.16:

Question 20:

$\int_{0}^{π / 2} \sin x \sin 2 x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \sin x \sin 2 x d x . Then, I = \int_{0}^{\frac{π}{2}} 2 \sin^{2} x \cos x d x \Rightarrow I = \int_{0}^{\frac{π}{2}} 2 (1 - \cos^{2} x) \cos x d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (2 \cos x - 2 \cos^{3} x) d x \Rightarrow I = {[2 \sin x - 2 (\sin x - \frac{\sin^{3} x}{3})]}_{0}^{\frac{π}{2}} \Rightarrow I = [2 - 2 (1 - \frac{1}{3})] - 0 \Rightarrow I = \frac{2}{3}$

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Question 21:

$\int_{π / 3}^{π / 4} {(\tan x + \cot x)}^{2} d x$

Answer:

$Let I = \int_{\frac{π}{3}}^{\frac{π}{4}} {(\tan x + \cot x)}^{2} d x . Then, I = \int_{\frac{π}{3}}^{\frac{π}{4}} (\tan^{2} x + \cot^{2} x + 2 \tan x \cot x) d x \Rightarrow I = \int_{\frac{π}{3}}^{\frac{π}{4}} (\tan^{2} x + \cot^{2} x + 2) d x \Rightarrow I = \int_{\frac{π}{3}}^{\frac{π}{4}} (\sec^{2} x - 1 + {cosec}^{2} x - 1 + 2) d x \Rightarrow I = \int_{\frac{π}{3}}^{\frac{π}{4}} (\sec^{2} x + {cosec}^{2} x) d x \Rightarrow I = {[\tan x - \cot x]}_{\frac{π}{3}}^{\frac{π}{4}} \Rightarrow I = (1 - 1) - (\sqrt{3} - \frac{1}{\sqrt{3}}) \Rightarrow I = \frac{- 2}{\sqrt{3}}$

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Question 22:

$\int_{0}^{π / 2} \cos^{4} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \cos^{4} x d x . Then, I = \int_{0}^{\frac{π}{2}} {(\cos^{2} x)}^{2} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{{(1 + \cos 2 x)}^{2}}{4} d x \Rightarrow I = \frac{1}{4} \int_{0}^{\frac{π}{2}} (1 + \cos^{2} 2 x + 2 \cos 2 x) d x \Rightarrow I = \frac{1}{4} \int_{0}^{\frac{π}{2}} (1 + 2 \cos 2 x + \frac{1 + \cos 4 x}{2}) d x \Rightarrow I = \frac{1}{4} \int_{0}^{\frac{π}{2}} (\frac{3 + 4 \cos 2 x + \cos 4 x}{2}) d x \Rightarrow I = \frac{1}{4} {[\frac{3 x}{2} + \frac{2 \sin 2 x}{2} + \frac{\sin 4 x}{8}]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{1}{4} [\frac{3 π}{4} + 0] \Rightarrow I = \frac{3 π}{16}$

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Question 23:

$\int_{0}^{π / 2} (a^{2} \cos^{2} x + b^{2} \sin^{2} x) d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} (a^{2} \cos^{2} x + b^{2} \sin^{2} x) d x . Then, I = \int_{0}^{\frac{π}{2}} (a^{2} \cos^{2} x + b^{2} (1 - \cos^{2} x)) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (b^{2} + (a^{2} - b^{2}) \cos^{2} x) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (b^{2} + \frac{(a^{2} - b^{2}) (1 + \cos 2 x)}{2}) d x \Rightarrow I = {[b^{2} x + \frac{a^{2} - b^{2}}{2} (x + \frac{\sin 2 x}{2})]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{b^{2} π}{2} + \frac{a^{2} - b^{2}}{2} \frac{π}{2} + 0 \Rightarrow I = \frac{π}{4} (a^{2} + b^{2})$

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Question 24:

$\int_{0}^{π / 2} \sqrt{1 + \sin x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \sqrt{1 + \sin x} d x . Then, I = \int_{0}^{\frac{π}{2}} \sqrt{1 + \sin x} \times \frac{\sqrt{1 - \sin x}}{\sqrt{1 - \sin x}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{1 - \sin^{2} x}}{\sqrt{1 - \sin x}} d x \Rightarrow I = = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\sqrt{1 - \sin x}} d x Let 1 - \sin x = u \Rightarrow - \cos x d x = d u ∴ I = \int \frac{- d u}{\sqrt{u}} \Rightarrow I = = [- 2 \sqrt{u}] \Rightarrow I = = {[- 2 \sqrt{1 - \sin x}]}_{0}^{\frac{π}{2}} \Rightarrow I = = 0 + 2 \Rightarrow I = = 2$

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Question 25:

$\int_{0}^{π / 2} \sqrt{1 + \cos x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \sqrt{1 + \cos x} d x . Then, I = \int_{0}^{\frac{π}{2}} \sqrt{1 + \cos x} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{1 - \cos^{2} x}}{\sqrt{1 - \cos x}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin x}{\sqrt{1 - \cos x}} d x Let 1 - \cos x = u \Rightarrow \sin x d x = d u ∴ I = \int \frac{d u}{\sqrt{u}} \Rightarrow I = [2 \sqrt{u}] \Rightarrow I = {[2 \sqrt{1 - \cos x}]}_{0}^{\frac{π}{2}} \Rightarrow I = 2 - 0 \Rightarrow I = 2$

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Question 26:

Evaluate the following definite integrals:

$\int_{0}^{\frac{π}{2}} x^{2} \sin x d x$ [CBSE 2014]

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x^{2} \sin x$

Applying integration by parts, we have

$I = x^{2} (- \cos x) |_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x (- \cos x) d x \Rightarrow I = (0 - 0) + 2 \int_{0}^{\frac{π}{2}} x \cos x d x (\cos \frac{π}{2} = 0)$

Again applying integration by parts, we have

$I = 0 + 2 [x \sin x |_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 1 \times \sin x d x] \Rightarrow I = 2 (\frac{π}{2} \sin \frac{π}{2} - 0) - 2 \int_{0}^{\frac{π}{2}} \sin x d x \Rightarrow I = 2 (\frac{π}{2} - 0) - 2 (- \cos x) |_{0}^{\frac{π}{2}} \Rightarrow I = π + 2 (\cos \frac{π}{2} - \cos 0) \Rightarrow I = π + 2 (0 - 1) \Rightarrow I = π - 2$

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Question 27:

$\int_{0}^{π / 2} x \cos x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x \cos x d x . Then, Integrating by parts I = {[x \sin x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 1 \sin x d x \Rightarrow I = {[x \sin x]}_{0}^{\frac{π}{2}} + {[\cos x]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{π}{2} - 1$

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Question 28:

$\int_{0}^{π / 2} x^{2} \cos x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x^{2} \cos x d x . Then, Integrating by parts I = {[x^{2} \sin x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x \sin x d x \Rightarrow I = {[x^{2} \sin x]}_{0}^{\frac{π}{2}} - {[- 2 x \cos x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} - 2 \cos x d x \Rightarrow I = {[x^{2} \sin x]}_{0}^{\frac{π}{2}} + {[2 x \cos x]}_{0}^{\frac{π}{2}} - {[2 \sin x]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{π^{2}}{4} - 2$

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Question 29:

$\int_{0}^{π / 4} x^{2} \sin x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{4}} x^{2} \sin x d x . Then, Integrating by parts I = {[- x^{2} \cos x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} - 2 x \cos x d x \Rightarrow I = {[- x^{2} \cos x]}_{0}^{\frac{π}{4}} + {[2 x \sin x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} 2 \sin x dx \Rightarrow I = {[- x^{2} \cos x]}_{0}^{\frac{π}{4}} + {[2 x \sin x]}_{0}^{\frac{π}{4}} + {[2 \cos x]}_{0}^{\frac{π}{4}} \Rightarrow I = \frac{- π^{2}}{16 \sqrt{2}} + \frac{π}{2 \sqrt{2}} + \frac{2}{\sqrt{2}} - 2 \Rightarrow I = \sqrt{2} + \frac{π}{2 \sqrt{2}} - \frac{π^{2}}{16 \sqrt{2}} - 2$

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Question 30:

$\int_{0}^{π / 2} x^{2} \cos 2 x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x^{2} \cos 2 x d x . Then, Integrating by parts I = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x \frac{\sin 2 x}{2} d x \Rightarrow I = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} - {[- x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} - 1 \frac{\cos 2 x}{2} d x \Rightarrow I = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} + {[x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} - {[\frac{\sin 2 x}{4}]}_{0}^{\frac{π}{2}} \Rightarrow I = 0 - \frac{π}{4} - 0 \Rightarrow I = - \frac{π}{4}$

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Question 31:

$\int_{0}^{π / 2} x^{2} \cos^{2} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x^{2} \cos^{2} x d x . Then, I = \int_{0}^{\frac{π}{2}} x^{2} (\frac{1 + \cos 2 x}{2}) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (\frac{x^{2}}{2} + \frac{x^{2} \cos 2 x}{2}) d x \Rightarrow I = {[\frac{x^{3}}{6}]}_{0}^{\frac{π}{2}} + {[\frac{x^{2} \sin 2 x}{4}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{x}{2} \sin 2 x d x \Rightarrow I = {[\frac{x^{3}}{6}]}_{0}^{\frac{π}{2}} + {[\frac{x^{2} \sin 2 x}{4}]}_{0}^{\frac{π}{2}} - {[\frac{- x \cos 2 x}{4}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} - 1 \frac{\cos 2 x}{2} d x \Rightarrow I = {[\frac{x^{3}}{6}]}_{0}^{\frac{π}{2}} + {[\frac{x^{2} \sin 2 x}{4}]}_{0}^{\frac{π}{2}} + {[\frac{x \cos 2 x}{4}]}_{0}^{\frac{π}{2}} - {[\frac{\sin 2 x}{4}]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{π^{3}}{48} - \frac{π}{8}$

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Question 32:

$\int_{1}^{2} \log x d x$

Answer:

$Let I = \int_{1}^{2} \log x d x . Then, I = \int_{1}^{2} 1 \log x d x Integrating by parts \Rightarrow I = {[x \log x]}_{1}^{2} - \int_{1}^{2} \frac{1}{x} x d x \Rightarrow I = {[x \log x]}_{1}^{2} - \int_{1}^{2} d x \Rightarrow I = {[x \log x]}_{1}^{2} - {[x]}_{1}^{2} \Rightarrow I = 2 \log 2 - 2 + 1 \Rightarrow I = 2 \log 2 - 1$

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Question 33:

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

Answer:

$Let I = \int_{1}^{3} \frac{\log x}{{(1 + x)}^{2}} d x . Then, I = {[\frac{- 1}{1 + x} \log x]}_{1}^{3} - \int_{1}^{3} \frac{1}{x} (\frac{- 1}{x + 1}) d x \Rightarrow I = {[\frac{- 1}{1 + x} \log x]}_{1}^{3} + \int_{1}^{3} \frac{1}{x (x + 1)} d x \Rightarrow I = {[\frac{- 1}{1 + x} \log x]}_{1}^{3} + \int_{1}^{3} (\frac{1}{x} - \frac{1}{x + 1}) d x \Rightarrow I = {[\frac{- 1}{1 + x} \log x]}_{1}^{3} + {[\log x - \log (x + 1)]}_{1}^{3} \Rightarrow I = \frac{- 1}{4} \log 3 + \log 3 - \log 4 + \log 2 \Rightarrow I = \frac{3}{4} \log 3 - \log 2$

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Question 34:

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

Answer:

$Let I = \int_{1}^{e} \frac{e^{x}}{x} (1 + x \log x) d x . Then, I = \int_{1}^{e} (\frac{e^{x}}{x} + e^{x} \log x) d x \Rightarrow I = \int_{1}^{e} \frac{e^{x}}{x} d x + \int_{1}^{e} e^{x} \log x d x Integrating first term by parts \Rightarrow I = {[\log x e^{x}]}_{1}^{e} - \int_{1}^{e} e^{x} \log x d x + \int_{1}^{e} e^{x} \log x d x \Rightarrow I = (\log e) e^{e} - 0 \Rightarrow I = e^{e}$

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Question 35:

$\int_{1}^{e} \frac{\log x}{x} d x$

Answer:

$Let I = \int_{1}^{e} \frac{\log x}{x} d x Let \log x = u \Rightarrow \frac{1}{x} d x = d u ∴ I = \int u d u \Rightarrow I = [\frac{u^{2}}{2}] \Rightarrow I = {[\frac{(\log x)^{2}}{2}]}_{1}^{e} \Rightarrow I = \frac{1}{2} - 0 \Rightarrow I = \frac{1}{2}$

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Question 36:

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

Answer:

$Let I = \int_{e}^{e^{2}} \{\frac{1}{\log x} - \frac{1}{{(\log x)}^{2}}\} d x . Then, I = \int_{e}^{e^{2}} 1 \frac{1}{\log x} d x - \int_{e}^{e^{2}} \frac{1}{{(\log x)}^{2}} d x Integrating by parts \Rightarrow I = \{{[\frac{x}{\log x}]}_{e}^{e^{2}} - \int_{e}^{e^{2}} \frac{- 1}{x {(\log x)}^{2}} x d x\} - \int_{e}^{e^{2}} \frac{1}{{(\log x)}^{2}} d x \Rightarrow I = {[\frac{x}{\log x}]}_{e}^{e^{2}} + \int_{e}^{e^{2}} \frac{1}{{(\log x)}^{2}} d x - \int_{e}^{e^{2}} \frac{1}{{(\log x)}^{2}} d x \Rightarrow I = {[\frac{x}{\log x}]}_{e}^{e^{2}} + 0 \Rightarrow I = \frac{e^{2}}{\log e^{2}} - \frac{e}{\log e} \Rightarrow I = \frac{e^{2}}{2 \log e} - \frac{e}{\log e} \Rightarrow I = \frac{e^{2}}{2} - e$

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Question 37:

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

Answer:

$Let I = \int_{1}^{2} \frac{x + 3}{x (x + 2)} d x . Then, I = \int_{1}^{2} (\frac{x}{x (x + 2)} + \frac{3}{x (x + 2)}) d x \Rightarrow I = \int_{1}^{2} \frac{d x}{(x + 2)} + \int_{1}^{2} \frac{3}{x (x + 2)} d x \Rightarrow I = {[\log (x + 2)]}_{1}^{2} + \frac{3}{2} \int_{1}^{2} (\frac{1}{x} - \frac{1}{x + 2}) d x \Rightarrow I = {[\log (x + 2)]}_{1}^{2} + \frac{3}{2} {[\log x - \log (x + 2)]}_{1}^{2} \Rightarrow I = \log 4 - \log 3 + \frac{3}{2} [\log 2 - \log 4 - 0 + \log 3] \Rightarrow I = \log 4 - \log 3 + \frac{3}{2} [- \log 2 + \log 3] \Rightarrow I = 2 \log 2 - \log 3 + \frac{3}{2} \log 3 - \frac{3}{2} \log 2 \Rightarrow I = \frac{1}{2} \log 2 + \frac{1}{2} \log 3 \Rightarrow I = \frac{1}{2} (\log 2 + \log 3) \Rightarrow I = \frac{1}{2} \log 6$

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Question 38:

$\int_{0}^{1} \frac{2 x + 3}{5 x^{2} + 1} d x$

Answer:

$Let I = \int_{0}^{1} \frac{2 x + 3}{5 x^{2} + 1} d x . Then, I = \int_{0}^{1} \frac{2 x}{5 x^{2} + 1} d x + \int_{0}^{1} \frac{3}{5 x^{2} + 1} d x \Rightarrow I = \frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2} + 1} d x + 3 \int_{0}^{1} \frac{1}{{(\sqrt{5} x)}^{2} + 1^{2}} d x \Rightarrow I = \frac{1}{5} {[\log (5 x^{2} + 1)]}_{0}^{1} + \frac{3}{\sqrt{5}} {[\tan^{- 1} (\sqrt{5} x)]}_{0}^{1} \Rightarrow I = \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{- 1} \sqrt{5}$

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Question 39:

$\int_{0}^{2} \frac{1}{4 + x - x^{2}} d x$

Answer:

$Let I = \int_{0}^{2} \frac{1}{4 + x - x^{2}} d x . Then, I = - \int_{0}^{2} \frac{1}{x^{2} - x - 4} d x \Rightarrow I = - \int_{0}^{2} \frac{1}{(x^{2} - x + \frac{1}{4}) - \frac{1}{4} - 4} d x = - \int_{0}^{2} \frac{1}{{(x - \frac{1}{2})}^{2} - \frac{17}{4}} d x = - \int_{0}^{2} \frac{1}{{(x - \frac{1}{2})}^{2} - {(\frac{\sqrt{17}}{2})}^{2}} d x = \int_{0}^{2} \frac{1}{- {(\frac{2 x - 1}{2})}^{2} + {(\frac{\sqrt{17}}{2})}^{2}} d x = \frac{1}{\sqrt{17}} {[\log (\frac{\sqrt{17} + 2 x - 1}{\sqrt{17} - 2 x + 1})]}_{0}^{2} = \frac{1}{\sqrt{17}} \{\log \frac{\sqrt{17} + 3}{\sqrt{17} - 3} - \log \frac{\sqrt{17} - 1}{\sqrt{17} + 1}\} = \frac{1}{\sqrt{17}} \{\log \frac{26 + 6 \sqrt{17}}{8} - \log \frac{18 - 2 \sqrt{17}}{16}\} = \frac{1}{\sqrt{17}} \{\log \frac{52 + 12 \sqrt{17}}{18 - 2 \sqrt{17}}\} = \frac{1}{\sqrt{17}} \{\log \frac{52 + 12 \sqrt{17}}{18 - 2 \sqrt{17}} \times \frac{18 + 2 \sqrt{17}}{18 + 2 \sqrt{17}}\} \Rightarrow I = \frac{1}{\sqrt{17}} \log \frac{1344 + 320 \sqrt{17}}{256} \Rightarrow I = \frac{1}{\sqrt{17}} \log \frac{21 + 5 \sqrt{17}}{4}$

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Question 40:

$\int_{0}^{1} \frac{1}{2 x^{2} + x + 1} d x$

Answer:

$Let I = \int_{0}^{1} \frac{1}{2 x^{2} + x + 1} d x . Then, I = \frac{1}{2} \int_{0}^{1} \frac{1}{x^{2} + \frac{x}{2} + \frac{1}{2}} d x I = \frac{1}{2} \int_{0}^{1} \frac{1}{(x^{2} + \frac{x}{2} + \frac{1}{16}) - \frac{1}{16} + \frac{1}{2}} d x \Rightarrow I = \frac{1}{2} \int_{0}^{1} \frac{1}{{(x + \frac{1}{4})}^{2} + \frac{7}{16}} d x \Rightarrow I = \frac{1}{2} \times \frac{4}{\sqrt{7}} {[\tan^{- 1} (\frac{x + \frac{1}{4}}{\frac{\sqrt{7}}{4}})]}_{0}^{1} \Rightarrow I = \frac{2}{\sqrt{7}} (\tan^{- 1} \frac{5}{\sqrt{7}} - \tan^{- 1} \frac{1}{\sqrt{7}})$

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Question 41:

$\int_{0}^{1} \sqrt{x (1 - x)} d x$

Answer:

$Let I = \int_{0}^{1} \sqrt{x (1 - x)} d x . Then, I = \int_{0}^{1} \sqrt{\frac{1}{4} - {(x - \frac{1}{2})}^{2}} d x \Rightarrow I = \frac{1}{2} \int_{0}^{1} \sqrt{1 - \frac{{(x - \frac{1}{2})}^{2}}{\frac{1}{4}}} d x \Rightarrow I = \frac{1}{2} \int_{0}^{1} \sqrt{1 - {(\frac{x - \frac{1}{2}}{\frac{1}{2}})}^{2}} d x Let (\frac{x - \frac{1}{2}}{\frac{1}{2}}) = \sin u \Rightarrow 2 d x = \cos u d u ∴ I = \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{1 - \sin^{2} u} \cos u d u \Rightarrow I = \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} u d u \Rightarrow I = \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{\cos 2 u + 1}{2}) d u \Rightarrow I = \frac{1}{8} {[\frac{\sin 2 u}{2} + u]}_{- \frac{π}{2}}^{\frac{π}{2}} \Rightarrow I = \frac{1}{8} [\frac{π}{2} + \frac{π}{2}] \Rightarrow I = \frac{π}{8}$

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Question 42:

$\int_{0}^{2} \frac{1}{\sqrt{3 + 2 x - x^{2}}} d x$

Answer:

$Let I = \int_{0}^{2} \frac{1}{\sqrt{3 + 2 x - x^{2}}} d x . Then, I = \int_{0}^{2} \frac{1}{\sqrt{- x^{2} + 2 x - 1 + 1 + 3}} d x \Rightarrow I = \int_{0}^{2} \frac{1}{\sqrt{- {(x - 1)}^{2} + 4}} d x \Rightarrow I = {[\sin^{- 1} \frac{(x - 1)}{2}]}_{0}^{2} \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} (- \frac{1}{2}) \Rightarrow I = 2 \sin^{- 1} \frac{1}{2} \Rightarrow I = 2 \times \frac{π}{6} = \frac{π}{3}$

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Question 43:

$\int_{0}^{4} \frac{1}{\sqrt{4 x - x^{2}}} d x$

Answer:

$Let I = \int_{0}^{4} \frac{1}{\sqrt{4 x - x^{2}}} d x . Then, I = \int_{0}^{4} \frac{1}{\sqrt{4 x - x^{2} - 4 + 4}} d x \Rightarrow I = \int_{0}^{4} \frac{1}{\sqrt{- {(x - 2)}^{2} + 4}} d x \Rightarrow I = {[\sin^{- 1} \frac{(x - 2)}{2}]}_{0}^{4} \Rightarrow I = (\sin^{- 1} 1 - \sin^{- 1} (- 1)) \Rightarrow I = 2 \sin^{- 1} 1 \Rightarrow I = 2 \frac{π}{2} = π$

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Question 44:

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

Answer:

$Let I = \int_{- 1}^{1} \frac{1}{x^{2} + 2 x + 5} d x . Then, I = \int_{- 1}^{1} \frac{1}{(x^{2} + 2 x + 1) + 4} d x \Rightarrow I = \int_{- 1}^{1} \frac{1}{{(x + 1)}^{2} + 2^{2}} d x \Rightarrow I = \frac{1}{2} {[\tan^{- 1} \frac{(x + 1)}{2}]}_{- 1}^{1} \Rightarrow I = \frac{1}{2} (\tan^{- 1} 1 - \tan^{- 1} 0) \Rightarrow I = \frac{1}{2} (\frac{π}{4}) \Rightarrow I = \frac{π}{8}$

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Question 45:

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

Answer:

$Let I = \int_{1}^{4} \frac{x^{2} + x}{\sqrt{2 x + 1}} d x . Let 2 x + 1 = u \Rightarrow x = \frac{u - 1}{2} \Rightarrow d x = \frac{d u}{2} ∴ I = \int \frac{{(\frac{u - 1}{2})}^{2} + \frac{u - 1}{2}}{\sqrt{u}} \frac{d u}{2} \Rightarrow I = \frac{1}{8} \int \frac{u^{2} + 1 - 2 u + 2 u - 2}{\sqrt{u}} d u = \frac{1}{8} \int \frac{(u^{2} - 1)}{\sqrt{u}} d u = \frac{1}{8} \int (u^{\frac{3}{2}} - u^{- \frac{1}{2}}) d u = \frac{1}{8} [\frac{2 u^{\frac{5}{2}}}{5} - \frac{2 u^{\frac{1}{2}}}{1}] = \frac{1}{8} {[\frac{2 {(2 x + 1)}^{\frac{5}{2}}}{5} - \frac{2 {(2 x + 1)}^{\frac{1}{2}}}{1}]}_{1}^{4} = \frac{1}{8} [\frac{2}{5} \times 243 - 6 - \frac{2}{5} \times 9 \sqrt{3} + 2 \sqrt{3}] \Rightarrow I = \frac{1}{8} [\frac{456}{5} - \frac{8 \sqrt{3}}{5}] \Rightarrow I = \frac{57 - \sqrt{3}}{5}$

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Question 46:

$\int_{0}^{1} x {(1 - x)}^{5} d x$

Answer:

$Let I = \int_{0}^{1} x {(1 - x)}^{5} d x . Then, I = \int_{0}^{1} (x - 1 + 1) {(1 - x)}^{5} d x \Rightarrow I = \int_{0}^{1} [- {(1 - x)}^{6} + {(1 - x)}^{5}] d x \Rightarrow I = {[\frac{{(1 - x)}^{7}}{7}]}_{0}^{1} - {[\frac{{(1 - x)}^{6}}{6}]}_{0}^{1} \Rightarrow I = - \frac{1}{7} + \frac{1}{6} \Rightarrow I = \frac{1}{42}$

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Question 47:

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

Answer:

$Let I = \int_{1}^{2} (\frac{x - 1}{x^{2}}) e^{x} d x . Then, I = \int_{1}^{2} (\frac{e^{x}}{x} - \frac{e^{x}}{x^{2}}) d x \Rightarrow I = \int_{1}^{2} \frac{e^{x}}{x} d x - \int_{1}^{2} \frac{e^{x}}{x^{2}} d x Integrating first term by parts I = \{{[\frac{e^{x}}{x}]}_{1}^{2} - \int_{1}^{2} \frac{- 1}{x^{2}} e^{x} d x\} - \int_{1}^{2} \frac{e^{x}}{x^{2}} d x \Rightarrow I = {[\frac{e^{x}}{x}]}_{1}^{2} + \int_{1}^{2} \frac{e^{x}}{x^{2}} d x - \int_{1}^{2} \frac{e^{x}}{x^{2}} d x \Rightarrow I = {[\frac{e^{x}}{x}]}_{1}^{2} \Rightarrow I = \frac{e^{2}}{2} - e$

Page No 19.17:

Question 48:

$\int_{0}^{1} (x e^{2 x} + \sin \frac{πx}{2}) d x$

Answer:

$Let I = \int_{0}^{1} (x e^{2 x} + \sin \frac{πx}{2}) d x . Then, I = \int_{0}^{1} x e^{2 x} d x + \int_{0}^{1} \sin \frac{π x}{2} d x Integrating first term by parts I = {[x \frac{e^{2 x}}{2}]}_{0}^{1} - \int_{0}^{1} 1 \frac{e^{2 x}}{2} d x + {[- \frac{\cos \frac{πx}{2}}{\frac{π}{2}}]}_{0}^{1} \Rightarrow I = {[x \frac{e^{2 x}}{2}]}_{0}^{1} - {[\frac{e^{2 x}}{4}]}_{0}^{1} - \frac{2}{π} {[\cos \frac{πx}{2}]}_{0}^{1} \Rightarrow I = \frac{e^{2}}{2} - \frac{e^{2}}{4} + \frac{1}{4} + \frac{2}{π} \Rightarrow I = \frac{e^{2}}{4} + \frac{1}{4} + \frac{2}{π}$

Page No 19.17:

Question 49:

$\int_{0}^{1} (x e^{x} + \cos \frac{πx}{4}) d x$

Answer:

$Let I = \int_{0}^{1} (x e^{x} + \cos \frac{π x}{4}) d x . Then, I = \int_{0}^{1} x e^{x} d x + \int_{0}^{1} \cos \frac{π x}{4} d x Integrating first term by parts I = \{{[x e^{x}]}_{0}^{1} - \int_{0}^{1} 1 e^{x} d x\} + {[\frac{\sin \frac{π x}{4}}{\frac{π}{4}}]}_{0}^{1} \Rightarrow I = {[x e^{x}]}_{0}^{1} - {[e^{x}]}_{0}^{1} + {[\frac{\sin \frac{π x}{4}}{\frac{π}{4}}]}_{0}^{1} \Rightarrow I = e - e + 1 + \frac{4}{π} \sin \frac{π}{4} \Rightarrow I = 1 + \frac{4}{π \sqrt{2}} \Rightarrow I = 1 + \frac{2 \sqrt{2}}{π}$

Disclaimer: The answer given in the book has some error. The solution here is created according to the question given in the book.

Page No 19.17:

Question 50:

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

Answer:

$Let I = \int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x . Then, I = \int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^{2} \frac{x}{2}}) d x [As, \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}, \cos A = 1 - 2 \sin^{2} \frac{A}{2}] \Rightarrow I = \int_{\frac{π}{2}}^{π} e^{x} (\frac{1}{2} {cosec}^{2} \frac{x}{2} - \cot \frac{x}{2}) d x \Rightarrow I = \int_{\frac{π}{2}}^{π} \frac{1}{2} e^{x} {cosec}^{2} \frac{x}{2} d x - \int_{\frac{π}{2}}^{π} e^{x} \cot \frac{x}{2} d x Integrating second term by parts I = \{- {[e^{x} \cot \frac{x}{2}]}_{\frac{π}{2}}^{π} - \int_{\frac{π}{2}}^{π} \frac{1}{2} e^{x} {cosec}^{2} \frac{x}{2} d x\} + \int_{\frac{π}{2}}^{π} \frac{1}{2} e^{x} {cosec}^{2} \frac{x}{2} d x \Rightarrow I = - [0 - e^{\frac{π}{2}}] \Rightarrow I = e^{\frac{π}{2}}$

Page No 19.17:

Question 51:

$\int_{0}^{2 π} e^{x / 2} \sin (\frac{x}{2} + \frac{π}{4}) d x$

Answer:

$Let I = \int_{0}^{2 π} e^{\frac{x}{2}} \sin (\frac{x}{2} + \frac{π}{4}) d x . Then, Integrating by parts I = {[- 2 e^{\frac{x}{2}} \cos (\frac{x}{2} + \frac{π}{4})]}_{0}^{2 π} - \int_{0}^{2 π} - \frac{2}{2} e^{\frac{x}{2}} \cos (\frac{x}{2} + \frac{π}{4}) d x Again, integrating second term by parts \Rightarrow I = {[- 2 e^{\frac{x}{2}} \cos (\frac{x}{2} + \frac{π}{4})]}_{0}^{2 π} + \{{[2 e^{\frac{x}{2}} \sin (\frac{x}{2} + \frac{π}{4})]}_{0}^{2 π} - \int_{0}^{2 π} \frac{2}{2} e^{\frac{x}{2}} \sin (\frac{x}{2} + \frac{π}{4}) d x\} \Rightarrow I = {[- 2 e^{\frac{x}{2}} \cos (\frac{x}{2} + \frac{π}{4})]}_{0}^{2 π} + {[2 e^{\frac{x}{2}} \sin (\frac{x}{2} + \frac{π}{4})]}_{0}^{2 π} - I \Rightarrow 2 I = \frac{2}{\sqrt{2}} e^{π} + \frac{2}{\sqrt{2}} - \frac{2}{\sqrt{2}} e^{π} - \frac{2}{\sqrt{2}} = 0 \Rightarrow I = 0$

Page No 19.17:

Question 52:

$\int_{0}^{2 π} e^{x} \cos (\frac{π}{4} + \frac{x}{2}) d x$

Answer:

$Let I = \int_{0}^{2 π} e^{x} \cos (\frac{π}{4} + \frac{x}{2}) d x . Then, Integrating by parts I = {[2 e^{x} \sin (\frac{π}{4} + \frac{x}{2})]}_{0}^{2 π} - \int_{0}^{2 π} 2 e^{x} \sin (\frac{π}{4} + \frac{x}{2}) d x Integrating second term by parts I = {[2 e^{x} \sin (\frac{π}{4} + \frac{x}{2})]}_{0}^{2 π} + \{{[4 e^{x} \cos (\frac{π}{4} + \frac{x}{2})]}_{0}^{2 π} + \int_{0}^{2 π} - 4 e^{x} \cos (\frac{π}{4} + \frac{x}{2}) d x\} \Rightarrow I = {[2 e^{x} \sin (\frac{π}{4} + \frac{x}{2})]}_{0}^{2 π} + {[4 e^{x} \cos (\frac{π}{4} + \frac{x}{2})]}_{0}^{2 π} - 4 I \Rightarrow 5 I = - 2 e^{2 π} \frac{1}{\sqrt{2}} - 2 \frac{1}{\sqrt{2}} - 4 e^{2 π} \frac{1}{\sqrt{2}} - 4 \frac{1}{\sqrt{2}} \Rightarrow 5 I = - 3 \sqrt{2} e^{2 π} - 3 \sqrt{2} \Rightarrow I = - \frac{3 \sqrt{2}}{5} (e^{2 π} + 1)$

Page No 19.17:

Question 53:

Evaluate $\int_{0}^{π} e^{2 x} \cdot \sin (\frac{π}{4} + x) d x$

Answer:

$Let I = \int_{0}^{π} e^{2 x} \sin (\frac{π}{4} + x) d x Integrating by parts, we get I = \frac{1}{2} {[e^{2 x} \sin (\frac{π}{4} + x)]}_{0}^{π} - \frac{1}{2} \int_{0}^{π} e^{2 x} \cos (\frac{π}{4} + x) d x Now, integrating the second term by parts, we get \Rightarrow I = \frac{1}{2} {[e^{2 x} \sin (\frac{π}{4} + x)]}_{0}^{π} - \frac{1}{2} \{{[\frac{1}{2} e^{2 x} \cos (\frac{π}{4} + x)]}_{0}^{π} + \frac{1}{2} \int_{0}^{π} e^{2 x} \sin (\frac{π}{4} + x) d x\} \Rightarrow I = \frac{1}{2} {[e^{2 x} \sin (\frac{π}{4} + x)]}_{0}^{π} - \frac{1}{4} {[e^{2 x} \cos (\frac{π}{4} + x)]}_{0}^{π} - \frac{1}{4} I \Rightarrow \frac{5}{4} I = \frac{1}{2} [e^{2 π} \sin (π + \frac{π}{4}) - \sin (\frac{π}{4})] - \frac{1}{4} [e^{2 π} \cos (π + \frac{π}{4}) - \cos (\frac{π}{4})] \Rightarrow \frac{5}{4} I = \frac{1}{2} [- e^{2 π} \times \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}] - \frac{1}{4} [- e^{2 π} \times \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}] \Rightarrow \frac{5}{4} I = - \frac{1}{2 \sqrt{2}} e^{2 π} - \frac{1}{2 \sqrt{2}} + \frac{1}{4 \sqrt{2}} e^{2 π} + \frac{1}{4 \sqrt{2}} \Rightarrow I = - \frac{1}{5 \sqrt{2}} (e^{2 π} + 1)$

Page No 19.17:

Question 54:

$\int_{0}^{1} \frac{1}{\sqrt{1 + x} - \sqrt{x}} d x$

Answer:

$Let I = \int_{0}^{1} \frac{1}{\sqrt{1 + x} - \sqrt{x}} d x . Then, I = \int_{0}^{1} (\frac{1}{\sqrt{1 + x} - \sqrt{x}} \times \frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}}) d x \Rightarrow I = \int_{0}^{1} \frac{\sqrt{1 + x} + \sqrt{x}}{1 + x - x} d x \Rightarrow I = \int_{0}^{1} (\sqrt{1 + x} + \sqrt{x}) d x \Rightarrow I = {[\frac{2}{3} {(1 + x)}^{\frac{3}{2}} + \frac{2}{3} x^{\frac{3}{2}}]}_{0}^{1} \Rightarrow I = \frac{2}{3} \times 2 \sqrt{2} + \frac{2}{3} - \frac{2}{3} \Rightarrow I = \frac{4 \sqrt{2}}{3}$

Page No 19.17:

Question 55:

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

Answer:

$Let I = \int_{1}^{2} \frac{x}{(x + 1) (x + 2)} d x . Then, I = \int_{1}^{2} (\frac{- 1}{(x + 1)} + \frac{2}{(x + 2)}) d x \Rightarrow I = - \int_{1}^{2} \frac{1}{(x + 1)} d x + 2 \int_{1}^{2} \frac{1}{(x + 2)} d x \Rightarrow I = {[- \log (x + 1) + 2 \log (x + 2)]}_{1}^{2} \Rightarrow I = - \log 3 + 2 \log 4 + \log 2 - 2 \log 3 \Rightarrow I = 5 \log 2 - 3 \log 3 \Rightarrow I = \log 2^{5} - \log 3^{3} \Rightarrow I = \log \frac{32}{27}$

Page No 19.17:

Question 56:

$\int_{0}^{π / 2} \sin^{3} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \sin^{3} x d x . Then, I = \int_{0}^{\frac{π}{2}} \sin x \sin^{2} x d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \sin x (1 - \cos^{2} x) d x Let u = \cos x, d u = - \sin x d x ∴ I = \int - (1 - u^{2}) d u \Rightarrow I = [\frac{u^{3}}{3} - u] \Rightarrow I = {[\frac{\cos^{3} x}{3} - \cos x]}_{0}^{\frac{π}{2}} \Rightarrow I = 0 - \frac{1}{3} + 1 \Rightarrow I = \frac{2}{3}$

Page No 19.17:

Question 57:

$\int_{0}^{π} (\sin^{2} \frac{x}{2} - \cos^{2} \frac{x}{2}) d x$

Answer:

$Let I = \int_{0}^{π} (\sin^{2} \frac{x}{2} - \cos^{2} \frac{x}{2}) d x . Then, I = - \int_{0}^{π} \cos x d x [∵ \cos A = \cos^{2} \frac{A}{2} - \sin^{2} \frac{A}{2}] \Rightarrow I = - {[\sin x]}_{0}^{π} \Rightarrow I = 0$

Page No 19.17:

Question 58:

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

Answer:

$Let I \int_{1}^{2} e^{2 x} (\frac{1}{x} - \frac{1}{2 x^{2}}) d x . Then, I = \int_{1}^{2} e^{2 x} \frac{1}{x} - \int_{1}^{2} e^{2 x} \frac{1}{2 x^{2}} d x Integrating first term by parts \Rightarrow I = \{{[\frac{e^{2 x}}{2 x}]}_{1}^{2} - \int_{1}^{2} - e^{2 x} \frac{1}{2 x^{2}}\} - \int_{1}^{2} e^{2 x} \frac{1}{2 x^{2}} d x \Rightarrow I = {[\frac{e^{2 x}}{2 x}]}_{1}^{2} \Rightarrow I = \frac{e^{4}}{4} - \frac{e^{2}}{2} \Rightarrow I = \frac{e^{4} - 2 e^{2}}{4}$

Page No 19.17:

Question 59:

Evaluate the following definite integrals:

$\int_{0}^{1} \frac{1}{\sqrt{(x - 1) (2 - x)}} d x$ [NCERT EXEMPLAR]

Answer:

Let I = $\int_{1}^{2} \frac{1}{\sqrt{(x - 1) (2 - x)}} d x$

Put $x = \cos^{2} θ + 2 \sin^{2} θ$

$∴ d x = 2 \cos θ (- \sin θ) d θ + 4 \sin θ \cos θ d θ = 2 \sin θ \cos θ d θ$

Also,

$x = \cos^{2} θ + 2 \sin^{2} θ \Rightarrow x = 1 + \sin^{2} θ \Rightarrow \sin θ = \sqrt{x - 1}$

When $x \to 1, \sin θ \to 0 or θ \to 0$

When $x \to 2, \sin θ \to 1 or θ \to \frac{π}{2}$

∴ I = $\int_{1}^{2} \frac{1}{\sqrt{(x - 1) (2 - x)}} d x$
$\Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{2 \sin θ \cos θ d θ}{\sqrt{(\cos^{2} θ + 2 \sin^{2} θ - 1) (2 - \cos^{2} θ - 2 \sin^{2} θ)}} \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{2 \sin θ \cos θ d θ}{\sqrt{\sin^{2} θ \cos^{2} θ}} (\sin^{2} θ + \cos^{2} θ = 1) \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{2 \sin θ \cos θ d θ}{\sin θ \cos θ} \Rightarrow I = 2 \int_{0}^{\frac{π}{2}} d θ \Rightarrow I = 2 θ |_{0}^{\frac{π}{2}}$
$\Rightarrow I = 2 (\frac{π}{2} - 0) = π$

Page No 19.18:

Question 60:

If $\int_{0}^{k} \frac{1}{2 + 8 x^{2}} d x = \frac{π}{16},$ find the value of k.

Answer:

$We have, \int_{0}^{k} \frac{1}{2 + 8 x^{2}} d x = \frac{π}{16} \Rightarrow \frac{1}{8} \int_{0}^{k} \frac{1}{\frac{1}{4} + x^{2}} d x = \frac{π}{16} \Rightarrow \frac{1}{4} {[\tan^{- 1} 2 x]}_{0}^{k} = \frac{π}{16} \Rightarrow \tan^{- 1} 2 k = \frac{π}{4} \Rightarrow 2 k = \tan \frac{π}{4} \Rightarrow 2 k = 1 \Rightarrow k = \frac{1}{2}$

Page No 19.18:

Question 61:

If $\int_{0}^{a} 3 x^{2} d x = 8,$ find the value of a.

Answer:

$We have, \int_{0}^{a} 3 x^{2} d x = 8 \Rightarrow {[3 \frac{x^{3}}{3}]}_{0}^{a} = 8 \Rightarrow a^{3} = 8 \Rightarrow a = 2$

Page No 19.18:

Question 62:

$\int_{π}^{\frac{3 π}{2}} \sqrt{1 - \cos 2 x} d x$

Answer:

$\int_{π}^{\frac{3 π}{2}} \sqrt{1 - \cos 2 x} d x$
$= \int_{π}^{\frac{3 π}{2}} \sqrt{2 \sin^{2} x} d x = \sqrt{2} \int_{π}^{\frac{3 π}{2}} |\sin x| d x = - \sqrt{2} \int_{π}^{\frac{3 π}{2}} \sin x d x (\sin x < 0 for π \leq x \leq 2 π)$
$= - \sqrt{2} (- \cos x) |_{π}^{\frac{3 π}{2}} = \sqrt{2} (\cos \frac{3 π}{2} - cosπ) = \sqrt{2} [0 - (- 1)] = \sqrt{2} \times 1 = \sqrt{2}$

Page No 19.18:

Question 63:

$\int_{0}^{2 π} \sqrt{1 + \sin \frac{x}{2}} d x$

Answer:

$I = \int_{0}^{2 π} \sqrt{1 + \sin \frac{x}{2}} d x = \int_{0}^{2 π} \sqrt{\cos^{2} \frac{x}{4} + \sin^{2} \frac{x}{4} + 2 \sin \frac{x}{4} \cos \frac{x}{4}} d x = \int_{0}^{2 π} \sqrt{{(\cos \frac{x}{4} + \sin \frac{x}{4})}^{2}} d x = \int_{0}^{2 π} |\cos \frac{x}{4} + \sin \frac{x}{4}| d x$

When $0 \leq x \leq 2 π$ , $0 \leq \frac{x}{4} \leq \frac{π}{2}$
$∴ \sin \frac{x}{4} \geq 0, \cos \frac{x}{4} \geq 0 \Rightarrow \cos \frac{x}{4} + \sin \frac{x}{4} \geq 0 \Rightarrow |\cos \frac{x}{4} + \sin \frac{x}{4}| = \cos \frac{x}{4} + \sin \frac{x}{4}$

$∴ I = \int_{0}^{2 π} (\cos \frac{x}{4} + \sin \frac{x}{4}) d x = {\frac{\sin \frac{x}{4}}{\frac{1}{4}}}_{0}^{2 π} + {\frac{(- \cos \frac{x}{4})}{\frac{1}{4}}}_{0}^{2 π} = 4 (\sin \frac{π}{2} - \sin 0) - 4 (\cos \frac{π}{2} - \cos 0) = 4 (1 - 0) - 4 (0 - 1) = 4 + 4 = 8$

Page No 19.18:

Question 64:

$\int_{0}^{\frac{π}{4}} {(\tan x + \cot x)}^{- 2} d x$

Answer:

$\int_{0}^{\frac{π}{4}} {(\tan x + \cot x)}^{- 2} d x = \int_{0}^{\frac{π}{4}} \frac{1}{{(\tan x + \cot x)}^{2}} d x = \int_{0}^{\frac{π}{4}} \frac{1}{{(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x})}^{2}} d x = \int_{0}^{\frac{π}{4}} \frac{1}{{(\frac{\sin^{2} x + \cos^{2} x}{\sin x \cos x})}^{2}} d x = \int_{0}^{\frac{π}{4}} \sin^{2} x \cos^{2} x d x$
$= \frac{1}{4} \int_{0}^{\frac{π}{4}} {(2 \sin x \cos x)}^{2} d x = \frac{1}{4} \int_{0}^{\frac{π}{4}} \sin^{2} 2 x d x = \frac{1}{4} \int_{0}^{\frac{π}{4}} (\frac{1 - \cos 4 x}{2}) d x = \frac{1}{8} \int_{0}^{\frac{π}{4}} d x - \frac{1}{8} \int_{0}^{\frac{π}{4}} \cos 4 x d x = \frac{1}{8} x_{0}^{\frac{π}{4}} - \frac{1}{8} {(\frac{\sin 4 x}{4})}_{0}^{\frac{π}{4}}$
$= \frac{1}{8} (\frac{π}{4} - 0) - \frac{1}{32} (sinπ - \sin 0) = \frac{π}{32} - \frac{1}{32} \times (0 - 0) = \frac{π}{32}$

Page No 19.18:

Question 65:

$\int_{0}^{1} x \log (1 + 2 x) d x$

Answer:

Let I = $\int_{0}^{1} x \log (1 + 2 x) d x$

Applying integration by parts, we have

$I = {\log (1 + 2 x) \frac{x^{2}}{2}}_{0}^{1} - \int_{0}^{1} (\frac{2}{1 + 2 x}) \times \frac{x^{2}}{2} d x = \frac{1}{2} (\log 3 - 0) - \int_{0}^{1} \frac{x^{2}}{1 + 2 x} d x = \frac{1}{2} \log 3 - \frac{1}{4} \int_{0}^{1} \frac{4 x^{2} - 1 + 1}{1 + 2 x} d x = \frac{1}{2} \log 3 - \frac{1}{4} \int_{0}^{1} \frac{(2 x + 1) (2 x - 1)}{1 + 2 x} d x - \frac{1}{4} \int_{0}^{1} \frac{1}{1 + 2 x} d x = \frac{1}{2} \log 3 - \frac{1}{4} \int_{0}^{1} (2 x - 1) d x - \frac{1}{4} \int_{0}^{1} \frac{1}{1 + 2 x} d x$
$= \frac{1}{2} \log 3 - \frac{1}{4} \times {\frac{{(2 x - 1)}^{2}}{2 \times 2}}_{0}^{1} - \frac{1}{4} \times {\frac{\log (1 + 2 x)}{2}}_{0}^{1} = \frac{1}{2} \log 3 - \frac{1}{16} (1 - 1) - \frac{1}{8} (\log 3 - \log 1) = \frac{1}{2} \log 3 - 0 - \frac{1}{8} \log 3 (\log 1 = 0) = \frac{3}{8} \log 3$

Page No 19.18:

Question 66:

$\int_{\frac{π}{6}}^{\frac{π}{3}} {(\tan x + \cot x)}^{2} d x$

Answer:

$\int_{\frac{π}{6}}^{\frac{π}{3}} {(\tan x + \cot x)}^{2} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} (\tan^{2} x + \cot^{2} x + 2 \tan x \cot x) d x = \int_{\frac{π}{6}}^{\frac{π}{3}} (\sec^{2} x - 1 + {cosec}^{2} x - 1 + 2) d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \sec^{2} x d x + \int_{\frac{π}{6}}^{\frac{π}{3}} {cosec}^{2} x d x$
$= {\tan x}_{\frac{π}{6}}^{\frac{π}{3}} + {(- \cot x)}_{\frac{π}{6}}^{\frac{π}{3}} = (\tan \frac{π}{3} - \tan \frac{π}{6}) - (\cot \frac{π}{3} - \cot \frac{π}{6}) = (\sqrt{3} - \frac{1}{\sqrt{3}}) - (\frac{1}{\sqrt{3}} - \sqrt{3}) = 2 \sqrt{3} - \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}}$

Page No 19.18:

Question 67:

$\int_{0}^{\frac{π}{4}} (a^{2} \cos^{2} x + b^{2} \sin^{2} x) d x$

Answer:

$\int_{0}^{\frac{π}{4}} (a^{2} \cos^{2} x + b^{2} \sin^{2} x) d x = \int_{0}^{\frac{π}{4}} [a^{2} (\frac{1 + \cos 2 x}{2}) + b^{2} (\frac{1 - \cos 2 x}{2})] d x = \int_{0}^{\frac{π}{4}} [(\frac{a^{2} + b^{2}}{2}) + (\frac{a^{2} - b^{2}}{2}) \cos 2 x] d x = (\frac{a^{2} + b^{2}}{2}) \int_{0}^{\frac{π}{4}} d x + (\frac{a^{2} - b^{2}}{2}) \int_{0}^{\frac{π}{4}} \cos 2 x d x$
$= (\frac{a^{2} + b^{2}}{2}) {\times x}_{0}^{\frac{π}{4}} + (\frac{a^{2} - b^{2}}{2}) \times {\frac{\sin 2 x}{2}}_{0}^{\frac{π}{4}} = (\frac{a^{2} + b^{2}}{2}) (\frac{π}{4} - 0) + (\frac{a^{2} - b^{2}}{4}) (\sin \frac{π}{2} - \sin 0) = (\frac{a^{2} + b^{2}}{2}) \frac{π}{4} + (\frac{a^{2} - b^{2}}{4}) (1 - 0) = (a^{2} + b^{2}) \frac{π}{8} + \frac{1}{4} (a^{2} - b^{2})$

Page No 19.18:

Question 68:

$\int_{0}^{1} \frac{1}{1 + 2 x + 2 x^{2} + 2 x^{3} + x^{4}} d x$

Answer:

$\int_{0}^{1} \frac{1}{1 + 2 x + 2 x^{2} + 2 x^{3} + x^{4}} d x = \int_{0}^{1} \frac{1}{{(x^{2} + 1)}^{2} + 2 x (x^{2} + 1)} d x = \int_{0}^{1} \frac{1}{(x^{2} + 1) (x^{2} + 1 + 2 x)} d x = \int_{0}^{1} \frac{1}{(x^{2} + 1) {(x + 1)}^{2}} d x$

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

Putting x = −1, we have

1 = 2B $\Rightarrow B = \frac{1}{2}$            .....(1)

Putting x = 0, we have

A + B + D = 1              .....(2)

Equating coefficient of x³ on both sides, we have

A + C = 0                    .....(3)

Equating coefficient of x²on both sides, we have

A + B + 2C + D = 0               .....(4)

⇒ 2C = −1               [Using (1)]

$\Rightarrow C = - \frac{1}{2}$

$∴ A = \frac{1}{2}$    [Using (3)]

Putting $A = \frac{1}{2}, B = \frac{1}{2}$ and $C = - \frac{1}{2}$ in (4), we have

D = 0

$∴ \int_{0}^{1} \frac{1}{{(x + 1)}^{2} (x^{2} + 1)} d x = \int_{0}^{1} \frac{\frac{1}{2}}{x + 1} d x + \int_{0}^{1} \frac{\frac{1}{2}}{{(x + 1)}^{2}} d x + \int_{0}^{1} \frac{- \frac{1}{2} x}{x^{2} + 1} = \frac{1}{2} {\log (x + 1)}_{0}^{1} + \frac{1}{2} \times {(- \frac{1}{x + 1})}_{0}^{1} - \frac{1}{4} \int_{0}^{1} \frac{2 x}{x^{2} + 1} d x = \frac{1}{2} (\log 2 - \log 1) - \frac{1}{2} (\frac{1}{2} - 1) - \frac{1}{4} {\log (x^{2} + 1)}_{0}^{1} = \frac{1}{2} \log 2 + \frac{1}{4} - \frac{1}{4} (\log 2 - \log 1) (\log 1 = 0)$
$= \frac{1}{2} \log 2 + \frac{1}{4} \log e - \frac{1}{4} \log 2 = \frac{1}{4} \log 2 + \frac{1}{4} \log e = \frac{1}{4} (\log 2 + \log e) = \frac{1}{4} \log (2 e)$

Page No 19.38:

Question 1:

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

Answer:

$Let x^{2} = t . Then, 2 x d x = d t When x = 2, t = 4 and x = 4, t = 16 . ∴ I = \int_{2}^{4} \frac{x}{x^{2} + 1} d x \Rightarrow I = \int_{4}^{16} \frac{1}{2} \frac{d t}{t + 1} \Rightarrow I = \frac{1}{2} {[\log (t + 1)]}_{4}^{16} \Rightarrow I = \frac{1}{2} \log 17 - \frac{1}{2} \log 5 \Rightarrow I = \frac{1}{2} \log \frac{17}{5}$

Page No 19.38:

Question 2:

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

Answer:

$Let 1 + \log x = t . Then, \frac{1}{x} d x = d t When x = 1, t = 1 and x = 2, t = (1 + \log 2) ∴ I = \int_{1}^{2} \frac{1}{x {(1 + \log x)}^{2}} d x \Rightarrow I = \int_{1}^{(1 + \log 2)} \frac{1}{t^{2}} d t \Rightarrow I = {[\frac{- 1}{t}]}_{1}^{(1 + \log 2)} \Rightarrow I = - \frac{1}{(1 + \log 2)} + 1 \Rightarrow I = \frac{\log 2}{\log 2 + \log e} \Rightarrow I = \frac{\log 2}{\log 2 e}$

Page No 19.38:

Question 3:

$\int_{1}^{2} \frac{3 x}{9 x^{2} - 1} d x$

Answer:

$Let x^{2} = t . Then, 2 x d x = d t When x = 1, t = 1 and x = 2, t = 4 ∴ I = \int_{1}^{2} \frac{3 x}{9 x^{2} - 1} d x \Rightarrow I = \frac{3}{2} \int_{1}^{4} \frac{d t}{9 t - 1} \Rightarrow I = \frac{3}{18} {[\log (9 t - 1)]}_{1}^{4} \Rightarrow I = \frac{3}{18} (\log 35 - \log 8) \Rightarrow I = \frac{(\log 35 - \log 8)}{6}$

Page No 19.38:

Question 4:

$\int_{0}^{π / 2} \frac{1}{5 \cos x + 3 \sin x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{1}{5 \cos x + 3 \sin x} d x . Then, I = \int_{0}^{\frac{π}{2}} \frac{1}{5 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) + 3 (\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x [∵ \sin A = (\frac{2 \tan \frac{A}{2}}{1 + \tan^{2} \frac{A}{2}}), \cos A = (\frac{1 - \tan^{2} \frac{A}{2}}{1 + \tan^{2} \frac{A}{2}})] \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{5 - 5 \tan^{2} \frac{x}{2} + 6 \tan \frac{x}{2}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} \frac{x}{2}}{5 - 5 \tan^{2} \frac{x}{2} + 6 \tan \frac{x}{2}} d x Let \tan \frac{x}{2} = t . Then, \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t Also, x = 0, t = 0 and x = \frac{π}{2}, t = 1 ∴ I = \int_{0}^{1} \frac{2 d t}{5 - 5 t^{2} + 6 t} \Rightarrow I = \frac{1}{5} \int_{0}^{1} \frac{2 d t}{1 - t^{2} + \frac{6}{5} t + \frac{36}{100} - \frac{36}{100}} = \frac{2}{5} \int_{0}^{1} \frac{d t}{- {(t - \frac{6}{10})}^{2} + \frac{136}{100}} = \frac{2}{5} \times \frac{10}{\sqrt{136}} {[- \log (\frac{t - \frac{6}{10} - \frac{\sqrt{136}}{10}}{t - \frac{6}{10} + \frac{\sqrt{136}}{10}})]}_{0}^{1} = \frac{1}{\sqrt{34}} [- \log (\frac{4 - 2 \sqrt{34}}{4 + 2 \sqrt{34}}) + \log (\frac{- 6 - 2 \sqrt{34}}{- 6 + 2 \sqrt{34}})] = \frac{1}{\sqrt{34}} \log (\frac{6 + 2 \sqrt{34}}{6 - 2 \sqrt{34}} \times \frac{4 + 2 \sqrt{34}}{4 - 2 \sqrt{34}}) = \frac{1}{\sqrt{34}} \log (\frac{160 + 20 \sqrt{34}}{160 - 20 \sqrt{34}}) = \frac{1}{\sqrt{34}} \log (\frac{8 + \sqrt{34}}{8 - \sqrt{34}})$

Page No 19.38:

Question 5:

$\int_{0}^{a} \frac{x}{\sqrt{a^{2} + x^{2}}} d x$

Answer:

$Let x = a \tan t . Then, d x = a \sec^{2} t d t When x = 0, t = 0 and x = a, t = \frac{π}{4} ∴ I = \int_{0}^{a} \frac{x}{\sqrt{a^{2} + x^{2}}} d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{a \tan t}{\sqrt{a^{2} + a^{2} \tan^{2} t}} a \sec^{2} t d t = \int_{0}^{\frac{π}{4}} \frac{(a \tan t) a \sec^{2} t}{a \sec t} d t = \int_{0}^{\frac{π}{4}} a \tan t \sec t d t = a {[\sec t]}_{0}^{\frac{π}{4}} = a (\sqrt{2} - 1)$

Page No 19.38:

Question 6:

$\int_{0}^{1} \frac{e^{x}}{1 + e^{2 x}} d x$

Answer:

$Let e^{x} = t . Then, e^{x} d x = d t When x = 0, t = 1 and x = 1, t = e ∴ I = \int_{0}^{1} \frac{e^{x}}{1 + e^{2 x}} d x \Rightarrow I = \int_{1}^{e} \frac{d t}{1 + t^{2}} \Rightarrow I = {[\tan^{- 1} x]}_{1}^{e} \Rightarrow I = \tan^{- 1} e - \tan^{- 1} 1 \Rightarrow I = \tan^{- 1} e - \frac{π}{4}$

Page No 19.38:

Question 7:

$\int_{0}^{1} x e^{x^{2}} d x$

Answer:

$Let I = \int_{0}^{1} x e^{x^{2}} d x . Let x^{2} = t . Then, 2 x d x = d t When x = 0, t = 0 and x = 1, t = 1 ∴ I = \frac{1}{2} \int_{0}^{1} e^{t} d t \Rightarrow I = \frac{1}{2} {(e^{t})}_{0}^{1} \Rightarrow I = \frac{1}{2} (e - 1)$

Page No 19.38:

Question 8:

$\int_{1}^{3} \frac{\cos (\log x)}{x} d x$

Answer:

$Let I = \int_{1}^{3} \frac{\cos (\log x)}{x} d x . Let \log x = t . Then, \frac{1}{x} d x = d t When x = 1, t = 0 and x = 3, t = \log 3 ∴ I = \int_{0}^{\log 3} \cos t d t = {[\sin t]}_{0}^{\log 3} = \sin (\log 3)$

Page No 19.38:

Question 9:

$\int_{0}^{1} \frac{2 x}{1 + x^{4}} d x$

Answer:

$Let I = \int_{0}^{1} \frac{2 x}{1 + x^{4}} d x . Let x^{2} = t . Then, 2 x d x = d t When x = 0, t = 0 and x = 1, t = 1 ∴ I = \int_{0}^{1} \frac{2 x}{1 + x^{4}} d x \Rightarrow I = \int_{0}^{1} \frac{1}{1 + t^{2}} d t \Rightarrow I = {[\tan^{- 1} t]}_{0}^{1} \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} 0 \Rightarrow I = \frac{π}{4}$

Page No 19.38:

Question 10:

$\int_{0}^{a} \sqrt{a^{2} - x^{2}} d x$

Answer:

$Let I = \int_{0}^{a} \sqrt{a^{2} - x^{2}} d x . Let x = a \sin t . Then, d x = a \cos t d t When x = 0, t = 0 and x = a, t = \frac{π}{2} ∴ I = \int_{0}^{a} \sqrt{a^{2} - x^{2}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \sqrt{(a^{2} - a^{2} \sin^{2} t)} a \cos t d t \Rightarrow I = \int_{0}^{\frac{π}{2}} a^{2} \cos^{2} t d t \Rightarrow I = a^{2} \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 t}{2} d t \Rightarrow I = \frac{a^{2}}{2} {[t + \frac{\sin 2 t}{2}]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{a^{2}}{2} (\frac{π}{2} - 0) \Rightarrow I = \frac{{πa}^{2}}{4}$

Page No 19.39:

Question 11:

$\int_{0}^{π / 2} \sqrt{\sin ϕ} \cos^{5} ϕ d ϕ$

Answer:

$\int_{0}^{\frac{π}{2}} \sqrt{\sin ϕ} \cos^{5} ϕ d ϕ Let \sin ϕ = t . Then, \cos ϕ d ϕ = d t When ϕ = 0, t = 0 and ϕ = \frac{π}{2}, t = 1 Also, \cos^{5} ϕ = \cos^{4} ϕ \cos ϕ = {(1 - \sin^{2} ϕ)}^{2} \cos ϕ ∴ I = \int_{0}^{\frac{π}{2}} \sqrt{\sin ϕ} \cos^{5} ϕ d ϕ \Rightarrow I = \int_{0}^{1} \sqrt{t} {(1 - t^{2})}^{2} d t \Rightarrow I = \int_{0}^{1} \sqrt{t} (1 + t^{4} - 2 t^{2}) d t \Rightarrow I = \int_{0}^{1} (\sqrt{t} + t^{\frac{9}{2}} - 2 t^{\frac{5}{2}}) d t \Rightarrow I = {[\frac{2 t^{\frac{3}{2}}}{3} + \frac{2 t^{\frac{11}{2}}}{11} - \frac{4 t^{\frac{7}{2}}}{7}]}_{0}^{1} \Rightarrow I = \frac{2}{3} + \frac{2}{11} - \frac{4}{7} \Rightarrow I = \frac{64}{231}$

Page No 19.39:

Question 12:

$\int_{0}^{π / 2} \frac{\cos x}{1 + \sin^{2} x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\cos x}{1 + \sin^{2} x} d x . Let \sin x = t . Then, \cos x d x = d t When x = 0, t = 0 and x = \frac{π}{2}, t = 1 ∴ I = \int_{0}^{\frac{π}{2}} \frac{\cos x}{1 + \sin^{2} x} d x \Rightarrow I = \int_{0}^{1} \frac{1}{1 + t^{2}} d t \Rightarrow I = {[\tan^{- 1} t]}_{0}^{1} \Rightarrow I = \frac{π}{4}$

Page No 19.39:

Question 13:

$\int_{0}^{π / 2} \frac{\sin θ}{\sqrt{1 + \cos θ}} d θ$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sqrt{1 + \cos θ}} d θ . Let \cos θ = t . Then, - \sin θ d θ = d t When θ = 0, t = 1 and θ = \frac{π}{2}, t = 0 ∴ I = \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sqrt{1 + \cos θ}} d θ = \int_{1}^{0} \frac{- d t}{\sqrt{1 + t}} = \int_{0}^{1} \frac{d t}{\sqrt{1 + t}} = 2 {[\sqrt{1 + t}]}_{0}^{1} = 2 (\sqrt{2} - 1)$

Page No 19.39:

Question 14:

$\int_{0}^{π / 3} \frac{\cos x}{3 + 4 \sin x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{3}} \frac{\cos x}{3 + 4 \sin x} d x . Let \sin x = t . Then, \cos x d x = d t When x = 0, t = 0 and x = \frac{π}{3}, t = \frac{\sqrt{3}}{2} ∴ I = \int_{0}^{\frac{π}{3}} \frac{\cos x}{3 + 4 \sin x} d x = \int_{0}^{\frac{\sqrt{3}}{2}} \frac{1}{3 + 4 t} d t = \frac{1}{4} {[\log (3 + 4 t)]}_{0}^{\frac{\sqrt{3}}{2}} = \frac{1}{4} (\log (3 + 2 \sqrt{3}) - \log 3) = \frac{1}{4} \log (\frac{3 + 2 \sqrt{3}}{3})$

Page No 19.39:

Question 15:

$\int_{0}^{1} \frac{\sqrt{\tan^{- 1} x}}{1 + x^{2}} d x$

Answer:

$Let I = \int_{0}^{1} \frac{\sqrt{\tan^{- 1} x}}{1 + x^{2}} d x . Let \tan^{- 1} x = t . Then, \frac{1}{1 + x^{2}} d x = d t When x = 0, t = 0 and x = 1, t = \frac{π}{4} ∴ I = \int_{0}^{1} \frac{\sqrt{\tan^{- 1} x}}{1 + x^{2}} d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \sqrt{t} d t \Rightarrow I = {[\frac{2 t^{\frac{3}{2}}}{3}]}_{0}^{\frac{π}{4}} \Rightarrow I = \frac{2}{3} {(\frac{π}{4})}^{\frac{3}{2}} \Rightarrow I = \frac{1}{12} π^{\frac{3}{2}}$

Page No 19.39:

Question 16:

$\int_{0}^{2} x \sqrt{x + 2} d x$

Answer:

$Let I = \int_{0}^{2} x \sqrt{x + 2} d x . Let x + 2 = t^{2} . Then, d x = 2 t d t When x = 0, t = \sqrt{2} and x = 2, t = 2 ∴ I = \int_{\sqrt{2}}^{2} (t^{2} - 2) t 2 t d t \Rightarrow I = 2 \int_{\sqrt{2}}^{2} (t^{4} - 2 t^{2}) d t \Rightarrow I = 2 {[\frac{t^{5}}{5} - \frac{2}{3} t^{3}]}_{\sqrt{2}}^{2} \Rightarrow I = 2 [(\frac{32}{3} - \frac{16}{3}) - (\frac{4 \sqrt{2}}{5} - \frac{4 \sqrt{2}}{3})] \Rightarrow I = 2 (\frac{16}{15} + \frac{8 \sqrt{2}}{15}) \Rightarrow I = \frac{16}{15} (2 + \sqrt{2})$

Page No 19.39:

Question 17:

$\int_{0}^{1} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x$

Answer:

$\int_{0}^{1} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x = \int_{0}^{1} 2 \tan^{- 1} x = 2 {[x \tan^{- 1} x]}_{0}^{1} - 2 \int_{0}^{1} \frac{x}{1 + x^{2}} d x = 2 {[x \tan^{- 1} x]}_{0}^{1} - {[\log (1 + x^{2})]}_{0}^{1} = 2 \frac{π}{4} - 0 - \log 2 + 0 = \frac{π}{2} - \log 2$

Page No 19.39:

Question 18:

$\int_{0}^{π / 2} \frac{\sin x \cos x}{1 + \sin^{4} x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 + \sin^{4} x} d x . Let \sin x = t . Then, \cos x d x = d t When x = 0, t = 0 and x = \frac{π}{2}, t = 1 ∴ I = \int_{0}^{1} \frac{t}{1 + t^{4}} d t Let t^{2} = u . Then, 2 t d t = d u So, I = \int_{0}^{1} \frac{t}{1 + t^{4}} d t \Rightarrow I = \frac{1}{2} \int_{0}^{1} \frac{1}{1 + u^{2}} d u \Rightarrow I = \frac{1}{2} {[\tan^{- 1} u]}_{0}^{1} \Rightarrow I = \frac{π}{8}$

Page No 19.39:

Question 19:

$\int_{0}^{π / 2} \frac{d x}{a \cos x + b \sin x} a, b > 0$

Answer:

$\int_{0}^{\frac{π}{2}} \frac{1}{a \cos x + b \sin x} d x = \int_{0}^{\frac{π}{2}} \frac{1}{a (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) + b (\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x = \int_{0}^{\frac{π}{2}} \frac{(1 + \tan^{2} \frac{x}{2})}{a - a \tan^{2} \frac{x}{2} + 2 b \tan \frac{x}{2}} d x = \int_{0}^{\frac{π}{2}} \frac{s e c^{2} \frac{x}{2}}{a - a t a n^{2} \frac{x}{2} + 2 b t a n \frac{x}{2}} d x L e t \tan \frac{x}{2} = t, T h e n, \frac{1}{2} s e c^{2} \frac{x}{2} d x = d t W h e n x = 0, t = 0, x = \frac{π}{2}, t = 1 Therefore the integral becomes I = \int_{0}^{1} \frac{2 dt}{a - {at}^{2} + 2 bt} = \int_{0}^{1} \frac{2 d t}{- a [t^{2} - \frac{2 b t}{a} - 1]} = \frac{2}{a} \int_{0}^{1} \frac{dt}{- [{(t - \frac{b}{a})}^{2} - 1 - \frac{b^{2}}{a^{2}}]} = \frac{2}{a} \int_{0}^{1} \frac{d t}{(\frac{b^{2}}{a^{2}} + 1) - {(t - \frac{b}{a})}^{2}} = \frac{2}{a} [\frac{1}{2 \sqrt{\frac{a^{2} + b^{2}}{a^{2}}}} {(\log |\frac{\sqrt{\frac{a^{2} + b^{2}}{a^{2}}} + (t - \frac{b}{a})}{\sqrt{\frac{a^{2} + b^{2}}{a^{2}}} - (t - \frac{b}{a})}|)}_{0}^{1}]$
$\frac{1}{\sqrt{a^{2} + b^{2}}} \log (\frac{a + b + \sqrt{a^{2} + b^{2}}}{a + b - \sqrt{a^{2} + b^{2}}})$

Page No 19.39:

Question 20:

$\int_{0}^{π / 2} \frac{1}{5 + 4 \sin x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{1}{5 + 4 \sin x} d x . Then, I = \int_{0}^{\frac{π}{2}} \frac{1}{5 + 4 (\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{5 (1 + \tan^{2} \frac{x}{2}) + 8 \tan \frac{x}{2}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} \frac{x}{2}}{5 \tan^{2} \frac{x}{2} + 8 \tan \frac{x}{2} + 5} d x Let \tan \frac{x}{2} = t . Then, \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t When x = 0, t = 0 and x = \frac{π}{2}, t = 1 ∴ I = 2 \int_{0}^{1} \frac{1}{5 t^{2} + 8 t + 5} d t \Rightarrow I = 2 \int_{0}^{1} \frac{1}{{(\sqrt{5} t)}^{2} + 8 t + 5 + {(\frac{4}{\sqrt{5}})}^{2} - {(\frac{4}{\sqrt{5}})}^{2}} d t \Rightarrow I = 2 \int_{0}^{1} \frac{1}{{(\sqrt{5} t + \frac{4}{\sqrt{5}})}^{2} + \frac{9}{5}} d t \Rightarrow I = \frac{2}{3} {[\tan^{- 1} (\frac{\sqrt{5} t + \frac{4}{\sqrt{5}}}{\frac{3}{\sqrt{5}}})]}_{0}^{1} \Rightarrow I = \frac{2}{3} [\tan^{- 1} 3 - \tan^{- 1} \frac{4}{3}] \Rightarrow I = \frac{2}{3} [\tan^{- 1} (\frac{3 - \frac{4}{3}}{1 + 3 \times \frac{4}{3}})] \Rightarrow I = \frac{2}{3} \tan^{- 1} \frac{1}{3}$

Page No 19.39:

Question 21:

$\int_{0}^{π} \frac{\sin x}{\sin x + \cos x} d x$

Answer:

$\int_{0}^{π} \frac{\sin x}{\sin x + \cos x} d x = \frac{1}{2} \int_{0}^{π} \frac{2 \sin x}{\sin x + \cos x} d x = \frac{1}{2} \int_{0}^{π} \frac{(\sin x + \cos x) - (\cos x - \sin x)}{\sin x + \cos x} d x = \frac{1}{2} \int_{0}^{π} d x - \frac{1}{2} \int_{0}^{π} \frac{\cos x - \sin x}{\sin x + \cos x} d x = \frac{1}{2} {[x]}_{0}^{π} - \frac{1}{2} {[\log |\sin x + \cos x|]}_{0}^{π} = \frac{1}{2} [π - 0] - \frac{1}{2} [\log 1 - \log 1] = \frac{π}{2}$

Page No 19.39:

Question 22:

$\int_{0}^{π} \frac{1}{3 + 2 \sin x + \cos x} d x$

Answer:

$Let I = \int_{0}^{π} \frac{1}{3 + 2 \sin x + \cos x} d x . Then, I = \int_{0}^{π} \frac{1}{3 + 2 (\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}) + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x \Rightarrow I = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{2 \tan^{2} \frac{x}{2} + 4 \tan \frac{x}{2} + 4} d x Let \tan \frac{x}{2} = t . Then, \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t When x = 0, t = 0 and x = π, t = \infty ∴ I = \int_{0}^{\infty} \frac{2 d t}{2 t^{2} + 4 t + 4} \Rightarrow I = \int_{0}^{\infty} \frac{d t}{{(t + 1)}^{2} + 1} \Rightarrow I = {[\tan^{- 1} (t + 1)]}_{0}^{\infty} \Rightarrow I = \frac{π}{2} - \frac{π}{4} \Rightarrow I = \frac{π}{4}$

Page No 19.39:

Question 23:

$\int_{0}^{1} \tan^{- 1} x d x$

Answer:

$Let I = \int_{0}^{1} \tan^{- 1} x d x . Then, I = \int_{0}^{1} 1 \tan^{- 1} x d x Integrating by parts I = {[x \tan^{- 1} x]}_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x^{2}} d x \Rightarrow I = {[x \tan^{- 1} x]}_{0}^{1} - \frac{1}{2} {[\log (x^{2} + 1)]}_{0}^{1} \Rightarrow I = \frac{π}{4} - 0 - \frac{1}{2} \log 2 + 0 \Rightarrow I = \frac{π}{4} - \frac{1}{2} \log 2$

Page No 19.39:

Question 24:

$\int_{0}^{\frac{1}{2}} \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x$

Answer:

Let I = $\int_{0}^{\frac{1}{2}} \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} d x$

Put $x = \sin θ$

$∴ d x = \cos θ d θ$

When $x \to 0, θ \to 0$

When $x \to \frac{1}{2}, θ \to \frac{π}{6}$

$∴ I = \int_{0}^{\frac{π}{6}} \frac{\sin θ \sin^{- 1} (\sin θ)}{\cos θ} \cos θ d θ = \int_{0}^{\frac{π}{6}} θ \sin θ d θ$

Applying integration by parts, we have

$I = {θ (- \cos θ)}_{0}^{\frac{π}{6}} - \int_{0}^{\frac{π}{6}} 1 \times (- \cos θ) d θ = - (\frac{π}{6} \cos \frac{π}{6} - 0) + \int_{0}^{\frac{π}{6}} \cos θ d θ = - \frac{π}{6} \times \frac{\sqrt{3}}{2} + {\sin θ}_{0}^{\frac{π}{6}} = - \frac{π}{4 \sqrt{3}} + (\sin \frac{π}{6} - \sin 0) = - \frac{π}{4 \sqrt{3}} + (\frac{1}{2} - 0) = \frac{1}{2} - \frac{π}{4 \sqrt{3}}$

Page No 19.39:

Question 25:

$\int_{0}^{π / 4} (\sqrt{\tan} x + \sqrt{\cot} x) d x$

Answer:

$Let I = \int_{0}^{\frac{π}{4}} (\sqrt{\tan x} + \sqrt{\cot x}) d x . Then, I = \int_{0}^{\frac{π}{4}} (\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}}) d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} d x \Rightarrow I = \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} d x \Rightarrow I = \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{\sqrt{1 - {(\sin x - \cos x)}^{2}}} d x Let \sin x - \cos x = t . Then, \cos x + \sin x d x = d t When x = 0, t = 1 and x = \frac{π}{4}, t = 0 ∴ I = \sqrt{2} \int_{- 1}^{0} \frac{d t}{\sqrt{1 - t^{2}}} \Rightarrow I = \sqrt{2} {[\sin^{- 1} t]}_{- 1}^{0} \Rightarrow I = \frac{π}{\sqrt{2}} \Rightarrow I =$

Page No 19.39:

Question 26:

$\int_{0}^{π / 4} \frac{\tan^{3} x}{1 + \cos 2 x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{4}} \frac{\tan^{3} x}{1 + \cos 2 x} d x . Then, I = \int_{0}^{\frac{π}{4}} \frac{\tan^{3} x}{2 \cos^{2} x} d x \Rightarrow I = \frac{1}{2} \int_{0}^{\frac{π}{4}} \tan^{3} x \sec^{2} x d x Let \tan x = t . Then, \sec^{2} x d x = d t When x = 0, t = 0 and x = \frac{π}{4}, t = 1 ∴ I = \frac{1}{2} \int_{0}^{1} t^{3} d t \Rightarrow I = \frac{1}{2} {[\frac{t^{4}}{4}]}_{0}^{1} \Rightarrow I = \frac{1}{2} (\frac{1}{4} - 0) \Rightarrow I = \frac{1}{8}$

Page No 19.39:

Question 27:

$\int_{0}^{π} \frac{1}{5 + 3 \cos x} d x$

Answer:

$Let I = \int_{0}^{π} \frac{1}{5 + 3 \cos x} d x . Then, I = \int_{0}^{π} \frac{1}{5 + 3 (\frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})} d x \Rightarrow I = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{5 + 5 \tan^{2} \frac{x}{2} - 3 \tan^{2} \frac{x}{2}} d x \Rightarrow I = \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{5 + 2 \tan^{2} \frac{x}{2}} d x Let \tan \frac{x}{2} = t . Then, \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t When x = 0, t = 0 and x = π, t = \infty ∴ I = \int_{0}^{\infty} \frac{d t}{5 + 2 t^{2}} \Rightarrow I = \frac{1}{2} \int_{0}^{\infty} \frac{d t}{\frac{5}{2} + t^{2}} \Rightarrow I = \frac{1}{2} {[\tan^{- 1} \frac{\sqrt{5} t}{\sqrt{2}}]}_{0}^{\infty} \Rightarrow I = \frac{1}{2} (\frac{π}{2} - 0) \Rightarrow I = \frac{π}{4}$

Page No 19.39:

Question 28:

$\int_{0}^{π / 2} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x . Then, Dividing the numerator and denominator by \cos^{2} x, we get I = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{a^{2} \tan^{2} x + b^{2}} d x Let \tan x = t . Then, \sec^{2} x d x = d t When x = 0, t = 0 and x = \frac{π}{2}, t = \infty ∴ I = \int_{0}^{\infty} \frac{1}{a^{2} t^{2} + b^{2}} d t \Rightarrow I = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{t^{2} + \frac{b^{2}}{a^{2}}} d t \Rightarrow I = \frac{1}{a^{2}} \times \frac{a}{b} {[\tan^{- 1} \frac{a t}{b}]}_{0}^{\infty} \Rightarrow I = \frac{1}{a b} \frac{π}{2} \Rightarrow I = \frac{π}{2 a b}$

Page No 19.39:

Question 29:

$\int_{0}^{π / 2} \frac{x + \sin x}{1 + \cos x} d x$

Answer:

$Let, I = \int_{0}^{π / 2} \frac{x + \sin x}{1 + \cos x} d x = \int_{0}^{π / 2} \frac{x + \sin x}{2 \cos^{2} \frac{x}{2}} d x = \int_{0}^{π / 2} [\frac{x}{2 \cos^{2} \frac{x}{2}} + \frac{\sin x}{2 \cos^{2} \frac{x}{2}}] d x = \frac{1}{2} \int_{0}^{π / 2} x s e c^{2} \frac{x}{2} d x + \int_{0}^{π / 2} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} d x = \frac{1}{2} {[x \frac{\tan \frac{x}{2}}{\frac{1}{2}}]}_{0}^{π / 2} - \frac{1}{2} \int_{0}^{π / 2} \frac{\tan \frac{x}{2}}{\frac{1}{2}} d x + \int_{0}^{π / 2} \tan \frac{x}{2} d x = {[x \tan \frac{x}{2}]}_{0}^{π / 2} - \int_{0}^{π / 2} \tan \frac{x}{2} d x + \int_{0}^{π / 2} \tan \frac{x}{2} d x = [\frac{π}{2} \tan \frac{π}{4}] = \frac{π}{2} \times 1 = \frac{π}{2}$

Page No 19.39:

Question 30:

$\int_{0}^{1} \frac{\tan^{- 1} x}{1 + x^{2}} d x$

Answer:

$Let I = \int_{0}^{1} \frac{\tan^{- 1} x}{1 + x^{2}} d x . Then, Let \tan^{- 1} x = t . Then, \frac{1}{1 + x^{2}} d x = d t When x = 0, t = 0 and x = 1, t = \frac{π}{4} ∴ I = \int_{0}^{\frac{π}{4}} t d t \Rightarrow I = {[\frac{t^{2}}{2}]}_{0}^{\frac{π}{4}} \Rightarrow I = \frac{π^{2}}{32}$

Page No 19.39:

Question 31:

$\int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{3 + \sin 2 x} d x$

Answer:

Let $I = \int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{3 + \sin 2 x} d x$
$= \int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{4 - (1 - \sin 2 x)} d x = \int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{4 - (\sin^{2} x + \cos^{2} x - 2 \sin x \cos x)} d x = \int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{4 - {(\sin x - \cos x)}^{2}} d x$

Put $\sin x - \cos x = z$

$∴ (\cos x + \sin x) d x = d z$

When $x \to 0, z \to - 1 (z = \sin 0 - \cos 0 = 0 - 1 = - 1)$

When $x \to \frac{π}{4}, z \to 0 (z = \sin \frac{π}{4} - \cos \frac{π}{4} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0)$

$∴ I = \int_{- 1}^{0} \frac{d z}{2^{2} - z^{2}} = {\frac{1}{2 \times 2} \log (\frac{2 + z}{2 - z})}_{- 1}^{0} = \frac{1}{4} (\log 1 - \log \frac{1}{3}) = \frac{1}{4} [0 - (\log 1 - \log 3)] = - \frac{1}{4} (0 - \log 3) = \frac{1}{4} \log 3$

Page No 19.39:

Question 32:

$\int_{0}^{1} x \tan^{- 1} x d x$

Answer:

$Let I = \int_{0}^{1} x \tan^{- 1} x d x . Then, Integrating by parts I = {[\frac{x^{2} \tan^{- 1} x}{2}]}_{0}^{1} - \frac{1}{2} \int_{0}^{1} \frac{x^{2}}{1 + x^{2}} d x \Rightarrow I = {[\frac{x^{2} \tan^{- 1} x}{2}]}_{0}^{1} - \frac{1}{2} \int_{0}^{1} (\frac{1 + x^{2}}{1 + x^{2}} - \frac{1}{1 + x^{2}}) d x \Rightarrow I = {[\frac{x^{2} \tan^{- 1} x}{2}]}_{0}^{1} - \frac{1}{2} {[x - \tan^{- 1} x]}_{0}^{1} \Rightarrow I = \frac{π}{8} - 0 - \frac{1}{2} (1 - \frac{π}{4} - 0) \Rightarrow I = \frac{π}{4} - \frac{1}{2}$

Page No 19.39:

Question 33:

$\int_{0}^{1} \frac{1 - x^{2}}{x^{4} + x^{2} + 1} d x$

Answer:

$Let, I = \int \frac{1 - x^{2}}{x^{4} + x^{2} + 1} d x = - \int \frac{x^{2} - 1}{x^{4} + x^{2} + 1} d x = - \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} d x = - \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + 2 + \frac{1}{x^{2}} - 1} d x = - \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 1} d x Let, x + \frac{1}{x} = t \Rightarrow (1 - \frac{1}{x^{2}}) d x = d t Then integral becomes, I = - \int \frac{1}{t^{2} - 1} d t = - \frac{1}{2} \log |\frac{t - 1}{t + 1}| = \frac{1}{2} \log |\frac{t + 1}{t - 1}| = \frac{1}{2} \log |\frac{x + \frac{1}{x} + 1}{x + \frac{1}{x} - 1}| = \frac{1}{2} \log |\frac{x^{2} + x + 1}{x^{2} - x + 1}| i . e ., \int \frac{1 - x^{2}}{x^{4} + x^{2} + 1} d x = \frac{1}{2} \log |\frac{x^{2} + x + 1}{x^{2} - x + 1}| \Rightarrow \int_{0}^{1} \frac{1 - x^{2}}{x^{4} + x^{2} + 1} d x = {[\frac{1}{2} \log |\frac{x^{2} + x + 1}{x^{2} - x + 1}|]}_{0}^{1} = \frac{1}{2} \log 3$

Page No 19.39:

Question 34:

$\int_{0}^{1} \frac{24 x^{3}}{{(1 + x^{2})}^{4}} d x$

Answer:

$Let I = \int_{0}^{1} \frac{24 x^{3}}{{(1 + x^{2})}^{4}} d x . Then, Let x^{2} = t . Then, 2 x d x = d t When x =, t = 0 and x = 1, t = 1 ∴ I = \int_{0}^{1} \frac{12 t}{{(1 + t)}^{4}} d t Integrating by parts I = 12 {[\frac{t}{- 3 {(1 + t)}^{3}}]}_{0}^{1} + 12 \int_{0}^{1} \frac{1}{3 {(1 + t)}^{3}} d t \Rightarrow I = 12 \{{[\frac{t}{- 3 {(1 + t)}^{3}}]}_{0}^{1} - {[\frac{1}{6 {(1 + t)}^{2}}]}_{0}^{1}\} \Rightarrow I = 12 \{- \frac{1}{24} - 0 - \frac{1}{24} + \frac{1}{6}\} \Rightarrow I = 12 \times \frac{1}{12} \Rightarrow I = 1$

Page No 19.39:

Question 35:

$\int_{4}^{12} x {(x - 4)}^{1 / 3} d x$

Answer:

$Let I = \int_{4}^{12} x {(x - 4)}^{\frac{1}{3}} d x . Let x - 4 = t . Then, d x = d t When x = 4, t = 0 and x = 12, t = 8 ∴ I = \int_{0}^{8} (t + 4) t^{\frac{1}{3}} d t \Rightarrow I = \int_{0}^{8} (t^{\frac{4}{3}} + 4 t^{\frac{1}{3}}) d t \Rightarrow I = {[\frac{3}{7} t^{\frac{7}{3}} + \frac{3}{1} t^{\frac{4}{3}}]}_{0}^{8} \Rightarrow I = \frac{384}{7} + 48 \Rightarrow I = \frac{720}{7}$

Page No 19.39:

Question 36:

$\int_{0}^{π / 2} x^{2} \sin x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x^{2} \sin x d x . Then, Integrating by parts I = {[- x^{2} \cos x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} - 2 x \cos x d x Again, integratting by parts \Rightarrow I = {[- x^{2} \cos x]}_{0}^{\frac{π}{2}} + \{2 {[x \sin x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 1 \sin x d x\} \Rightarrow I = {[- x^{2} \cos x]}_{0}^{\frac{π}{2}} + 2 {[x \sin x]}_{0}^{\frac{π}{2}} - {[- \cos x]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{π^{2}}{4} 0 - 0 + 2 \frac{π}{2} - 0 + 0 - 2 \Rightarrow I = π - 2$

Page No 19.39:

Question 37:

$\int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x$

Answer:

$Let I = \int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x . Then, I = \int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} \times \frac{\sqrt{1 - x}}{\sqrt{1 - x}} d x \Rightarrow I = \int_{0}^{1} \frac{1 - x}{\sqrt{1 - x^{2}}} d x \Rightarrow I = \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} d x - \int_{0}^{1} \frac{x}{\sqrt{1 - x^{2}}} d x \Rightarrow I = {[\sin^{- 1} x]}_{0}^{1} + \frac{1}{2} \int_{0}^{1} \frac{- 2 x}{\sqrt{1 - x^{2}}} d x \Rightarrow I = {[\sin^{- 1} x]}_{0}^{1} + \frac{1}{2} {[2 \sqrt{1 - x^{2}}]}_{0}^{1} \Rightarrow I = \frac{π}{2} - 0 + 0 - 1 \Rightarrow I = \frac{π}{2} - 1$

Page No 19.39:

Question 38:

$\int_{0}^{1} \frac{1 - x^{2}}{{(1 + x^{2})}^{2}} d x$

Answer:

$Let I = \int_{0}^{1} \frac{1 - x^{2}}{{(1 + x^{2})}^{2}} d x . Then, I = \int_{0}^{1} \frac{(\frac{1}{x^{2}} - 1)}{{(x + \frac{1}{x})}^{2}} d x Let x + \frac{1}{x} = t . Then, 1 - \frac{1}{x^{2}} d x = d t When x = 0, t = \infty and x = 1, t = 2 ∴ I = \int_{\infty}^{2} \frac{- d t}{t^{2}} \Rightarrow I = {[\frac{1}{t}]}_{\infty}^{2} \Rightarrow I = \frac{1}{2} - 0 \Rightarrow I = \frac{1}{2}$

Page No 19.39:

Question 39:

$\int_{- 1}^{1} 5 x^{4} \sqrt{x^{5} + 1} d x$

Answer:

$Let I = \int_{- 1}^{1} 5 x^{4} \sqrt{x^{5} + 1} d x . Then, Let x^{5} + 1 = t . Then, 5 x^{4} d x = d t When x = - 1, t = 0 and x = 1, t = 2 ∴ I = \int_{0}^{2} \sqrt{t} d t \Rightarrow I = {[\frac{2}{3} t^{\frac{3}{2}}]}_{0}^{2} \Rightarrow I = \frac{2}{3} \sqrt{8} \Rightarrow I = \frac{4 \sqrt{2}}{3}$

Page No 19.39:

Question 40:

$\int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{1 + 3 \sin^{2} x} d x$ [CBSE 2015]

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{1 + 3 \sin^{2} x} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{1 + 3 (1 - \cos^{2} x)} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{4 - 3 \cos^{2} x} d x = - \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{4 - 3 \cos^{2} x - 4}{4 - 3 \cos^{2} x} d x$
$= - \frac{1}{3} \int_{0}^{\frac{π}{2}} d x + \frac{4}{3} \int_{0}^{\frac{π}{2}} \frac{1}{4 - 3 \cos^{2} x} d x = - \frac{1}{3} x_{0}^{\frac{π}{2}} + \frac{4}{3} \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{4 \sec^{2} x - 3} d x (Dividing numerator and denominator by \cos^{2} x) = - \frac{1}{3} (\frac{π}{2} - 0) + \frac{4}{3} \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{4 (1 + \tan^{2} x) - 3} d x = - \frac{π}{6} + \frac{4}{3} \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{4 \tan^{2} x + 1} d x$

Put tanx = z

$∴ \sec^{2} x d x = d z$

When $x \to 0, z \to 0$

When $x \to \frac{π}{2}, z \to \infty$

$∴ I = - \frac{π}{6} + \frac{4}{3} \int_{0}^{\infty} \frac{d z}{4 z^{2} + 1} = - \frac{π}{6} + \frac{4}{3} \int_{0}^{\infty} \frac{d z}{{(2 z)}^{2} + 1} = - \frac{π}{6} + \frac{4}{3} \times {\frac{\tan^{- 1} 2 z}{2}}_{0}^{\infty} = - \frac{π}{6} + \frac{2}{3} (\tan^{- 1} \infty - \tan^{- 1} 0) = - \frac{π}{6} + \frac{2}{3} (\frac{π}{2} - 0) = - \frac{π}{6} + \frac{π}{3} = \frac{π}{6}$

Page No 19.39:

Question 41:

$\int_{0}^{π / 4} \sin^{3} 2 t \cos 2 t d t$

Answer:

$Let I = \int_{0}^{\frac{π}{4}} \sin^{3} 2 t \cos 2 t d t . Then, Let \sin 2 t = u . Then, 2 \cos 2 t d t = d u When t = 0, u = 0 and t = \frac{π}{4}, u = 1 ∴ I = \frac{1}{2} \int_{0}^{1} u^{3} d u \Rightarrow I = \frac{1}{2} {[\frac{u^{4}}{4}]}_{0}^{1} \Rightarrow I = \frac{1}{2} (\frac{1}{4} - 0) \Rightarrow I = \frac{1}{8}$

Page No 19.39:

Question 42:

$\int_{0}^{π} 5 {(5 - 4 \cos θ)}^{1 / 4} \sin θ d θ$

Answer:

$Let I = \int_{0}^{π} 5 {(5 - 4 \cos θ)}^{\frac{1}{4}} \sin θ d θ . Let (5 - 4 \cos θ) = t . Then, 4 \sin θ d θ = d t When θ = 0, t = 1 and θ = π, t = 9 ∴ I = \frac{5}{4} \int_{1}^{9} t^{\frac{1}{4}} d t \Rightarrow I = \frac{5}{4} {[\frac{4 t^{\frac{5}{4}}}{5}]}_{1}^{9} \Rightarrow I = (9 \sqrt{3} - 1)$

Page No 19.39:

Question 43:

$\int_{0}^{π / 6} \cos^{- 3} 2 θ \sin 2 θ d θ$

Answer:

$Let I = \int_{0}^{\frac{π}{6}} \cos^{- 3} 2 θ \sin 2 θ d θ . Then, I = \int_{0}^{\frac{π}{6}} \frac{\sin 2 θ}{\cos^{3} 2 θ} d θ Let \cos 2 θ = t . Then, - 2 \sin 2 θ d θ = d t When θ = 0, t = 1 and θ = \frac{π}{6}, t = \frac{1}{2} ∴ I = \frac{- 1}{2} \int_{1}^{\frac{1}{2}} \frac{d t}{t^{3}} \Rightarrow I = \frac{1}{2} {[\frac{1}{2 t^{2}}]}_{1}^{\frac{1}{2}} \Rightarrow I = \frac{1}{2} (2 - \frac{1}{2}) \Rightarrow I = \frac{3}{4}$

Page No 19.39:

Question 44:

$\int_{0}^{{(π)}^{2 / 3}} \sqrt{x} \cos^{2} x^{3 / 2} d x$

Answer:

$Let I = \int_{0}^{{(π)}^{\frac{2}{3}}} \sqrt{x} \cos^{2} x^{\frac{3}{2}} d x . Then, Let x^{\frac{3}{2}} = t . Then, \frac{3}{2} \sqrt{x} d x = d t When x = 0, t = 0 and x = {(π)}^{\frac{2}{3}}, t = π ∴ I = \frac{2}{3} \int_{0}^{π} \cos^{2} t d t \Rightarrow I = \frac{2}{3} \int_{0}^{π} \frac{1 + \cos 2 x}{2} d x \Rightarrow I = \frac{1}{3} {[x + \frac{\sin 2 x}{2}]}_{0}^{π} \Rightarrow I = \frac{1}{3} (π + 0) \Rightarrow I = \frac{π}{3}$

Page No 19.40:

Question 45:

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

Answer:

$Let I = \int_{1}^{2} \frac{1}{x {(1 + \log x)}^{2}} d x . Then, Let (1 + \log x) = t . Then, \frac{1}{x} d x = d t When x = 1, t = 1 and x = 2, t = 1 + \log 2 ∴ I = \int_{1}^{(1 + \log 2)} \frac{1}{t^{2}} d t \Rightarrow I = {[\frac{- 1}{t}]}_{1}^{(1 + \log 2)} \Rightarrow I = - \frac{1}{1 + \log 2} + 1 \Rightarrow I = \frac{\log 2}{1 + \log 2}$

Page No 19.40:

Question 46:

$\int_{0}^{π / 2} \cos^{5} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \cos^{5} x d x . Then, I = \int_{0}^{\frac{π}{2}} \cos^{4} x \cos x d x \Rightarrow I = \int_{0}^{\frac{π}{2}} {(1 - \sin^{2} x)}^{2} \cos x d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (1 - 2 \sin^{2} x + \sin^{4} x) \cos x d x Let \sin x = t . Then, \cos d x = d u When x = 0, t = 0 and x = \frac{π}{2}, t = 1 ∴ I = \int_{0}^{1} (1 - 2 t^{2} + t^{4}) d t \Rightarrow I = {[t - \frac{2 t^{3}}{3} + \frac{t^{5}}{5}]}_{0}^{1} \Rightarrow I = 1 - \frac{2}{3} + \frac{1}{5} \Rightarrow I = \frac{8}{15}$

Page No 19.40:

Question 47:

$\int_{4}^{9} \frac{\sqrt{x}}{{(30 - x^{3 / 2})}^{2}} d x$

Answer:

$Let I = \int_{4}^{9} \frac{\sqrt{x}}{{(30 - x^{\frac{3}{2}})}^{2}} d x . Then, Let (30 - x^{\frac{3}{2}}) = t . Then, - \frac{3}{2} \sqrt{x} d x = d t When, x = 4, t = 22 and x = 9, t = 3 ∴ I = \int_{22}^{3} - \frac{2}{3} \frac{1}{t^{2}} d t \Rightarrow I = \frac{2}{3} {[\frac{1}{t}]}_{22}^{3} \Rightarrow I = \frac{2}{3} (\frac{1}{3} - \frac{1}{22}) \Rightarrow I = \frac{19}{99}$

Page No 19.40:

Question 48:

$\int_{0}^{π} \sin^{3} x (1 + 2 \cos x) {(1 + \cos x)}^{2} d x$

Answer:

$Let I = \int_{0}^{π} \sin^{3} x (1 + 2 \cos x) {(1 + \cos x)}^{2} d x . Then, I = \int_{0}^{π} \sin x \sin^{2} x (1 + 2 \cos x) {(1 + \cos x)}^{2} d x \Rightarrow I = \int_{0}^{π} \sin x (1 - \cos^{2} x) (1 + 2 \cos x) {(1 + \cos x)}^{2} d x \Rightarrow I = \int_{0}^{π} \sin x (1 - \cos x) (1 + 2 \cos x) {(1 + \cos x)}^{3} d x Let \cos x = t . Then, - \sin x d x = d t When x = 0, t = 1 and x = π, t = - 1 ∴ I = - \int_{1}^{- 1} (1 - t) (1 + 2 t) {(1 + t)}^{3} d t \Rightarrow I = \int_{- 1}^{1} (1 + t - 2 t^{2}) (1 + t^{3} + 3 t + 3 t^{2}) d t \Rightarrow I = \int_{- 1}^{1} (1 + t^{3} + 3 t + 3 t^{2} + t + t^{4} + 3 t^{2} + 3 t^{3} - 2 t^{2} - 2 t^{5} - 6 t^{3} - 6 t^{4}) d t \Rightarrow I = \int_{- 1}^{1} (1 + 4 t + 4 t^{2} - 2 t^{3} - 5 t^{4} - 2 t^{5}) d t \Rightarrow I = {[t + 2 t^{2} + \frac{4 t^{3}}{3} - \frac{t^{4}}{2} - t^{5} - \frac{t^{6}}{3}]}_{- 1}^{1} \Rightarrow I = 1 + 2 + \frac{4}{3} - \frac{1}{2} - 1 - \frac{1}{3} + 1 - 2 + \frac{4}{3} + \frac{1}{2} - 1 + \frac{1}{3} \Rightarrow I = \frac{8}{3}$

Page No 19.40:

Question 49:

$\int_{0}^{π / 2} 2 \sin x \cos x \tan^{- 1} (\sin x) d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} 2 \sin x \cos x \tan^{- 1} (\sin x) d x . Then, Let \sin x = t . Then, \cos x d x = d t When x = 0, t = 0 and x = \frac{π}{2}, t = 1 ∴ I = \int_{0}^{1} 2 t \tan^{- 1} t d t \Rightarrow I = 2 {[\frac{t^{2} \tan^{- 1} t}{2}]}_{0}^{1} - 2 \int_{0}^{1} \frac{t^{2}}{1 + t^{2}} d t \Rightarrow I = 2 {[\frac{t^{2} \tan^{- 1} t}{2}]}_{0}^{1} - 2 \int_{0}^{1} (\frac{1 + t^{2}}{1 + t^{2}} - \frac{1}{1 + t^{2}}) d t \Rightarrow I = 2 {[\frac{t^{2} \tan^{- 1} t}{2}]}_{0}^{1} - {[t - \tan^{- 1} t +]}_{0}^{1} \Rightarrow I = 1 \tan^{- 1} 1 - 0 - 1 + \tan^{- 1} 1 + 0 \Rightarrow I = \frac{π}{4} - 1 + \frac{π}{4} \Rightarrow I = \frac{π}{2} - 1$

Page No 19.40:

Question 50:

$\int_{0}^{π / 2} \sin 2 x \tan^{- 1} (\sin x) d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \sin 2 x \tan^{- 1} (\sin x) d x . Then, I = \int_{0}^{\frac{π}{2}} 2 \sin x \cos x \tan^{- 1} (\sin x) d x Let \sin x = t . Then, \cos x d x = d t When x = 0, t = 0 and x = \frac{π}{2}, t = 1 ∴ I = 2 \int_{0}^{1} t \tan^{- 1} t d t \Rightarrow I = 2 {[\frac{t^{2}}{2} \tan^{- 1} t]}_{0}^{1} - 2 \int_{0}^{1} \frac{t}{1 + t^{2}} d t \Rightarrow I = 2 {[\frac{t^{2}}{2} \tan^{- 1} t]}_{0}^{1} - {[\log (1 + t^{2})]}_{0}^{1} \Rightarrow I = \frac{2 π}{4} - 1 \Rightarrow I = \frac{π}{2} - 1$

Page No 19.40:

Question 51:

$\int_{0}^{1} {(\cos^{- 1} x)}^{2} d x$

Answer:

$Let I = \int_{0}^{1} {(\cos^{- 1} x)}^{2} d x . Then, I = \int_{0}^{1} 1 {(\cos^{- 1} x)}^{2} d x Integrating by parts \Rightarrow I = {[x {(\cos^{- 1} x)}^{2}]}_{0}^{1} - \int_{0}^{1} 2 x \cos^{- 1} x \frac{- 1}{\sqrt{1 - x^{2}}} d x Again, integrating second term by parts \Rightarrow I = {[x {(\cos^{- 1} x)}^{2}]}_{0}^{1} + \{2 {[\sqrt{1 - x^{2}} \cos^{- 1} x]}_{0}^{1} - 2 \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} \sqrt{1 - x^{2}} d x\} \Rightarrow I = {[x {(\cos^{- 1} x)}^{2}]}_{0}^{1} + 2 {[\sqrt{1 - x^{2}} \cos^{- 1} x]}_{0}^{1} - 2 {[x]}_{0}^{1} \Rightarrow I = 0 + \frac{2 π}{2} - 2 \Rightarrow I = π - 2$

Page No 19.40:

Question 52:

$\int_{0}^{a} \sin^{- 1} \sqrt{\frac{x}{a + x}} d x$

Answer:

$Let, I = \int_{0}^{a} \sin^{- 1} \sqrt{\frac{x}{a + x}} d x Let, x = a \tan^{2} θ \Rightarrow θ = \tan^{- 1} \sqrt{\frac{x}{a}} When, x \to x; θ \to 0 a n d x \to a; θ \to \frac{π}{4} and d x = 2 a \tan θ s e c^{2} θ d θ Then, I = \int_{0}^{\frac{π}{4}} \sin^{- 1} \sqrt{\frac{a \tan^{2} θ}{a + a \tan^{2} θ}} 2 a \tan θ s e c^{2} θ d θ \Rightarrow I = {2 a \int}_{0}^{\frac{π}{4}} \sin^{- 1} (\sin θ) \tan θ s e c^{2} θ d θ \Rightarrow I = {2 a \int}_{0}^{\frac{π}{4}} θ \tan θ s e c^{2} θ d θ Let, \tan θ = t \Rightarrow θ = \tan^{- 1} t \Rightarrow s e c^{2} θ d θ = d t when, θ \to 0; t \to 0 a n d θ \to \frac{π}{4}; t \to 1 Then, I = 2 a \int_{0}^{1} \tan^{- 1} t t d t = 2 a \int_{0}^{1} \tan^{- 1} t t d t = 2 a {[\tan^{- 1} t \frac{t^{2}}{2}]}_{0}^{1} - \frac{2 a}{2} \int_{0}^{1} \frac{t^{2}}{1 + t^{2}} d t = 2 a [\frac{π}{4} \times \frac{1}{2} - 0] - a \int_{0}^{1} [1 - \frac{1}{1 + t^{2}}] d t = 2 a [\frac{π}{8}] - a {[t - \tan^{- 1} t]}_{0}^{1} = \frac{πa}{4} - a [1 - \frac{π}{4}] = \frac{πa}{4} - a + \frac{π a}{4} = \frac{πa}{2} - a = a (\frac{π}{2} - 1)$

Page No 19.40:

Question 53:

$\int_{π / 3}^{π / 2} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{3 / 2}} d x$

Answer:

$\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{\frac{3}{2}}} d x = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{\frac{3}{2}}} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} d x = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 - \cos^{2} x}}{{(1 - \cos x)}^{2}} d x = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sin x}{{(1 - \cos x)}^{2}} d x Let 1 - \cos x = t, Then \sin x d x = d t When x = \frac{π}{3}, t = \frac{1}{2} and x = \frac{π}{2}, t = 1 Therefore the integral becomes = \int_{\frac{1}{2}}^{1} \frac{dt}{t^{2}} = {[- \frac{1}{t}]}_{\frac{1}{2}}^{1} = - 1 + 2 = 1$

Page No 19.40:

Question 54:

$\int_{0}^{a} x \sqrt{\frac{a^{2} - x^{2}}{a^{2} + x^{2}}} d x$

Answer:

$Let, I = \int_{0}^{a} x \sqrt{\frac{a^{2} - x^{2}}{a^{2} + x^{2}}} d x Consider, x^{2} = a^{2} \cos 2 θ \Rightarrow 2 x d x = - 2 a^{2} \sin 2 θ d θ \Rightarrow x d x = - a^{2} \sin 2 θ d θ When, x \to 0; θ \to \frac{π}{4} and x \to a; θ \to 0 Now, integral becomes, I = \int_{\frac{π}{4}}^{0} - a^{2} \sin 2 θ \sqrt{\frac{a^{2} - a^{2} \cos 2 θ}{a^{2} + a^{2} \cos 2 θ}} d θ = \int_{\frac{π}{4}}^{0} - a^{2} \sin 2 θ \tan θ d θ = a^{2} \int_{0}^{\frac{π}{4}} 2 \sin θ \cos θ \frac{\sin θ}{\cos θ} d θ = a^{2} \int_{0}^{\frac{π}{4}} 2 \sin^{2} θ d θ = a^{2} \int_{0}^{\frac{π}{4}} [1 - \cos 2 θ] d θ = a^{2} {[θ - \frac{\sin 2 θ}{2}]}_{0}^{\frac{π}{4}} = a^{2} [\frac{π}{4} - \frac{1}{2}]$

Page No 19.40:

Question 55:

$\int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} d x$

Answer:

$Let, I = \int_{- a}^{a} \sqrt{\frac{a - x}{a + x}} d x Consider, x = a \cos 2 y Then y = \frac{1}{2} \cos^{- 1} (\frac{x}{a}) \Rightarrow dx = - 2 a \sin 2 y dy When, x \to - a; y \to \frac{π}{2} and x \to a; y \to 0 Now, integral becomes, I = \int_{\frac{π}{2}}^{0} - 2 a \sin 2 y \sqrt{\frac{a - a \cos 2 y}{a + a \cos 2 y}} dy = \int_{0}^{\frac{π}{2}} 2 a \sin 2 y \tan y dy = 2 a \int_{0}^{\frac{π}{2}} 2 \sin y \cos y \frac{\sin y}{\cos y} dy = 2 a \int_{0}^{\frac{π}{2}} 2 \sin^{2} y dy = 2 a \int_{0}^{\frac{π}{2}} (1 - \cos 2 y) dy = 2 a {[y - \frac{\sin 2 y}{2}]}_{0}^{\frac{π}{2}} = 2 a {[\frac{π}{2} - \frac{\sin 2 y}{2}]}_{0}^{\frac{π}{2}} = πa$

Page No 19.40:

Question 56:

$\int_{0}^{π / 2} \frac{\sin x \cos x}{\cos^{2} x + 3 \cos x + 2} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{\cos^{2} x + 3 \cos x + 2} d x . Then, Let \cos x = t . Then, - \sin x d x = d t When x = 0, t = 1 and x = \frac{π}{2}, t = 0 ∴ I = - \int_{1}^{0} \frac{t d t}{t^{2} + 3 t + 2} \Rightarrow I = \int_{1}^{0} \frac{- t d t}{(t + 2) (t + 1)} \Rightarrow I = \int_{1}^{0} (\frac{1}{(t + 1)} - \frac{2}{(t + 2)}) d t \Rightarrow I = {[\log (t + 1) - 2 \log (t + 2)]}_{1}^{0} \Rightarrow I = {[\log \frac{(t + 1)}{{(t + 2)}^{2}}]}_{0}^{1} \Rightarrow I = {[\log (\frac{1}{4}) - \log (\frac{2}{9})]}_{0}^{1} \Rightarrow I = \log \frac{9}{8}$

Page No 19.40:

Question 57:

$\int_{0}^{\frac{π}{2}} \frac{\tan x}{1 + m^{2} \tan^{2} x} d x$

Answer:

Let I = $\int_{0}^{\frac{π}{2}} \frac{\tan x}{1 + m^{2} \tan^{2} x} d x$
$= \int_{0}^{\frac{π}{2}} \frac{\frac{\sin x}{\cos x}}{1 + m^{2} \frac{\sin^{2} x}{\cos^{2} x}} d x = \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{\cos^{2} x + m^{2} \sin^{2} x} d x$

Put $\cos^{2} x + m^{2} \sin^{2} x = z$

$∴ 2 \cos x (- \sin x) d x + m^{2} \times 2 \sin x \cos x d x = d z \Rightarrow 2 (m^{2} - 1) \sin x \cos x d x = d z \Rightarrow \sin x \cos x d x = \frac{d z}{2 (m^{2} - 1)}$

When $x \to 0, z \to 1$ $(z = \cos^{2} x + m^{2} \sin^{2} x = 1 + m^{2} \times 0 = 1)$

When $x \to \frac{π}{2}, z \to m^{2}$ $(z = \cos^{2} x + m^{2} \sin^{2} x = 0 + m^{2} \times 1 = m^{2})$

$∴ I = \frac{1}{2 (m^{2} - 1)} \int_{1}^{m^{2}} \frac{d z}{z} = \frac{1}{2 (m^{2} - 1)} {\log z}_{1}^{m^{2}} = \frac{1}{2 (m^{2} - 1)} (\log m^{2} - \log 1) = \frac{1}{2 (m^{2} - 1)} (2 \log |m| - 0) = \frac{\log |m|}{m^{2} - 1}$

Page No 19.40:

Question 58:

$\int_{0}^{\frac{1}{2}} \frac{1}{(1 + x^{2}) \sqrt{1 - x^{2}}} d x$

Answer:

Let I = $\int_{0}^{\frac{1}{2}} \frac{1}{(1 + x^{2}) \sqrt{1 - x^{2}}} d x$

Put $x = \sin θ$

$∴ d x = \cos θ d θ$

When $x \to 0, θ \to 0$

When $x \to \frac{1}{2}, θ \to \frac{π}{6}$

$∴ I = \int_{0}^{\frac{π}{6}} \frac{1}{(1 + \sin^{2} θ) \cos θ} \times \cos θ d θ = \int_{0}^{\frac{π}{6}} \frac{1}{1 + \sin^{2} θ} d θ$

Dividing numerator and denominator by $\cos^{2} θ$ , we have

$I = \int_{0}^{\frac{π}{6}} \frac{\sec^{2} θ}{\sec^{2} θ + \tan^{2} θ} d θ = \int_{0}^{\frac{π}{6}} \frac{\sec^{2} θ}{1 + 2 \tan^{2} θ} d θ$

Now, put $\tan θ = u$

$∴ \sec^{2} θ d θ = d u$

When $θ \to 0, u \to 0$

When $θ \to \frac{π}{6}, u \to \frac{1}{\sqrt{3}}$
$∴ I = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{d u}{1 + 2 u^{2}} = \int_{0}^{\frac{1}{\sqrt{3}}} \frac{d u}{1 + {(\sqrt{2} u)}^{2}} = {\frac{\tan^{- 1} \sqrt{2} u}{\sqrt{2}}}_{0}^{\frac{1}{\sqrt{3}}} = \frac{1}{\sqrt{2}} (\tan^{- 1} \frac{\sqrt{2}}{\sqrt{3}} - 0) = \frac{1}{\sqrt{2}} \tan^{- 1} \sqrt{\frac{2}{3}}$

Page No 19.40:

Question 59:

$\int_{\frac{1}{3}}^{1} \frac{{(x - x^{3})}^{\frac{1}{3}}}{x^{4}} d x$

Answer:

Let I = $\int_{\frac{1}{3}}^{1} \frac{{(x - x^{3})}^{\frac{1}{3}}}{x^{4}} d x$
$= \int_{\frac{1}{3}}^{1} \frac{{[x^{3} (\frac{x}{x^{3}} - 1)]}^{\frac{1}{3}}}{x^{4}} d x = \int_{\frac{1}{3}}^{1} \frac{x {(\frac{1}{x^{2}} - 1)}^{\frac{1}{3}}}{x^{4}} d x = \int_{\frac{1}{3}}^{1} \frac{{(\frac{1}{x^{2}} - 1)}^{\frac{1}{3}}}{x^{3}} d x$

Put $(\frac{1}{x^{2}} - 1) = z$

$∴ - \frac{2}{x^{3}} d x = d z \Rightarrow \frac{d x}{x^{3}} = - \frac{d z}{2}$

When $x \to \frac{1}{3}, z \to 8$

When $x \to 1, z \to 0$

$∴ I = - \frac{1}{2} \int_{8}^{0} z^{\frac{1}{3}} d z = - \frac{1}{2} \times {\frac{z^{\frac{4}{3}}}{\frac{4}{3}}}_{8}^{0} = - \frac{3}{8} [0 - {(8)}^{\frac{4}{3}}] = - \frac{3}{8} \times (- 16) = 6$

Page No 19.40:

Question 60:

$\int_{0}^{\frac{π}{4}} \frac{\sin^{2} x \cos^{2} x}{{(\sin^{3} x + \cos^{3} x)}^{2}} d x$

Answer:

Let $I = \int_{0}^{\frac{π}{4}} \frac{\sin^{2} x \cos^{2} x}{{(\sin^{3} x + \cos^{3} x)}^{2}} d x$
$= \int_{0}^{\frac{π}{4}} \frac{\sin^{2} x \cos^{2} x}{\cos^{6} x {(\tan^{3} x + 1)}^{2}} d x = \int_{0}^{\frac{π}{4}} \frac{\tan^{2} x \sec^{2} x}{{(\tan^{3} x + 1)}^{2}} d x$

Put $\tan^{3} x + 1 = z$

$∴ 3 \tan^{2} x \sec^{2} x d x = d z \Rightarrow \tan^{2} x \sec^{2} x d x = \frac{d z}{3}$

When $x \to 0, z \to 1$

When $x \to \frac{π}{4}, z \to 2$

$∴ I = \frac{1}{3} \int_{1}^{2} \frac{d z}{z^{2}} = \frac{1}{3} \times - {\frac{1}{z}}_{1}^{2} = - \frac{1}{3} (\frac{1}{2} - 1) = - \frac{1}{3} \times (- \frac{1}{2}) = \frac{1}{6}$

Page No 19.40:

Question 61:

$\int_{0}^{\frac{π}{2}} \sqrt{\cos x - \cos^{3} x} (\sec^{2} x - 1) \cos^{2} x d x$

Answer:

Let I = $\int_{0}^{\frac{π}{2}} \sqrt{\cos x - \cos^{3} x} (\sec^{2} x - 1) \cos^{2} x d x$
$= \int_{0}^{\frac{π}{2}} \sqrt{\cos x (1 - \cos^{2} x)} (- \tan^{2} x) \cos^{2} x d x = - \int_{0}^{\frac{π}{2}} \sqrt{\cos x (\sin^{2} x)} \sin^{2} x d x = - \int_{0}^{\frac{π}{2}} \sqrt{\cos x} |\sin x| \sin^{2} x d x = - \int_{0}^{\frac{π}{2}} \sqrt{\cos x} (1 - \cos^{2} x) \sin x d x (|\sin x| = \sin x for 0 \leq x \leq \frac{π}{2})$

Put cosx = z²

$∴ - \sin x d x = 2 z d z$

When $x \to 0, z \to 1$

When $x \to \frac{π}{2}, z \to 0$

$∴ I = - \int_{1}^{0} z (1 - z^{4}) 2 z d z = - 2 \int_{1}^{0} z^{2} d z + 2 \int_{1}^{0} z^{6} d z = - 2 \times {\frac{z^{3}}{3}}_{1}^{0} + 2 \times {\frac{z^{7}}{7}}_{1}^{0} = - \frac{2}{3} (0 - 1) + \frac{2}{7} (0 - 1) = \frac{2}{3} - \frac{2}{7} = \frac{8}{21}$

Page No 19.40:

Question 62:

$\int_{0}^{\frac{π}{2}} \frac{\cos x}{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{n}} d x$

Answer:

Let I = $\int_{0}^{\frac{π}{2}} \frac{\cos x}{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{n}} d x$
$= \int_{0}^{\frac{π}{2}} \frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{n}} d x = \int_{0}^{\frac{π}{2}} \frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) (\cos \frac{x}{2} - \sin \frac{x}{2})}{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{n}} d x = \int_{0}^{\frac{π}{2}} \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})}{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{n - 1}} d x$

Put $\cos \frac{x}{2} + \sin \frac{x}{2} = z$

$∴ (- \sin \frac{x}{2} \times \frac{1}{2} + \cos \frac{x}{2} \times \frac{1}{2}) d x = d z \Rightarrow (\cos \frac{x}{2} - \sin \frac{x}{2}) d x = 2 d z$

When $x \to 0, z \to 1$

When $x \to \frac{π}{2}, z \to \sqrt{2} (z = \cos \frac{π}{4} + \sin \frac{π}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2})$

$∴ I = 2 \int_{1}^{\sqrt{2}} \frac{d z}{z^{n - 1}} = 2 \times {\frac{z^{2 - n}}{2 - n}}_{1}^{\sqrt{2}} = \frac{2}{(2 - n)} [{(\sqrt{2})}^{2 - n} - 1] = \frac{2}{(2 - n)} (2^{\frac{2 - n}{2}} - 1) = \frac{2}{(2 - n)} (2^{1 - \frac{n}{2}} - 1)$

Page No 19.55:

Question 1:

(i) $\int_{1}^{4} f (x) d x, where f (x) = \{\begin{cases} 4 x + 3, if 1 \leq x \leq 2 \\ 3 x + 5, if 2 \leq x \leq 4 \end{cases}$

(ii) $\int_{0}^{9} f (x) d x, where f (x) \{\begin{matrix} \sin x & , & 0 \leq x \leq π / 2 \\ 1 & , & π / 2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9 \end{matrix}$

(iii) $\int_{1}^{4} f (x) d x, where f (x) = \{\begin{matrix} 7 x + 3 & , & if 1 \leq x \leq 3 \\ 8 x & , & if 3 \leq x \leq 4 \end{matrix}$

(iv) $\int_{- 1}^{2} \frac{|x|}{x} d x$

Answer:

(iv) $\int_{- 1}^{2} \frac{|x|}{x} d x$

$Let I = \int_{- 1}^{2} \frac{|x|}{x} d x We know, |x| = \{\begin{cases} x, x \geq 0 \\ - x, x < 0 \end{cases} Thus, I = \int_{- 1}^{0} \frac{|x|}{x} d x + \int_{0}^{2} \frac{|x|}{x} d x = \int_{- 1}^{0} \frac{- x}{x} d x + \int_{0}^{2} \frac{x}{x} d x = \int_{- 1}^{0} - 1 d x + \int_{0}^{2} 1 d x = {[- x]}_{- 1}^{0} + {[x]}_{0}^{2} = (- 0 - 1) + (2 - 0) = - 1 + 2 = 1$

Hence, $\int_{- 1}^{2} \frac{|x|}{x} d x$ = 1.

Page No 19.56:

Question 2:

Evaluate the following integrals:

$\int_{- 4}^{4} |x + 2| d x$

Answer:

$\int_{- 4}^{4} |x + 2| d x We know that, |x + 2| = \{\begin{cases} - (x + 2), - 4 \leq x \leq - 2 \\ x + 2, - 2 < x \leq 4 \end{cases} ∴ I = \int_{- 4}^{4} |x + 2| d x \Rightarrow I = \int_{- 4}^{- 2} - (x + 2) d x + \int_{- 2}^{4} (x + 2) d x \Rightarrow I = {[- \frac{x^{2}}{2} - 2 x]}_{- 4}^{- 2} + {[\frac{x^{2}}{2} + 2 x]}_{- 2}^{4} \Rightarrow I = - 2 + 4 - 8 - 8 + 8 + 8 - 2 + 4 \Rightarrow I = 20$

Page No 19.56:

Question 3:

Evaluate the following integrals:
$\int_{- 3}^{3} |x + 1| d x$

Answer:

$I = \int_{- 3}^{3} |x + 1| d x We know that, |x + 1| = \{\begin{cases} - (x + 1), - 3 \leq x \leq - 1 \\ x + 1, - 1 < x \leq 3 \end{cases} ∴ I = \int_{- 3}^{- 1} - (x + 1) d x + \int_{- 1}^{3} [x + 1] d x \Rightarrow I = {[- \frac{{(x + 1)}^{2}}{2}]}_{- 3}^{- 1} + {[\frac{{(x + 1)}^{2}}{2}]}_{- 1}^{3} \Rightarrow I = 0 + 2 + 8 - 0 \Rightarrow I = 10$

Page No 19.56:

Question 4:

Evaluate the following integrals:
$\int_{- 1}^{1} |2 x + 1| d x$

Answer:

$\int_{- 1}^{1} |2 x + 1| d x We know that, |2 x + 1| = \{\begin{cases} - (2 x + 1), - 1 \leq x \leq - \frac{1}{2} \\ (2 x + 1), - \frac{1}{2} < x \leq 1 \end{cases} ∴ I = \int_{- 1}^{\frac{- 1}{2}} - (2 x + 1) d x + \int_{- \frac{1}{2}}^{1} (2 x + 1) d x \Rightarrow I = - {[x^{2} + x]}_{- 1}^{\frac{- 1}{2}} + {[x^{2} + x]}_{- \frac{1}{2}}^{1} \Rightarrow I = - \frac{1}{4} + \frac{1}{2} + 1 - 1 + 1 + 1 - \frac{1}{4} + \frac{1}{2} \Rightarrow I = \frac{5}{2}$

Page No 19.56:

Question 5:

Evaluate the following integrals:
$\int_{- 2}^{2} |2 x + 3| d x$

Answer:

$\int_{- 2}^{2} |2 x + 3| d x We know that, |2 x + 3| = \{\begin{cases} - (2 x + 3), - 2 \leq x \leq - \frac{3}{2} \\ (2 x + 3), - \frac{3}{2} < x \leq 2 \end{cases} ∴ I = \int_{- 2}^{\frac{- 3}{2}} - (2 x + 3) d x + \int_{- \frac{3}{2}}^{2} (2 x + 3) d x \Rightarrow I = - {[x^{2} + 3 x]}_{- 2}^{\frac{- 3}{2}} + {[x^{2} + 3 x]}_{- \frac{3}{2}}^{2} \Rightarrow I = - \frac{9}{4} + \frac{9}{2} + 4 - 6 + 4 + 6 - \frac{9}{4} + \frac{9}{2} \Rightarrow I = \frac{25}{2}$

Page No 19.56:

Question 6:

Evaluate the following integrals:
$\int_{0}^{2} |x^{2} - 3 x + 2| d x$

Answer:

$\int_{0}^{2} |x^{2} - 3 x + 2| d x We know that, |x^{2} - 3 x + 2| = \{\begin{cases} - (x^{2} - 3 x + 2), (x - 1) (x - 2) \leq 0 or, 1 \leq x \leq 2 \\ (x^{2} - 3 x + 2), x^{2} - 3 x + 2 \geq 0 or, x \in (- \infty, 1) \cup (2, \infty) \end{cases} ∴ I = \int_{0}^{2} (x^{2} - 3 x + 2) d x \Rightarrow I = \int_{0}^{1} (x^{2} - 3 x + 2) d x - \int_{1}^{2} (x^{2} - 3 x + 2) d x \Rightarrow I = {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{0}^{1} - {[\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x]}_{1}^{2} \Rightarrow I = \frac{1}{3} - \frac{3}{2} + 2 - [\frac{8}{3} - 6 + 4 - \frac{1}{3} + \frac{3}{2} - 2] \Rightarrow I = \frac{1}{3} - \frac{3}{2} + 2 - \frac{8}{3} + 6 - 2 + \frac{1}{3} - \frac{3}{2} \Rightarrow I = 1$

Page No 19.56:

Question 7:

Evaluate the following integrals:
$\int_{0}^{3} |3 x - 1| d x$

Answer:

$\int_{0}^{3} |3 x - 1| d x We know that, |3 x - 1| = \{\begin{cases} - (3 x - 1), 0 \leq x \leq \frac{1}{3} \\ (3 x - 1), \frac{1}{3} < x \leq 3 \end{cases} ∴ I = = \int_{0}^{\frac{1}{3}} - (3 x + 1) d x + \int_{\frac{1}{3}}^{0} (3 x + 1) d x \Rightarrow I = {[\frac{- 3 x^{2}}{2} - x]}_{0}^{\frac{1}{3}} + {[\frac{3 x^{2}}{2} + x]}_{\frac{1}{3}}^{3} \Rightarrow I = - \frac{1}{6} + \frac{1}{3} - 0 + \frac{27}{2} + 3 - \frac{1}{6} - \frac{1}{3} \Rightarrow I = \frac{65}{6}$

Page No 19.56:

Question 8:

Evaluate the following integrals:
$\int_{- 6}^{6} |x + 2| d x$

Answer:

$\int_{- 6}^{6} |x + 2| d x We know that, |x + 2| = \{\begin{cases} - (x + 2), - 6 \leq x \leq - 2 \\ x + 2, - 2 < x \leq 6 \end{cases} ∴ I = \int_{- 6}^{6} |x + 2| d x \Rightarrow I = \int_{- 6}^{- 2} - (x + 2) d x + \int_{- 2}^{6} (x + 2) d x \Rightarrow I = {[\frac{- x^{2}}{2} - 2 x]}_{- 6}^{- 2} + {[\frac{x^{2}}{2} + 2 x]}_{- 2}^{6} \Rightarrow I = - 2 + 4 + 18 - 12 + 18 + 12 - 2 + 4 \Rightarrow I = 40$

Page No 19.56:

Question 9:

Evaluate the following integrals:
$\int_{- 2}^{2} |x + 1| d x$

Answer:

$\int_{- 2}^{2} |x + 1| d x We know that, |x + 1| = \{\begin{cases} - (x + 1), - 2 \leq x \leq - 1 \\ x + 1, - 1 < x \leq 2 \end{cases} ∴ I = \int_{- 2}^{2} |x + 1| d x \Rightarrow I = \int_{- 2}^{- 1} - (x + 1) d x + \int_{- 1}^{2} (x + 1) d x \Rightarrow I = {[\frac{- x^{2}}{2} - x]}_{- 2}^{- 1} + {[\frac{x^{2}}{2} + x]}_{- 1}^{2} \Rightarrow I = \frac{- 1}{2} + 1 + 2 - 2 + 2 + 2 - \frac{1}{2} + 1 \Rightarrow I = 5$

Page No 19.56:

Question 10:

Evaluate the following integrals:
$\int_{1}^{2} |x - 3| d x$

Answer:

$\int_{1}^{2} |x - 3| d x We know that, |x + 1| = \{\begin{cases} - (x + 1), 1 \leq x \leq 3 \\ (x + 1), x > 3 \end{cases} ∴ I = \int_{1}^{2} |x - 3| d x \Rightarrow I = \int_{1}^{2} - (x - 3) d x \Rightarrow I = {[\frac{- x^{2}}{2} - 3 x]}_{1}^{2} \Rightarrow I = - 2 - 6 + \frac{1}{2} + 3 \Rightarrow I = \frac{3}{2}$

Page No 19.56:

Question 11:

Evaluate the following integrals:
$\int_{0}^{π / 2} |\cos 2 x| d x$

Answer:

$\int_{0}^{\frac{π}{2}} |\cos 2 x| d x We know that, |\cos 2 x| = \{\begin{cases} - \cos 2 x, \frac{π}{4} \leq x \leq \frac{π}{2} \\ \cos 2 x, 0 < x \leq \frac{π}{4} \end{cases} ∴ I = \int_{- 2}^{2} |\cos 2 x| d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \cos 2 x d x - \int_{\frac{π}{4}}^{\frac{π}{2}} \cos 2 x d x \Rightarrow I = {[\frac{\sin 2 x}{2}]}_{0}^{\frac{π}{4}} - {[\frac{\sin 2 x}{2}]}_{\frac{π}{4}}^{\frac{π}{2}} \Rightarrow I = \frac{1}{2} - 0 - 0 + \frac{1}{2} \Rightarrow I = 1$

Page No 19.56:

Question 12:

Evaluate the following integrals:
$\int_{0}^{2 π} |\sin x| d x$

Answer:

$\int_{0}^{2 π} |\sin x| d x We know that, |\sin x| = \{\begin{cases} - \sin x, π \leq x \leq 2 π \\ \sin x, 0 < x \leq π \end{cases} ∴ I = \int_{0}^{2 π} |\sin x| d x \Rightarrow I = \int_{0}^{π} \sin x d x + \int_{π}^{2 π} - \sin x d x \Rightarrow I = - {[\cos x]}_{0}^{π} + {[\cos x]}_{π}^{2 π} \Rightarrow I = 1 + 1 + 1 - (- 1) \Rightarrow I = 4$

Page No 19.56:

Question 13:

Evaluate the following integrals:
$\int_{- π / 4}^{π / 4} |\sin x| d x$

Answer:

$\int_{- \frac{π}{4}}^{\frac{π}{4}} |\sin x| d x We know that, |\sin x| = \{\begin{cases} - \sin x, - \frac{π}{4} \leq x \leq 0 \\ \sin x, 0 < x \leq \frac{π}{4} \end{cases} ∴ I = \int_{- \frac{π}{4}}^{\frac{π}{4}} |\sin x| d x \Rightarrow I = \int_{- \frac{π}{4}}^{0} - \sin x d x + \int_{0}^{\frac{π}{4}} \sin x d x \Rightarrow I = {[\cos x]}_{\frac{- π}{4}}^{0} - {[\cos x]}_{0}^{\frac{- π}{4}} \Rightarrow I = 1 - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + 1 \Rightarrow I = 2 - \frac{2}{\sqrt{2}} \Rightarrow I = 2 - \sqrt{2}$

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Question 14:

Evaluate the following integrals:
$\int_{2}^{8} |x - 5| d x$

Answer:

$\int_{2}^{8} |x - 5| d x We know that, |x - 5| = \{\begin{cases} - (x - 5), 2 \leq x \leq 5 \\ x - 5, 5 < x \leq 8 \end{cases} ∴ I = \int_{2}^{8} |x - 5| d x \Rightarrow I = \int_{2}^{5} - (x - 5) d x + \int_{5}^{8} (x - 5) d x \Rightarrow I = - {[\frac{x^{2}}{2} - 5 x]}_{2}^{5} + {[\frac{x^{2}}{2} - 5 x]}_{5}^{8} \Rightarrow I = \frac{- 25}{2} + 25 + 2 - 10 + 32 - 40 - \frac{25}{2} + 25 \Rightarrow I = 9$

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Question 15:

Evaluate the following integrals:
$\int_{- π / 2}^{π / 2} \{\sin |x| + \cos |x|\} d x$

Answer:

$\int_{\frac{- π}{2}}^{\frac{π}{2}} \{\sin |x| + \cos |x|\} d x Since, f (- x) = \sin |- x| + \cos |- x| = \sin |x| + \cos |x| = f (x) S o, f (x) is an even function . ∴ I = 2 \int_{0}^{\frac{π}{2}} (\sin x + \cos x) dx \Rightarrow I = 2 {[- \cos x + \sin x]}_{0}^{\frac{π}{2}} \Rightarrow I = 2 (0 + 1 + 1 - 0) \Rightarrow I = 4$

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Question 16:

Evaluate the following integrals:
$\int_{0}^{4} |x - 1| d x$

Answer:

$\int_{0}^{4} |x - 1| d x We know that, |x - 1| = \{\begin{cases} - (x - 1), 0 \leq x \leq 1 \\ x - 1, 1 < x \leq 4 \end{cases} ∴ I = \int_{0}^{4} |x - 1| d x \Rightarrow I = \int_{0}^{1} - (x - 1) d x + \int_{1}^{4} (x - 1) d x \Rightarrow I = {[- \frac{x^{2}}{2} + x]}_{0}^{1} + {[\frac{x^{2}}{2} - x]}_{1}^{4} \Rightarrow I = \frac{- 1}{2} + 1 - 0 + 8 - 4 - \frac{1}{2} + 1 \Rightarrow I = 5$

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Question 17:

Evaluate the following integrals:
$\int_{1}^{4} \{|x - 1| + |x - 2| + |x - 4|\} d x$

Answer:

$I = \int_{1}^{4} \{|x - 1| + |x - 2| + |x - 4|\} d x \Rightarrow I = \int_{1}^{4} |x - 1| d x + \int_{1}^{4} |x - 2| d x + \int_{1}^{4} |x - 4| d x We know that, |x - 1| = \{\begin{cases} - (x - 1), x \leq 1 \\ x - 1, 1 < x \leq 4 \end{cases} |x - 2| = \{\begin{cases} - (x - 2), 1 \leq x \leq 2 \\ x - 2, 2 < x \leq 4 \end{cases} |x - 4| = \{\begin{cases} - (x - 4), 1 \leq x \leq 4 \\ x - 4, x > 4 \end{cases} ∴ I = \int_{1}^{4} (x - 1) d x - \int_{1}^{2} (x - 2) d x + \int_{2}^{4} (x - 2) d x - \int_{1}^{4} (x - 4) d x \Rightarrow I = {[\frac{x^{2}}{2} - x]}_{1}^{4} - {[\frac{x^{2}}{2} - 2 x]}_{1}^{2} + {[\frac{x^{2}}{2} - 2 x]}_{2}^{4} - {[\frac{x^{2}}{2} - 4 x]}_{1}^{4} \Rightarrow I = 8 - 4 - \frac{1}{2} + 1 - (2 - 4 - \frac{1}{2} + 2) + 8 - 8 - 2 + 4 - (8 - 16 - \frac{1}{2} + 4) \Rightarrow I = \frac{23}{2}$

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Question 18:

Evaluate the following integrals:
$\int_{- 5}^{0} f (x) d x, where f (x) = |x| + |x + 2| + |x + 5|$

Answer:

$I = \int_{- 5}^{0} \{|x| + |x + 2| + |x + 5|\} d x \Rightarrow I = \int_{- 5}^{0} |x| d x + \int_{- 5}^{0} |x + 2| d x + \int_{- 5}^{0} |x + 5| d x We know that, |x| = \{\begin{cases} - x, - 5 \leq x \leq 0 \\ x, x > 0 \end{cases} |x + 2| = \{\begin{cases} - (x + 2), - 5 \leq x \leq - 2 \\ x + 2, - 2 < x \leq 0 \end{cases} |x + 5| = \{\begin{cases} - (x + 5), - 5 \leq x \leq 0 \\ x + 5, x > - 5 \end{cases} ∴ I = - \int_{- 5}^{0} x d x - \int_{- 5}^{- 2} (x + 2) d x + \int_{- 2}^{0} (x + 2) d x + \int_{- 5}^{0} (x + 5) d x \Rightarrow I = - {[\frac{x^{2}}{2}]}_{- 5}^{0} - {[\frac{x^{2}}{2} + 2 x]}_{- 5}^{- 2} + {[\frac{x^{2}}{2} + 2 x]}_{- 2}^{0} + {[\frac{x^{2}}{2} + 5 x]}_{- 5}^{0} \Rightarrow I = \frac{25}{2} - (2 - 4 - \frac{25}{2} + 10) - 2 + 4 + (- \frac{25}{2} + 25) \Rightarrow I = \frac{63}{2}$

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Question 19:

Evaluate the following integrals:
$\int_{0}^{4} (|x| + |x - 2| + |x - 4|) d x$

Answer:

$I = \int_{0}^{4} \{|x| + |x - 2| + |x - 4|\} d x \Rightarrow I = \int_{0}^{4} |x| d x + \int_{0}^{4} |x - 2| d x + \int_{0}^{4} |x - 4| d x We know that, |x| = \{\begin{cases} - x, - 5 \leq x \leq 0 \\ x, x > 0 \end{cases} |x - 2| = \{\begin{cases} - (x - 2), 0 \leq x \leq 2 \\ x - 2, 2 < x \leq 4 \end{cases} |x - 4| = \{\begin{cases} - (x - 4), 0 \leq x \leq 4 \\ x - 4, x > 4 \end{cases} ∴ I = \int_{0}^{4} x d x - \int_{0}^{2} (x - 2) d x + \int_{2}^{4} (x - 2) d x - \int_{0}^{4} (x - 4) d x \Rightarrow I = {[\frac{x^{2}}{2}]}_{0}^{4} - {[\frac{x^{2}}{2} - 2 x]}_{0}^{2} + {[\frac{x^{2}}{2} - 2 x]}_{2}^{4} - {[\frac{x^{2}}{2} - 4 x]}_{0}^{4} \Rightarrow I = 8 - (2 - 4) + 8 - 8 - 2 + 4 - (8 - 16) \Rightarrow I = 20$

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Question 20:

$\int_{- 1}^{2} (|x + 1| + |x| + |x - 1|) d x$

Answer:

We know that

$|x + 1| = \{\begin{cases} x + 1, & if x + 1 \geq 0 \\ - (x + 1), & if x + 1 < 0 \end{cases} = \{\begin{cases} x + 1, & if x \geq - 1 \\ - (x + 1), & if x < - 1 \end{cases}$

$|x| = \{\begin{cases} x, & if x \geq 0 \\ - x, & if x < 0 \end{cases}$

$|x - 1| = \{\begin{cases} x - 1, & if x - 1 \geq 0 \\ - (x - 1), & if x - 1 < 0 \end{cases} = \{\begin{cases} x - 1, & if x \geq 1 \\ - (x - 1), & if x < 1 \end{cases}$

When $- 1 \leq x \leq 0,$

$|x + 1| + |x| + |x - 1| = x + 1 + (- x) + [- (x - 1)] = 2 - x$

When $0 \leq x \leq 1,$

$|x + 1| + |x| + |x - 1| = x + 1 + x + [- (x - 1)] = x + 2$

When $1 \leq x \leq 2,$

$|x + 1| + |x| + |x - 1| = x + 1 + x + x - 1 = 3 x$

$∴ \int_{- 1}^{2} (|x + 1| + |x| + |x - 1|) d x = \int_{- 1}^{0} (2 - x) d x + \int_{0}^{1} (x + 2) d x + \int_{1}^{2} 3 x d x = {\frac{{(2 - x)}^{2}}{2 \times (- 1)}}_{- 1}^{0} + {\frac{{(x + 2)}^{2}}{2}}_{0}^{1} + 3 \times {\frac{x^{2}}{2}}_{1}^{2} = - \frac{1}{2} (4 - 9) + \frac{1}{2} (9 - 4) + \frac{3}{2} (4 - 1) = \frac{5}{2} + \frac{5}{2} + \frac{9}{2} = \frac{19}{2}$

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Question 21:

$\int_{- 2}^{2} x e^{|x|} d x$

Answer:

Consider $f (x) = x e^{|x|}$ .

Now,

$f (- x) = (- x) e^{|- x|} = - x e^{|x|} = - f (x)$

⇒ f(x) is an odd function.

$∴ \int_{- 2}^{2} x e^{|x|} d x = 0$ $[\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}]$

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Question 22:

$\int_{- \frac{π}{4}}^{\frac{π}{2}} \sin x |\sin x| d x$

Answer:

$\int_{- \frac{π}{4}}^{\frac{π}{2}} \sin x |\sin x| d x = \int_{- \frac{π}{4}}^{0} \sin x |\sin x| d x + \int_{0}^{\frac{π}{2}} \sin x |\sin x| d x = \int_{- \frac{π}{4}}^{0} \sin x (- \sin x) d x + \int_{0}^{\frac{π}{2}} \sin x \sin x d x (|\sin x| = \{\begin{cases} \sin x, & 0 \leq x \leq \frac{π}{2} \\ - \sin x, & - \frac{π}{4} \leq x \leq 0 \end{cases}) = - \int_{- \frac{π}{4}}^{0} \sin^{2} x d x + \int_{0}^{\frac{π}{2}} \sin^{2} x d x$
$= - \int_{- \frac{π}{4}}^{0} \frac{1 - \cos 2 x}{2} d x + \int_{0}^{\frac{π}{2}} \frac{1 - \cos 2 x}{2} d x = - \frac{1}{2} \int_{- \frac{π}{4}}^{0} d x + \frac{1}{2} \int_{- \frac{π}{4}}^{0} \cos 2 x d x + \frac{1}{2} \int_{0}^{\frac{π}{2}} d x - \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos 2 x d x = - \frac{1}{2} \times x_{- \frac{π}{4}}^{0} + \frac{1}{2} \times {\frac{\sin 2 x}{2}}_{- \frac{π}{4}}^{0} + \frac{1}{2} \times x_{0}^{\frac{π}{2}} - \frac{1}{2} \times {\frac{\sin 2 x}{2}}_{0}^{\frac{π}{2}} = - \frac{1}{2} (0 + \frac{π}{4}) + \frac{1}{4} (0 + \sin \frac{π}{2}) + \frac{1}{2} \times (\frac{π}{2} - 0) - \frac{1}{4} (sinπ - 0) = - \frac{π}{8} + \frac{1}{4} (0 + 1) + \frac{π}{4} - \frac{1}{4} (0 - 0) = \frac{π}{8} + \frac{1}{4}$

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Question 23:

$\int_{0}^{π} \cos x |\cos x| d x$

Answer:

Consider $f (x) = \cos x |\cos x|$

Now,

$f (π - x) = \cos (π - x) |\cos (π - x)| = - \cos x |- \cos x| = - \cos x |\cos x| = - f (x)$

$∴ \int_{0}^{π} \cos x |\cos x| d x = 0 [\int_{0}^{2 a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (2 a - x) = f (x) \\ 0, & if f (2 a - x) = - f (x) \end{cases}]$

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Question 24:

$\int_{- \frac{π}{2}}^{\frac{π}{2}} (2 \sin |x| + \cos |x|) d x$

Answer:

Consider $f (x) = 2 \sin |x| + \cos |x|$

Now,

$f (- x) = 2 \sin |- x| + \cos |- x| = 2 \sin |x| + \cos |x| = f (x)$

$∴ \int_{- \frac{π}{2}}^{\frac{π}{2}} (2 \sin |x| + \cos |x|) d x = 2 \int_{0}^{\frac{π}{2}} (2 \sin |x| + \cos |x|) d x [\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}] = 2 \int_{0}^{\frac{π}{2}} (2 \sin x + \cos x) d x [|x| = \{\begin{cases} x, & if x \geq 0 \\ - x, & if x < 0 \end{cases}] = 4 \int_{0}^{\frac{π}{2}} \sin x d x + 2 \int_{0}^{\frac{π}{2}} \cos x d x$
$= 4 \times {(- \cos x)}_{0}^{\frac{π}{2}} + 2 \times {\sin x}_{0}^{\frac{π}{2}} = - 4 (\cos \frac{π}{2} - \cos 0) + 2 (\sin \frac{π}{2} - \sin 0) = - 4 (0 - 1) + 2 (1 - 0) = 4 + 2 = 6$

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Question 25:

$\int_{- \frac{π}{2}}^{π} \sin^{- 1} (\sin x) d x$

Answer:

$\int_{- \frac{π}{2}}^{π} \sin^{- 1} (\sin x) d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{- 1} (\sin x) d x + \int_{\frac{π}{2}}^{π} \sin^{- 1} (\sin x) d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} x d x + \int_{\frac{π}{2}}^{π} (π - x) d x [\frac{π}{2} \leq x \leq π \Rightarrow - π \leq - x \leq - \frac{π}{2} \Rightarrow 0 \leq π - x \leq \frac{π}{2}] = {\frac{x^{2}}{2}}_{- \frac{π}{2}}^{\frac{π}{2}} + {\frac{{(π - x)}^{2}}{2 \times (- 1)}}_{\frac{π}{2}}^{π} = \frac{1}{2} (\frac{π^{2}}{4} - \frac{π^{2}}{4}) - \frac{1}{2} (0 - \frac{π^{2}}{4})$
$= 0 + \frac{π^{2}}{8} = \frac{π^{2}}{8}$

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Question 26:

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{- \frac{π}{2}}{\sqrt{\cos x \sin^{2} x}} d x$

Answer:

$Let I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{- \frac{π}{2}}{\sqrt{\cos x \sin^{2} x}} d x = - \frac{π}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1}{\sqrt{\cos x \sin^{2} x}} d x = - \frac{π}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1}{\sqrt{\cos x} |\sin x|} d x = - \frac{π}{2} \times 2 \int_{0}^{\frac{π}{2}} \frac{1}{\sqrt{\cos x} |\sin x|} d x [f (- x) = \sqrt{\cos (- x)} |\sin (- x)| = \sqrt{\cos x} |- \sin x| = \sqrt{\cos x} |\sin x| = f (x)]$
$= - π \int_{0}^{\frac{π}{2}} \frac{1}{\sqrt{\cos x} \sin x} dx (|\sin x| = \sin x, 0 \leq x \leq \frac{π}{2}) = - π \int_{0}^{\frac{π}{2}} \frac{\sin x}{\sqrt{\cos x} (1 - \cos^{2} x)} d x$

Put cosx = z²

$∴ - \sin x d x = 2 z d z$

When $x \to 0, z \to 1$

When $x \to \frac{π}{2}, z \to 0$

$∴ I = 2 π \int_{1}^{0} \frac{z d z}{z (1 - z^{4})} = 2 π \int_{1}^{0} \frac{d z}{1 - z^{4}} = 2 π \int_{1}^{0} \frac{d z}{(1 - z) (1 + z) (1 + z^{2})}$
Now,

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

Putting z = 1, we get

$A = \frac{1}{4}$

Putting z = −1, we get

$B = \frac{1}{4}$

Putting z = 0, we get

$1 = A + B + D \Rightarrow D = 1 - \frac{1}{4} - \frac{1}{4} = \frac{1}{2}$

Equating coefficient of z³ on both sides, we get

$A - B + C = 0 \Rightarrow \frac{1}{4} - \frac{1}{4} + C = 0 \Rightarrow C = 0$

$∴ I = 2 π \int_{1}^{0} \frac{d z}{(1 - z) (1 + z) (1 + z^{2})} = 2 π \int_{1}^{0} \frac{\frac{1}{4}}{1 - z} d z + 2 π \int_{1}^{0} \frac{\frac{1}{4}}{1 + z} d z + 2 π \int_{1}^{0} \frac{\frac{1}{2}}{1 + z^{2}} d z = \frac{2 π}{4} {\times \frac{\log (1 - z)}{- 1}}_{1}^{0} + \frac{2 π}{4} {\times \log (1 + z)}_{1}^{0} + \frac{2 π}{2} \times {\tan^{- 1} z}_{1}^{0} = - \frac{π}{2} (\log 1 - \log 0) + \frac{π}{2} (\log 1 - \log 2) + π (\tan^{- 1} 0 - \tan^{- 1} 1) = - \frac{π}{2} [0 - (- \infty)] + \frac{π}{2} (0 - \log 2) + π (0 - \frac{π}{4}) = - \infty - \frac{π}{2} \log 2 - \frac{π^{2}}{4} = - \infty$

Disclaimer: The answer does not matches with the answer provided for the question.

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Question 27:

$\int_{0}^{2} 2 x [x] d x$

Answer:

$\int_{0}^{2} 2 x [x] d x = \int_{0}^{1} 2 x [x] d x + \int_{1}^{2} 2 x [x] d x = \int_{0}^{1} 2 x \times 0 d x + \int_{1}^{2} 2 x \times 1 d x [[x] = \{\begin{cases} 0, & 0 \leq x < 1 \\ 1, & 1 \leq x < 2 \end{cases}] = 0 + 2 \int_{1}^{2} x d x = 2 \times {\frac{x^{2}}{2}}_{1}^{2} = 4 - 1 = 3$

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Question 28:

$\int_{0}^{2 π} \cos^{- 1} (\cos x) d x$

Answer:

$\int_{0}^{2 π} \cos^{- 1} (\cos x) d x = \int_{0}^{π} \cos^{- 1} (\cos x) d x + \int_{π}^{2 π} \cos^{- 1} (\cos x) d x = \int_{0}^{π} x d x + \int_{π}^{2 π} (2 π - x) d x [π \leq x \leq 2 π \Rightarrow - 2 π \leq - x \leq - π \Rightarrow 0 \leq 2 π - x \leq π]$
$= {\frac{x^{2}}{2}}_{0}^{π} + {\frac{{(2 π - x)}^{2}}{2 \times (- 1)}}_{π}^{2 π} = \frac{1}{2} (π^{2} - 0) - \frac{1}{2} (0 - π^{2}) = \frac{π^{2}}{2} + \frac{π^{2}}{2} = π^{2}$

Page No 19.61:

Question 1:

Evaluate each of the following integrals:

$\int_{0}^{2 π} \frac{e^{\sin x}}{e^{\sin x} + e^{- \sin x}} d x$

Answer:

Let I = $\int_{0}^{2 π} \frac{e^{\sin x}}{e^{\sin x} + e^{- \sin x}} d x$ .....(1)

Then,

$I = \int_{0}^{2 π} \frac{e^{\sin (2 π - x)}}{e^{\sin (2 π - x)} + e^{- \sin (2 π - x)}} d x (\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x) = \int_{0}^{2 π} \frac{e^{- \sin x}}{e^{- \sin x} + e^{\sin x}} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{0}^{2 π} \frac{e^{\sin x} + e^{- \sin x}}{e^{\sin x} + e^{- \sin x}} d x \Rightarrow 2 I = \int_{0}^{2 π} d x \Rightarrow 2 I = x_{0}^{2 π} \Rightarrow 2 I = 2 π - 0 \Rightarrow I = π$

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Question 2:

Evaluate each of the following integrals:

$\int_{0}^{2 π} \log (\sec x + \tan x) d x$

Answer:

Let I = $\int_{0}^{2 π} \log (\sec x + \tan x) d x$ .....(1)

Then,

$I = \int_{0}^{2 π} \log [\sec (2 π - x) + \tan (2 π - x)] d x [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{2 π} \log (\sec x - \tan x) d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{0}^{2 π} [\log (\sec x + \tan x) + \log (\sec x - \tan x)] d x \Rightarrow 2 I = \int_{0}^{2 π} \log (\sec^{2} x - \tan^{2} x) d x \Rightarrow 2 I = \int_{0}^{2 π} \log 1 d x (1 + \tan^{2} x = \sec^{2} x) \Rightarrow 2 I = 0 (\log 1 = 0) \Rightarrow I = 0$

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Question 3:

Evaluate each of the following integrals:

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} d x$

Answer:

Let I = $\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} d x$ .....(1)

Then,

$I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\tan (\frac{π}{3} + \frac{π}{6} - x)}}{\sqrt{\tan (\frac{π}{3} + \frac{π}{6} - x)} + \sqrt{\cot (\frac{π}{3} + \frac{π}{6} - x)}} d x [\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\tan (\frac{π}{2} - x)}}{\sqrt{\tan (\frac{π}{2} - x)} + \sqrt{\cot (\frac{π}{2} - x)}} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\tan x} + \sqrt{\cot x}}{\sqrt{\tan x} + \sqrt{\cot x}} d x \Rightarrow 2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} d x \Rightarrow 2 I = x_{\frac{π}{6}}^{\frac{π}{3}} \Rightarrow 2 I = \frac{π}{3} - \frac{π}{6} = \frac{π}{6} \Rightarrow I = \frac{π}{12}$

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Question 4:

Evaluate each of the following integrals:

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$

Answer:

Let I = $\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x$ .....(1)
Then,
$I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\sin (\frac{π}{3} + \frac{π}{6} - x)}}{\sqrt{\sin (\frac{π}{3} + \frac{π}{6} - x)} + \sqrt{\cos (\frac{π}{3} + \frac{π}{6} - x)}} d x [\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\sin (\frac{π}{2} - x)}}{\sqrt{\sin (\frac{π}{2} - x)} + \sqrt{\cos (\frac{π}{2} - x)}} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x \Rightarrow 2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} d x \Rightarrow 2 I = x_{\frac{π}{6}}^{\frac{π}{3}} \Rightarrow 2 I = \frac{π}{3} - \frac{π}{6} = \frac{π}{6} \Rightarrow I = \frac{π}{12}$

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Question 5:

Evaluate each of the following integrals:

$\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{\tan^{2} x}{1 + e^{x}} d x$

Answer:

Let I = $\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{\tan^{2} x}{1 + e^{x}} d x$ .....(1)
Then,
$I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{\tan^{2} [\frac{π}{4} + (- \frac{π}{4}) - x]}{1 + e^{[\frac{π}{4} + (- \frac{π}{4}) - x]}} d x [\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{\tan^{2} (- x)}{1 + e^{- x}} d x = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{\tan^{2} x}{1 + \frac{1}{e^{x}}} d x = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{e^{x} \tan^{2} x}{e^{x} + 1} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\frac{\tan^{2} x}{1 + e^{x}} + \frac{e^{x} \tan^{2} x}{1 + e^{x}}) d x \Rightarrow 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{(1 + e^{x}) \tan^{2} x}{1 + e^{x}} d x \Rightarrow 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \tan^{2} x d x \Rightarrow 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\sec^{2} x - 1) d x$
$\Rightarrow 2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \sec^{2} x d x - \int_{- \frac{π}{4}}^{\frac{π}{4}} d x \Rightarrow 2 I = {\tan x}_{- \frac{π}{4}}^{\frac{π}{4}} - x_{- \frac{π}{4}}^{\frac{π}{4}} \Rightarrow 2 I = [\tan \frac{π}{4} - \tan (- \frac{π}{4})] - [\frac{π}{4} - (- \frac{π}{4})] \Rightarrow 2 I = (1 + 1) - (\frac{2 π}{4}) \Rightarrow 2 I = 2 - \frac{π}{2} \Rightarrow I = 1 - \frac{π}{4}$

Disclaimer: This answer does not matches with the given answer in the book.

Page No 19.61:

Question 6:

Evaluate each of the following integrals:

$\int_{- a}^{a} \frac{1}{1 + a^{x}} d x$ , a > 0

Answer:

Let I = $\int_{- a}^{a} \frac{1}{1 + a^{x}} d x$ .....(1)
Then,
$I = \int_{- a}^{a} \frac{1}{1 + a^{[a + (- a) - x]}} d x = \int_{- a}^{a} \frac{1}{1 + a^{- x}} d x = \int_{- a}^{a} \frac{a^{x}}{a^{x} + 1} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{- a}^{a} \frac{1 + a^{x}}{1 + a^{x}} d x \Rightarrow 2 I = \int_{- a}^{a} d x \Rightarrow 2 I = x_{- a}^{a} \Rightarrow 2 I = a - (- a) = 2 a \Rightarrow I = a$

Page No 19.61:

Question 7:

Evaluate each of the following integrals:

$\int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{1}{1 + e^{\tan x}} d x$

Answer:

Let I = $\int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{1}{1 + e^{\tan x}} d x$ .....(1)
Then,
$I = \int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{1}{1 + e^{\tan [\frac{π}{3} + (- \frac{π}{3}) - x]}} d x [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{1}{1 + e^{\tan (- x)}} d x = \int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{1}{1 + e^{- \tan x}} d x = \int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{e^{\tan x}}{e^{\tan x} + 1} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{1 + e^{\tan x}}{1 + e^{\tan x}} d x \Rightarrow 2 I = \int_{- \frac{π}{3}}^{\frac{π}{3}} d x \Rightarrow 2 I = x_{- \frac{π}{3}}^{\frac{π}{3}} \Rightarrow 2 I = \frac{π}{3} - (- \frac{π}{3}) = \frac{2 π}{3} \Rightarrow I = \frac{π}{3}$

Page No 19.61:

Question 8:

Evaluate each of the following integrals:

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\cos^{2} x}{1 + e^{x}} d x$

Answer:

Let I = $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\cos^{2} x}{1 + e^{x}} d x$ .....(1)
Then,
$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\cos^{2} [\frac{π}{2} + (- \frac{π}{2}) - x]}{1 + e^{[\frac{π}{2} + (- \frac{π}{2}) - x]}} d x [\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\cos^{2} (- x)}{1 + e^{- x}} d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{e^{x} \cos^{2} x}{e^{x} + 1} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{\cos^{2} x}{1 + e^{x}} + \frac{e^{x} \cos^{2} x}{1 + e^{x}}) d x \Rightarrow 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\cos^{2} x (1 + e^{x})}{1 + e^{x}} d x \Rightarrow 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} x d x \Rightarrow 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1 + \cos 2 x}{2}) d x$
$\Rightarrow 2 I = \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} d x + \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos 2 x d x \Rightarrow 2 I = \frac{1}{2} \times x_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{1}{2} \times {\frac{\sin 2 x}{2}}_{- \frac{π}{2}}^{\frac{π}{2}} \Rightarrow 2 I = \frac{1}{2} [\frac{π}{2} - (- \frac{π}{2})] + \frac{1}{4} [sinπ - \sin (- π)] \Rightarrow 2 I = \frac{1}{2} \times π + \frac{1}{4} (0 + 0) [\sin (- π) = - sinπ = 0] \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4}$

Page No 19.61:

Question 9:

Evaluate each of the following integrals:

$\int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x^{11} - 3 x^{9} + 5 x^{7} - x^{5} + 1}{\cos^{2} x} d x$

Answer:

$Let I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x^{11} - 3 x^{9} + 5 x^{7} - x^{5} + 1}{\cos^{2} x} d x = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x^{11} - 3 x^{9} + 5 x^{7} - x^{5}}{\cos^{2} x} d x + \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{\cos^{2} x} d x = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x^{11} - 3 x^{9} + 5 x^{7} - x^{5}}{\cos^{2} x} d x + \int_{- \frac{π}{4}}^{\frac{π}{4}} \sec^{2} x d x = I_{1} + I_{2}$

Now,

Consider $f (x) = \frac{x^{11} - 3 x^{9} + 5 x^{7} - x^{5}}{\cos^{2} x}$ .

$∴ f (- x) = \frac{{(- x)}^{11} - 3 {(- x)}^{9} + 5 {(- x)}^{7} - {(- x)}^{5}}{\cos^{2} (- x)} = \frac{- x^{11} + 3 x^{9} - 5 x^{7} + x^{5}}{\cos^{2} x} = - \frac{x^{11} - 3 x^{9} + 5 x^{7} - x^{5}}{\cos^{2} x} = - f (x)$

$\Rightarrow I_{1} = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{x^{11} - 3 x^{9} + 5 x^{7} - x^{5}}{\cos^{2} x} d x = 0 [\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}]$

Let $g (x) = \sec^{2} x$ .

$∴ g (- x) = \sec^{2} (- x) = \sec^{2} x = g (x)$

$\Rightarrow I_{2} = \int_{- \frac{π}{4}}^{\frac{π}{4}} \sec^{2} x d x = 2 \int_{0}^{\frac{π}{4}} \sec^{2} x d x [\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}] = 2 \times {\tan x}_{0}^{\frac{π}{4}} = 2 (\tan \frac{π}{4} - \tan 0) = 2 \times (1 - 0) = 2$

$∴ I = I_{1} + I_{2} = 0 + 2 = 2$

Page No 19.61:

Question 10:

Evaluate each of the following integrals:

$\int_{a}^{b} \frac{x^{\frac{1}{n}}}{x^{\frac{1}{n}} + {(a + b - x)}^{\frac{1}{n}}} d x, n \in N, n \geq 2$

Answer:

Let I = $\int_{a}^{b} \frac{x^{\frac{1}{n}}}{x^{\frac{1}{n}} + {(a + b - x)}^{\frac{1}{n}}} d x . . . . . (1)$
Then,
$I = \int_{a}^{b} \frac{{(a + b - x)}^{\frac{1}{n}}}{{(a + b - x)}^{\frac{1}{n}} + {[a + b - (a + b - x)]}^{\frac{1}{n}}} d x [\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{a}^{b} \frac{{(a + b - x)}^{\frac{1}{n}}}{{(a + b - x)}^{\frac{1}{n}} + x^{\frac{1}{n}}} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{a}^{b} \frac{x^{\frac{1}{n}} + {(a + b - x)}^{\frac{1}{n}}}{x^{\frac{1}{n}} + {(a + b - x)}^{\frac{1}{n}}} d x \Rightarrow 2 I = \int_{a}^{b} d x \Rightarrow 2 I = x_{a}^{b} = (b - a) \Rightarrow I = \frac{b - a}{2}$

Page No 19.61:

Question 11:

$\int_{0}^{π / 2} (2 \log \cos x - \log \sin 2 x) d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} (2 \log \cos x - \log \sin 2 x) d x = \int_{0}^{\frac{π}{2}} [2 \log \cos x - \log (2 \sin x \cos x)] d x = \int_{0}^{\frac{π}{2}} (2 \log \cos x - \log 2 - \log \sin x - \log \cos x) d x = \int_{0}^{\frac{π}{2}} (\log \cos x - \log 2 - \log \sin x) d x = \int_{0}^{\frac{π}{2}} \log \cos x d x - \int_{0}^{\frac{π}{2}} \log 2 d x - \int_{0}^{\frac{π}{2}} \log \sin x d x = \int_{0}^{\frac{π}{2}} \log \cos x d x - \int_{0}^{\frac{π}{2}} \log 2 d x - \int_{0}^{\frac{π}{2}} \log \sin (\frac{π}{2} - x) d x [Using \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{\frac{π}{2}} \log \cos x d x - \int_{0}^{\frac{π}{2}} \log 2 d x - \int_{0}^{\frac{π}{2}} \log \cos x d x = - \log 2 {[x]}_{0}^{\frac{π}{2}} = - \frac{π}{2} \log 2$

Page No 19.61:

Question 12:

$\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x$

Answer:

$Let I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x . . . (i) \Rightarrow I = \int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{x}} d x [Using, \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \Rightarrow I = \int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{a} \frac{\sqrt{x} + \sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} d x = \int_{0}^{a} d x = {[x]}_{0}^{a} = a Hence I = \frac{a}{2}$

Page No 19.61:

Question 13:

$\int_{0}^{5} \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} d x$

Answer:

$Let I = \int_{0}^{5} \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}} d x . . . (i) I = \int_{0}^{5} \frac{\sqrt[4]{9 - x}}{\sqrt[4]{9 - x} - \sqrt[4]{x + 4}} d x [Using \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] I = - \int_{0}^{5} \frac{\sqrt[4]{9 - x}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{5} \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}} - \frac{\sqrt[4]{9 - x}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}} d x = \int_{0}^{5} \frac{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}}{\sqrt[4]{x + 4} - \sqrt[4]{9 - x}} d x = \int_{0}^{5} d x = {[x]}_{0}^{5} = 5 Hence I = \frac{5}{2}$

Page No 19.61:

Question 14:

$\int_{0}^{7} \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} d x$

Answer:

$Let I = \int_{0}^{7} \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7 - x}} d x . . . (i) = \int_{0}^{7} \frac{\sqrt[3]{7 - x}}{\sqrt[3]{7 - x} + \sqrt[3]{x}} d x (Using \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x) = \int_{0}^{7} \frac{\sqrt[3]{7 - x}}{\sqrt[3]{x} + \sqrt[3]{7 - x}} d x . . . (ii) Adding (i) and (i i) we get 2 I = \int_{0}^{7} \frac{\sqrt[3]{x} + \sqrt[3]{7 - x}}{\sqrt[3]{x} + \sqrt[3]{7 - x}} d x = \int_{0}^{7} d x = {[x]}_{0}^{7} = 7 Hence I = \frac{7}{2}$

Page No 19.61:

Question 15:

$\int_{π / 6}^{π / 3} \frac{1}{1 + \sqrt{\tan x}} d x$

Answer:

$Let I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{\tan x}} d x . . . (i) = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{\tan (\frac{π}{3} + \frac{π}{6} - x)}} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{c o t x}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} (\frac{1}{1 + \sqrt{\tan x}} + \frac{1}{1 + \sqrt{c o t x}}) d x = \int_{\frac{π}{6}}^{\frac{π}{3}} (\frac{1 + \sqrt{c o t x} + 1 + \sqrt{\tan x}}{1 + \sqrt{c o t x} + \sqrt{\tan x} + \sqrt{\tan x c o t x}}) d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{2 + \sqrt{c o t x} + \sqrt{\tan x}}{2 + \sqrt{c o t x} + \sqrt{\tan x}} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} d x = {[x]}_{\frac{π}{6}}^{\frac{π}{3}} = \frac{π}{3} - \frac{π}{6} ∴ 2 I = \frac{π}{6} Hence I = \frac{π}{12}$

Page No 19.61:

Question 16:

If $f (a + b - x) = f (x)$ , then prove that $\int_{a}^{b} x f (x) d x = \frac{a + b}{2} \int_{a}^{b} f (x) d x$ .

Answer:

$\int_{a}^{b} x f (x) d x = \int_{a}^{b} (a + b - x) f (a + b - x) d x [\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{a}^{b} (a + b - x) f (x) d x [f (a + b - x) = f (x)] ∴ \int_{a}^{b} x f (x) d x = \int_{a}^{b} (a + b) f (x) d x - \int_{a}^{b} x f (x) d x$
$\Rightarrow \int_{a}^{b} x f (x) d x + \int_{a}^{b} x f (x) d x = (a + b) \int_{a}^{b} f (x) d x \Rightarrow 2 \int_{a}^{b} x f (x) d x = (a + b) \int_{a}^{b} f (x) d x \Rightarrow \int_{a}^{b} x f (x) d x = \frac{a + b}{2} \int_{a}^{b} f (x) d x$

Page No 19.95:

Question 1:

$\int_{0}^{π / 2} \frac{1}{1 + \tan x}$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan (\frac{π}{2} - x)} d x [Using \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} (\frac{1}{1 + \tan x} + \frac{1}{1 + c o t x}) d x = \int_{0}^{\frac{π}{2}} [\frac{1 + c o t x + 1 + \tan x}{(1 + \tan x) (1 + c o t x)}] d x = \int_{0}^{\frac{π}{2}} \frac{2 + \tan x + c o t x}{1 + \tan x + c o t x + \tan x c o t x} d x = \int_{0}^{\frac{π}{2}} \frac{2 + \tan x + c o t x}{2 + \tan x + c o t x} d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} 2 I = \frac{π}{2} ∴ I = \frac{π}{4}$

Page No 19.95:

Question 2:

$\int_{0}^{π / 2} \frac{1}{1 + \cot x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t (\frac{π}{2} - x)} d x [Using \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t x} + \frac{1}{1 + \tan x} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \tan x + 1 + c o t x}{(1 + c o t x) (1 + \tan x)} d x = \int_{0}^{\frac{π}{2}} \frac{2 + \tan x + c o t x}{1 + \tan x + c o t x + \tan x c o t x} d x = \int_{0}^{\frac{π}{2}} \frac{2 + \tan x + c o t x}{2 + \tan x + c o t x} d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence, I = \frac{π}{4}$

Page No 19.95:

Question 3:

$\int_{0}^{π / 2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{tanx}} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cot (\frac{π}{2} - x)}}{\sqrt{\cot ((\frac{π}{2} - x))} + \sqrt{\tan (\frac{π}{2} - x)}} d x [Using \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) d x] = \int_{0}^{\frac{π}{2}} \frac{\sqrt{tanx}}{\sqrt{tanx} + \sqrt{cotx}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} (\frac{\sqrt{cotx}}{\sqrt{cotx} + \sqrt{tanx}} + \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}}) d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence, I = \frac{π}{4}$

Page No 19.95:

Question 4:

$\int_{0}^{π / 2} \frac{\sin^{3 / 2} x}{\sin^{3 / 2} x + \cos^{3 / 2} x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (\frac{π}{2} - x)}{\sin^{n} (\frac{π}{2} - x) + \cos^{n} (\frac{π}{2} - x)} d x Using \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\cos^{n} x + \sin^{n} x} dx = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x} dx . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} + \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x} dx = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x + \cos^{n} x}{\sin^{n} x + \cos^{n} x} dx = \int_{0}^{\frac{π}{2}} dx = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4} i . e ., \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x = \frac{π}{4} ∴ \int_{0}^{\frac{π}{2}} \frac{\sin^{3 / 2} x}{\sin^{3 / 2} x + \cos^{3 / 2} x} d x = \frac{π}{4}$

Page No 19.95:

Question 5:

$\int_{0}^{π / 2} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (\frac{π}{2} - x)}{\sin^{n} (\frac{π}{2} - x) + \cos^{n} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\cos^{n} x + \sin^{n} x} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} + \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x} d x = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x + \cos^{n} x}{\sin^{n} x + \cos^{n} x} d x = \int_{0}^{\frac{π}{2}} 1 d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 19.95:

Question 6:

$\int_{0}^{π / 2} \frac{1}{1 + \sqrt{\tan x}} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \sqrt{\tan x}} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \sqrt{\tan (\frac{π}{2} - x)}} d x Using \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \sqrt{c o t x}} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \sqrt{\tan x}} + \frac{1}{1 + \sqrt{c o t x}} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \sqrt{c o t x} + 1 + \sqrt{\tan x}}{(1 + \sqrt{\tan x}) (1 + \sqrt{c o t x})} dx = \int_{0}^{\frac{π}{2}} \frac{1 + \sqrt{c o t x} + 1 + \sqrt{\tan x}}{1 + \sqrt{c o t x} + \sqrt{\tan x} + \sqrt{tanx cotx}} dx = \int_{0}^{\frac{π}{2}} \frac{2 + \sqrt{c o t x} + \sqrt{\tan x}}{2 + \sqrt{c o t x} + \sqrt{\tan x}} dx = \int_{0}^{\frac{π}{2}} dx = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 19.95:

Question 7:

$\int_{0}^{a} \frac{1}{x + \sqrt{a^{2} - x^{2}}} d x$

Answer:

$We have, I = \int_{0}^{a} \frac{1}{x + \sqrt{a^{2} - x^{2}}} d x Putting x = a \sin θ \Rightarrow d x = a \cos θ d θ When x \to 0; θ \to 0 And x \to a; θ \to \frac{π}{2} ∴ I = \int_{0}^{\frac{π}{2}} \frac{a \cos θ}{a \sin θ + \sqrt{a^{2} - {(a \sin θ)}^{2}}} d θ = \int_{0}^{\frac{π}{2}} \frac{a \cos θ}{a \sin θ + a \cos θ} d θ I = \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ + \cos θ} d θ . . . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos (\frac{π}{2} - θ)}{\sin (\frac{π}{2} - θ) + \cos (\frac{π}{2} - θ)} d θ = \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\cos θ + \sin θ} d θ I = \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sin θ + \cos θ} d θ . . . . . (2) By adding (1) and (2), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\cos θ + \sin θ}{\sin θ + \cos θ} d θ \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} d θ \Rightarrow 2 I = {[θ]}_{0}^{\frac{π}{2}} \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4}$

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Question 8:

$\int_{0}^{\infty} \frac{\log x}{1 + x^{2}} d x$

Answer:

$We have, I = \int_{0}^{\infty} \frac{\log x}{1 + x^{2}} d x Putting x = \tan θ \Rightarrow d x = \sec^{2} θ d θ When x \to 0; θ \to 0 and x \to \infty; θ \to \frac{π}{2} Now, integral becomes,$

$I = \int_{0}^{\frac{π}{2}} \frac{\log (\tan θ)}{1 + \tan^{2} θ} \sec^{2} θ d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \log (\tan θ) d θ . . . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{2}} \log [\tan (\frac{π}{2} - θ)] d θ [∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \Rightarrow I = \int_{0}^{\frac{π}{2}} \log (\cot θ) d θ . . . . . (2) Adding (1) and (2), we get$

$2 I = \int_{0}^{\frac{π}{2}} \log (\tan θ) d θ + \int_{0}^{\frac{π}{2}} \log (\cot θ) d θ = \int_{0}^{\frac{π}{2}} [\log (\tan θ) + \log (\cot θ)] d θ = \int_{0}^{\frac{π}{2}} [\log (\tan θ \times \cot θ)] d θ = \int_{0}^{\frac{π}{2}} (\log 1) d θ = \int_{0}^{\frac{π}{2}} (0) d θ \Rightarrow 2 I = 0 \Rightarrow I = 0 ∴ \int_{0}^{\infty} \frac{\log x}{1 + x^{2}} d x = 0$

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Question 9:

$\int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} d x$

Answer:

$We have, I = \int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} d x Putting x = \tan θ \Rightarrow d x = \sec^{2} θ d θ When x \to 0; θ \to 0 and x \to 1; θ \to \frac{π}{4} Now, integral becomes$

$I = \int_{0}^{\frac{π}{4}} \frac{\log (1 + \tan θ)}{\sec^{2} θ} \sec^{2} θ d θ \Rightarrow I = \int_{0}^{\frac{π}{4}} [\log (1 + \tan θ)] d θ . . . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{4}} [\log \{1 + \tan (\frac{π}{4} - θ)\}] d θ [∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{\frac{π}{4}} [\log \{1 + \frac{\tan \frac{π}{4} - \tan θ}{1 + \tan \frac{π}{4} \tan θ}\}] d θ = \int_{0}^{\frac{π}{4}} [\log \{1 + \frac{1 - \tan θ}{1 + \tan θ}\}] d θ = \int_{0}^{\frac{π}{4}} [\log \{\frac{2}{1 + \tan θ}\}] d θ I = \int_{0}^{\frac{π}{4}} [\log 2 - \log (1 + \tan θ)] d θ . . . . . (2)$

$Adding (1) and (2), we get 2 I = \int_{0}^{\frac{π}{4}} (\log 2) d θ \Rightarrow 2 I = (\log 2) {[θ]}_{0}^{\frac{π}{4}} \Rightarrow 2 I = \frac{π}{4} \log 2 \Rightarrow I = \frac{π}{8} \log 2 ∴ \int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} d x = \frac{π}{8} \log 2$

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Question 10:

$\int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x$

Answer:

$We have, I = \int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x Putting x = \tan θ \Rightarrow d x = \sec^{2} θ d θ When x \to 0; θ \to 0 and x \to \infty; θ \to \frac{π}{2} Now, integral becomes$

$I = \int_{0}^{\frac{π}{2}} \frac{\tan θ}{(1 + \tan θ) \sec^{2} θ} \sec^{2} θ d θ = \int_{0}^{\frac{π}{2}} \frac{\tan θ}{1 + \tan θ} d θ = \int_{0}^{\frac{π}{2}} \frac{\frac{\sin θ}{co s θ}}{1 + \frac{\sin θ}{\cos θ}} d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sin θ + \cos θ} d θ . . . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - θ)}{\sin (\frac{π}{2} - θ) + \cos (\frac{π}{2} - θ)} d θ [∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\cos θ + \sin θ} d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ + \cos θ} d θ . . . . . (2)$

$Adding (1) and (2), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin θ + \cos θ}{\sin θ + \cos θ} d θ \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} d θ \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4} ∴ \int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x = \frac{π}{4}$

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Question 11:

$\int_{0}^{π} \frac{x \tan x}{\sec x cosec x} d x$

Answer:

$Let I = \int_{0}^{π} \frac{x \tan x}{s e c x \cos e c x} d x . . . (i) = \int_{0}^{π} \frac{(π - x) \tan (π - x)}{s e c (π - x) \cos e c (π - x)} d x [Using \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{π} \frac{- (π - x) \tan x}{- s e c x \cos e c x} d x = \int_{0}^{π} \frac{(π - x) \tan x}{s e c x \cos e c x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{π} \frac{x \tan x}{s e c x \cos e c x} + \frac{(π - x) \tan x}{s e c x \cos e c x} d x = \int_{0}^{π} (x + π - x) \frac{\tan x}{s e c x \cos e c x} d x = \int_{0}^{π} \frac{πtan x}{s e c x \cos e c x} d x = \int_{0}^{π} {πsin}^{2} x d x = π \int_{0}^{π} (1 - \cos^{2} x) d x = π {[x]}_{0}^{π} - \frac{π}{2} \int_{0}^{π} (1 + \cos 2 x) dx = \frac{π}{2} {[x]}_{0}^{π} - \frac{π}{2} {[\frac{\sin 2 x}{2}]}_{0}^{π} = \frac{π^{2}}{2} Hence, I = \frac{π^{2}}{4}$

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Question 12:

$\int_{0}^{π} x \sin x \cos^{4} x d x$

Answer:

$Let I = \int_{0}^{π} x \sin x \cos^{4} x d x . . . (i) = \int_{0}^{π} (π - x) \sin (π - x) \cos^{4} (π - x) d x = \int_{0}^{π} (π - x) \sin x \cos^{4} x d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{π} (x + π - x) \sin x \cos^{4} x d x = π \int_{0}^{π} \sin x \cos^{4} x d x Let \cos x = t, Then - sinx dx = dt, When x = 0, t = 1, x = π, t = - 1 Therefore, 2 I = - π \int_{1}^{- 1} t^{4} dt = π \int_{- 1}^{1} t^{4} dt = π {[\frac{t^{5}}{5}]}_{- 1}^{1} = \frac{π}{5} + \frac{π}{5} = \frac{2 π}{5} Hence I = \frac{π}{5}$

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Question 13:

$\int_{0}^{π} x \sin^{3} x d x$

Answer:

$Let I = \int_{0}^{π} x \sin^{3} x d x . . . (i) = \int_{0}^{π} (π - x) \sin^{3} (π - x) d x = \int_{0}^{π} (π - x) \sin^{3} x d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{π} (x + π - x) \sin^{3} x d x = \int_{0}^{π} π \sin^{3} x d x = \int_{0}^{π} π \frac{3 \sin x - \sin 3 x}{4} d x = \frac{π}{4} \int_{0}^{π} (3 \sin x - \sin 3 x) d x = \frac{π}{4} {[- 3 \cos x + \frac{\cos 3 x}{3}]}_{0}^{π} = \frac{π}{4} [- 3 \cos π + 3 \cos 0 + \frac{\cos 3 π}{3} - \frac{\cos 0}{3}] = \frac{π}{4} [3 + 3 + \frac{- 1}{3} - \frac{1}{3}] = \frac{π}{2} [3 - \frac{1}{3}] = \frac{π}{2} \times \frac{8}{3} = \frac{4 π}{3} ∴ I = \frac{2 π}{3}$

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Question 14:

$\int_{0}^{π} x \log \sin x d x$

Answer:

$\int_{0}^{π} x \log \sin x d x Let I = \int_{0}^{π} x \log (\sin x) d x . . . . . (i) I = \int_{0}^{π} (π - x) \log \sin (π - x) d x I = \int_{0}^{π} (π - x) \log (\sin x) d x . . . . . (ii) Adding (i) and (ii) 2 I = π \int_{0}^{π} \log \sin x d x = 2 π \int_{0}^{\frac{π}{2}} \log \sin x d x I = π \int_{0}^{\frac{π}{2}} \log \sin x d x . . . . . (iii) Let \int_{0}^{\frac{π}{2}} \log \sin x d x = I_{2} I_{2} = \int_{0}^{\frac{π}{2}} \log \sin (\frac{π}{2} - x) d x = \int_{0}^{\frac{π}{2}} \log \cos x d x 2 I_{2} = \int_{0}^{\frac{π}{2}} (\log \sin x + \log \cos x) d x = \int_{0}^{\frac{π}{2}} \log (\sin x \cos x) d x = \int_{0}^{\frac{π}{2}} \log (\sin 2 x) d x - \int_{0}^{\frac{π}{2}} \log 2 d x Let 2 x = t 2 dx = d t when, x = 0 \Rightarrow t = 0 x = 0 \Rightarrow t = π 2 I_{2} = \frac{1}{2} \int_{0}^{π} \log (\sin t) d t - \frac{π}{2} \log 2 2 I_{2} = \frac{2}{2} \int_{0}^{\frac{π}{2}} \log (\sin t) d t - \frac{π}{2} \log 2 2 I_{2} = I_{2} - \frac{π}{2} \log 2 I_{2} = - \frac{π}{2} \log 2 From (iii), I = π \int_{0}^{\frac{π}{2}} \log sinx dx = π I_{2} I = π (- \frac{π}{2} \log 2) I = \frac{- π^{2} \log 2}{2}$

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Question 15:

$\int_{0}^{π} \frac{x \sin x}{1 + \sin x} d x$

Answer:

$Let I = \int_{0}^{π} \frac{x \sin x}{1 + \sin x} d x . . . (i) = \int_{0}^{π} \frac{(π - x) \sin (π - x)}{1 + \sin (π - x)} d x = \int_{0}^{π} \frac{(π - x) \sin x}{1 + \sin x} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{π} (x + π - x) \frac{\sin x}{1 + \sin x} d x = \int_{0}^{π} \frac{π \sin x}{1 + \sin x} d x = π \int_{0}^{π} \frac{1 + sinx - 1}{1 + sinx} dx = π \int_{0}^{π} dx - π \int_{0}^{π} \frac{1}{1 + sinx} dx = π \int_{0}^{π} dx - π \int_{0}^{π} \frac{(1 - sinx)}{(1 + sinx) (1 - sinx)} dx = π \int_{0}^{π} dx - π \int_{0}^{π} \frac{(1 - sinx)}{1 - \sin^{2} x} dx = π \int_{0}^{π} dx - π \int_{0}^{π} \frac{(1 - sinx)}{\cos^{2} x} dx = π \int_{0}^{π} dx - π \int_{0}^{π} (\sec^{2} x - \sec x \tan x) dx = π {[x]}_{0}^{π} - π {[tanx - secx]}_{0}^{π} = π^{2} - π (0 + 1 - 0 + 1) = π^{2} - 2 π Hence I = π (\frac{π}{2} - 1)$

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Question 16:

$\int_{0}^{π} \frac{x}{1 + \cos α \sin x} d x, 0 < α < π$

Answer:

$We have, I = \int_{0}^{π} \frac{x}{1 + \cos α \sin x} d x . . . . . (1) = \int_{0}^{π} \frac{π - x}{1 + \cos α \sin (π - x)} d x = \int_{0}^{π} \frac{π - x}{1 + \cos α \sin x} d x . . . . . (2) Adding (1) and (2) we get, 2 I = \int_{0}^{π} \frac{x + π - x}{1 + \cos α \sin x} d x \Rightarrow I = \frac{π}{2} \int_{0}^{π} \frac{1}{1 + \cos α sinx} d x$

$= \frac{π}{2} \int_{0}^{π} \frac{1}{1 + \cos α sinx} = \frac{π}{2} \int_{0}^{π} \frac{1}{1 + \cos α \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \frac{π}{2} \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} + 2 \cos α \tan \frac{x}{2}} d x = \frac{π}{2} \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} + 2 \cos α \tan \frac{x}{2}} d x$
$Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} x d x = d t When x \to 0; t \to 0 and x \to π; t \to \infty ∴ I = \frac{π}{2} \int_{0}^{\infty} \frac{2}{1 + t^{2} + 2 \cos α t} d t = \frac{π}{2} \int_{0}^{\infty} \frac{2}{{(t + \cos α)}^{2} - \cos^{2} α + 1} d t = π \int_{0}^{\infty} \frac{1}{{(t + \cos α)}^{2} + \sin^{2} α} d t = π {[\frac{1}{\sin α} \tan^{- 1} (\frac{t + \cos α}{\sin α})]}_{0}^{1} = \frac{π}{sinα} [\tan^{- 1} (\infty) - \tan^{- 1} (\cot α)] = \frac{π}{sinα} [\frac{π}{2} - \tan^{- 1} (\tan (\frac{π}{2} - α))] = \frac{πα}{sinα}$

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Question 17:

$\int_{0}^{π} x \cos^{2} x d x$

Answer:

$Let I = \int_{0}^{π} x \cos^{2} x d x . . . (i) = \int_{0}^{π} (π - x) \cos^{2} (π - x) d x = \int_{0}^{π} (π - x) \cos^{2} x d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{π} (x + π - x) \cos^{2} x d x = \int_{0}^{π} π \cos^{2} x d x = π \int_{0}^{π} \frac{1 + \cos 2 x}{2} dx = \frac{π}{2} \int_{0}^{π} (1 + \cos 2 x) dx = \frac{π}{2} {[x + \frac{\sin 2 x}{2}]}_{0}^{π} = \frac{π}{2} (π - 0) Hence I = \frac{π^{2}}{4}$

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Question 18:

Evaluate the following integrals:

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \cot^{\frac{3}{2}} x} d x$

Answer:

Let I = $\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \cot^{\frac{3}{2}} x} d x$ .....(1)
Then,

$I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \cot^{\frac{3}{2}} (\frac{π}{3} + \frac{π}{6} - x)} d x [\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \cot^{\frac{3}{2}} (\frac{π}{2} - x)} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \tan^{\frac{3}{2}} x} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{\cot^{\frac{3}{2}} x}{\cot^{\frac{3}{2}} x + 1} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1 + \cot^{\frac{3}{2}} x}{1 + \cot^{\frac{3}{2}} x} d x \Rightarrow 2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} d x \Rightarrow 2 I = x_{\frac{π}{6}}^{\frac{π}{3}} \Rightarrow 2 I = \frac{π}{3} - \frac{π}{6} = \frac{π}{6} \Rightarrow I = \frac{π}{12}$

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Question 19:

Evaluate the following integrals:

$\int_{0}^{\frac{π}{2}} \frac{\tan^{7} x}{\tan^{7} x + \cot^{7} x} d x$

Answer:

Let I = $\int_{0}^{\frac{π}{2}} \frac{\tan^{7} x}{\tan^{7} x + \cot^{7} x} d x$ .....(1)
Then,
$I = \int_{0}^{\frac{π}{2}} \frac{\tan^{7} (\frac{π}{2} - x)}{\tan^{7} (\frac{π}{2} - x) + \cot^{7} (\frac{π}{2} - x)} d x [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{\frac{π}{2}} \frac{\cot^{7} x}{\cot^{7} x + \tan^{7} x} d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{0}^{\frac{π}{2}} \frac{\tan^{7} x + \cot^{7} x}{\tan^{7} x + \cot^{7} x} d x \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} d x \Rightarrow 2 I = x_{0}^{\frac{π}{2}} \Rightarrow 2 I = \frac{π}{2} - 0 = \frac{π}{2} \Rightarrow I = \frac{π}{4}$

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Question 20:

Evaluate the following integrals:

$\int_{2}^{8} \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}} d x$

Answer:

Let I = $\int_{2}^{8} \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}} d x$ .....(1)
Then,
$I = \int_{2}^{8} \frac{\sqrt{10 - (2 + 8 - x)}}{\sqrt{2 + 8 - x} + \sqrt{10 - (2 + 8 - x)}} d x [\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{2}^{8} \frac{\sqrt{x}}{\sqrt{10 - x} + \sqrt{x}} d x . . . . . (2)$

Adding (1) and (2), we have

$2 I = \int_{2}^{8} \frac{\sqrt{10 - x} + \sqrt{x}}{\sqrt{x} + \sqrt{10 - x}} d x \Rightarrow 2 I = \int_{2}^{8} d x \Rightarrow 2 I = x_{2}^{8} \Rightarrow 2 I = 8 - 2 = 6 \Rightarrow I = 3$

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Question 21:

Evaluate the following integrals:

$\int_{0}^{π} x \sin x \cos^{2} x d x$ [NCERT EXEMPLAR]

Answer:

Let I = $\int_{0}^{π} x \sin x \cos^{2} x d x$ .....(1)
Then,
$I = \int_{0}^{π} (π - x) \sin (π - x) \cos^{2} (π - x) d x [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{π} (π - x) \sin x \cos^{2} x d x . . . . . (2)$

Adding (1) and (2), we have

$2 I = \int_{0}^{π} (π - x + x) \sin x \cos^{2} x d x \Rightarrow 2 I = π \int_{0}^{π} \sin x \cos^{2} x d x \Rightarrow 2 I = - π \int_{0}^{π} \cos^{2} x (- \sin x) d x \Rightarrow 2 I = - π \times {\frac{\cos^{3} x}{3}}_{0}^{π} [\int {[f (x)]}^{n} f' (x) d x = \frac{{[f (x)]}^{n + 1}}{n + 1} + C] \Rightarrow 2 I = - \frac{π}{3} (\cos^{3} π - \cos^{2} 0)$
$\Rightarrow 2 I = - \frac{π}{3} (- 1 - 1) = \frac{2 π}{3} \Rightarrow I = \frac{π}{3}$

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Question 22:

$\int_{0}^{π / 2} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \sin (\frac{π}{2} - x) \cos (\frac{π}{2} - x)}{\sin^{4} (\frac{π}{2} - x) + \cos^{4} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \cos x \sin x}{\cos^{4} x + \sin^{4} x} d x = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{\frac{π}{2}} (x + \frac{π}{2} - x) \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{\sin^{4} x + \cos^{4} x} d x Let \sin^{2} x = t, Then 2 sinx cosx dx = dt When x = 0, t = 0, x = \frac{π}{2}, t = 1 Therefore 2 I = \frac{π}{4} \int_{0}^{1} \frac{dt}{t^{2} + {(1 - t)}^{2}} = \frac{π}{8} \int_{0}^{1} \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{1}{4}} = \frac{π}{8} \times 2 {[t a n^{- 1} (2 t - 1)]}_{0}^{1} = \frac{π}{4} (\frac{π}{4} + \frac{π}{4}) Hence I = \frac{π^{2}}{16}$

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Question 23:

$\int_{- π / 2}^{π / 2} \sin^{3} x d x$

Answer:

$Let I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \sin^{3} x d x = \int_{\frac{- π}{2}}^{\frac{π}{2}} \sin x (1 - \cos^{2} x) d x = \int_{\frac{- π}{2}}^{\frac{π}{2}} \sin x d x - \int_{\frac{- π}{2}}^{\frac{π}{2}} \sin x \cos^{2} x d x = {[- \cos x]}_{\frac{- π}{2}}^{\frac{π}{2}} + {[\frac{\cos^{3} x}{3}]}_{- \frac{π}{2}}^{\frac{π}{2}} = 0 + 0 = 0$

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Question 24:

$\int_{- π / 2}^{π / 2} \sin^{4} x d x$

Answer:

$Let I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{4} x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} {(\sin^{2} x)}^{2} d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} {(\frac{1 - \cos 2 x}{2})}^{2} d x = \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} (1 - 2 \cos 2 x + \cos^{2} 2 x) d x = \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} d x - \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos 2 x d x + \frac{1}{8} \int_{- \frac{π}{2}}^{\frac{π}{2}} (1 + \cos 4 x) d x = \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} d x - \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos 2 x d x + \frac{1}{8} \int_{- \frac{π}{2}}^{\frac{π}{2}} d x + \frac{1}{8} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos 4 x d x = \frac{3}{8} \int_{- \frac{π}{2}}^{\frac{π}{2}} d x - \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos 2 x d x + \frac{1}{8} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos 4 x d x = \frac{3}{8} {[x]}_{- \frac{π}{2}}^{\frac{π}{2}} - \frac{1}{4} {[\sin 2 x]}_{- \frac{π}{2}}^{\frac{π}{2}} + \frac{1}{32} {[\sin 4 x]}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{3}{8} (\frac{π}{2} + \frac{π}{2}) - \frac{1}{4} (0 - 0) + \frac{1}{32} (0 - 0) Hence I = \frac{3 π}{8}$

Page No 19.96:

Question 25:

(i) $\int_{- 1}^{1} \log (\frac{2 - x}{2 + x}) d x$

(ii) $\int_{- π}^{π} (1 - x^{2}) \sin x \cos^{2} x d x$

Answer:

(ii) $\int_{- π}^{π} (1 - x^{2}) \sin x \cos^{2} x d x$

$Let I = \int_{- π}^{π} (1 - x^{2}) \sin x \cos^{2} x d x We know, \int_{- a}^{a} f (x) d x = \{\begin{cases} 0, if f (- x) = - f (x) \\ 2 \int_{0}^{a} f (x) d x, if f (- x) = f (x) \end{cases} Here, f (x) = (1 - x^{2}) \sin x \cos^{2} x f (- x) = (1 - {(- x)}^{2}) \sin (- x) \cos^{2} (- x) = (1 - x^{2}) \sin (- x) \cos^{2} x = - (1 - x^{2}) \sin x \cos^{2} x = - f (x) Thus, \int_{- π}^{π} (1 - x^{2}) \sin x \cos^{2} x d x = 0$

Hence, $\int_{- π}^{π} (1 - x^{2}) \sin x \cos^{2} x d x$ = 0.

Page No 19.96:

Question 26:

$\int_{- π / 4}^{π / 4} \sin^{2} x d x$

Answer:

$Let I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \sin^{2} x d x Here f (x) = \sin^{2} x f (- x) = \sin^{2} (- x) = \sin^{2} x = f (x) Hence \sin^{2} x i s an even function Therefore, I = 2 \int_{0}^{\frac{π}{4}} \sin^{2} x d x = 2 \int_{0}^{\frac{π}{4}} (\frac{1 - \cos 2 x}{2}) dx = \int_{0}^{\frac{π}{4}} (1 - \cos 2 x) dx = {[x - \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{4}} = \frac{π}{4} - \frac{1}{2}$ →

Page No 19.96:

Question 27:

$\int_{0}^{π} \log (1 - \cos x) d x$

Answer:

$Let, I = \int_{0}^{π} \log (1 - \cos x) d x = \int_{0}^{π} \log (2 \sin^{2} \frac{x}{2}) d x = \int_{0}^{π} \log 2 d x + 2 \int_{0}^{π} \log \sin \frac{x}{2} d x Let, t = \frac{x}{2} in the secong integral . then d t = \frac{1}{2} d x When x \to 0; t \to 0 and x \to π; t \to \frac{π}{2} I = \log 2 {[x]}_{0}^{π} + 4 \int_{0}^{\frac{π}{2}} \log \sin t d t = πlog 2 + 4 \times (- \frac{π}{2} \log 2) [Where, \int_{0}^{\frac{π}{2}} \log \sin t dt = - \frac{π}{2} \log 2] = - π \log 2$

Page No 19.96:

Question 28:

$\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$

Answer:

$L e t I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \log (\frac{2 - \sin x}{2 + \sin x}) d x Here, f (x) = l og (\frac{2 - \sin x}{2 + \sin x}) f (- x) = l og (\frac{2 - \sin (- x)}{2 + \sin (- x)}) = l og (\frac{2 + \sin x}{2 - \sin x}) = - l og (\frac{2 - \sin x}{2 + \sin x}) = - f (x) Hence f (x) is an odd function ∴ I = 0$

Page No 19.96:

Question 29:

Evaluate the following integrals:

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

Answer:

Let I = $\int_{- π}^{π} \frac{2 x (1 + \sin x)}{1 + \cos^{2} x} d x$
Then,
$I = \int_{- π}^{π} \frac{2 x}{1 + \cos^{2} x} d x + \int_{- π}^{π} \frac{2 x \sin x}{1 + \cos^{2} x} d x = I_{1} + I_{2}$

Consider $f (x) = \frac{2 x}{1 + \cos^{2} x}$ .
Now,
$f (- x) = \frac{2 (- x)}{1 + \cos^{2} (π - x)} = - \frac{2 x}{1 + {(- \cos x)}^{2}} = - \frac{2 x}{1 + \cos^{2} x} = - f (x)$
$∴ I_{1} = \int_{- π}^{π} \frac{2 x}{1 + \cos^{2} x} d x = 0 [\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}]$

Again, consider $g (x) = \frac{2 x \sin x}{1 + \cos^{2} x}$ .

$g (- x) = \frac{2 (- x) \sin (- x)}{1 + \cos^{2} (- x)} = \frac{2 x \sin x}{1 + \cos^{2} x} = g (x) [\sin (- x) = - \sin x and \cos (- x) = \cos x]$
$∴ I_{2} = \int_{- π}^{π} \frac{2 x \sin x}{1 + \cos^{2} x} d x = 2 \times 2 \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x [\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}] = 4 \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x . . . . . (1)$
Then,
$I_{2} = 4 \int_{0}^{π} \frac{(π - x) \sin (π - x)}{1 + \cos^{2} (π - x)} d x = 4 \int_{0}^{π} \frac{(π - x) \sin x}{1 + \cos^{2} x} d x . . . . . (2) [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$

Adding (1) and (2), we get

$2 I_{2} = 4 \int_{0}^{π} \frac{π \sin x}{1 + \cos^{2} x} d x \Rightarrow 2 I_{2} = 4 π \int_{0}^{π} \frac{\sin x}{1 + \cos^{2} x} d x$

Put cosx = z

$\Rightarrow - \sin x d x = d z$

When $x \to 0, z \to 1$

When $x \to π, z \to - 1$

$∴ 2 I_{2} = - 4 π \int_{1}^{- 1} \frac{d z}{1 + z^{2}} \Rightarrow 2 I_{2} = - 4 π \times {\tan^{- 1} z}_{1}^{- 1} \Rightarrow 2 I_{2} = - 4 π [\tan^{- 1} (- 1) - \tan^{- 1} 1] \Rightarrow 2 I_{2} = - 4 π (- \frac{π}{4} - \frac{π}{4}) = 2 π^{2} \Rightarrow I_{2} = π^{2}$

$∴ I = I_{1} + I_{2} = 0 + π^{2} = π^{2}$

Page No 19.96:

Question 30:

Evaluate the following integrals:

$\int_{- a}^{a} \log (\frac{a - \sin θ}{a + \sin θ}) d θ$

Answer:

Let I = $\int_{- a}^{a} \log (\frac{a - \sin θ}{a + \sin θ}) d θ$

Consider $f (θ) = \log (\frac{a - \sin θ}{a + \sin θ})$ .

$f (- θ) = \log (\frac{a - \sin (- θ)}{a + \sin (- θ)}) = \log (\frac{a + \sin θ}{a - \sin θ}) [\sin (- x) = - \sin x] = \log {(\frac{a - \sin θ}{a + \sin θ})}^{- 1} = - \log (\frac{a - \sin θ}{a + \sin θ}) [\log a^{b} = b \log a] = - f (θ)$

$∴ f (- θ) = - f (θ) \Rightarrow I = \int_{- a}^{a} \log (\frac{a - \sin θ}{a + \sin θ}) d θ = 0 [\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}]$

Page No 19.96:

Question 31:

Evaluate the following integrals:

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

Answer:

Let I = $\int_{- 2}^{2} \frac{3 x^{3} + 2 |x| + 1}{x^{2} + |x| + 1} d x$

$= \int_{- 2}^{2} \frac{3 x^{3}}{x^{2} + |x| + 1} d x + \int_{- 2}^{2} \frac{2 |x| + 1}{x^{2} + |x| + 1} d x = I_{1} + I_{2}$

Consider $f (x) = \int_{- 2}^{2} \frac{3 x^{3}}{x^{2} + |x| + 1} d x$ .

$f (- x) = \int_{- 2}^{2} \frac{3 {(- x)}^{3}}{{(- x)}^{2} + |- x| + 1} d x = \int_{- 2}^{2} \frac{- 3 x^{3}}{x^{2} + |x| + 1} d x = - \int_{- 2}^{2} \frac{3 x^{3}}{x^{2} + |x| + 1} d x = - f (x)$

$∴ I_{1} = \int_{- 2}^{2} \frac{3 x^{3}}{x^{2} + |x| + 1} d x = 0 [\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}]$

Now, consider $g (x) = \int_{- 2}^{2} \frac{2 |x| + 1}{x^{2} + |x| + 1} d x$ .

$g (- x) = \int_{- 2}^{2} \frac{2 |- x| + 1}{{(- x)}^{2} + |- x| + 1} d x = \int_{- 2}^{2} \frac{2 |x| + 1}{x^{2} + |x| + 1} d x = g (x)$

$∴ I_{2} = \int_{- 2}^{2} \frac{2 |x| + 1}{x^{2} + |x| + 1} d x = 2 \int_{0}^{2} \frac{2 |x| + 1}{x^{2} + |x| + 1} d x [\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}] = 2 \int_{0}^{2} \frac{2 x + 1}{x^{2} + x + 1} d x [|x| = \{\begin{cases} x, & x \geq 0 \\ - x, & x < 0 \end{cases}] = 2 \times {\log (x^{2} + x + 1)}_{0}^{2} [\int \frac{f' (x)}{f (x)} d x = \log f (x) + C] = 2 \times (\log 7 - \log 1) = 2 \times (\log 7 - 0) = 2 \log 7$

$∴ I = I_{1} + I_{2} = 0 + 2 \log 7 = 2 \log 7$

Page No 19.96:

Question 32:

Evaluate the following integrals:

$\int_{- \frac{3 π}{2}}^{- \frac{π}{2}} \{\sin^{2} (3 π + x) + {(π + x)}^{3}\} d x$

Answer:

Let I = $\int_{- \frac{3 π}{2}}^{- \frac{π}{2}} \{\sin^{2} (3 π + x) + {(π + x)}^{3}\} d x$

Put $π + x = z$

$\Rightarrow d x = d z$

When $x \to - \frac{3 π}{2}, z \to - \frac{π}{2}$

When $x \to - \frac{π}{2}, z \to \frac{π}{2}$

$∴ I = \int_{- \frac{π}{2}}^{\frac{π}{2}} [\sin^{2} (2 π + z) + z^{3}] d z = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\sin^{2} z + z^{3}) d z [\sin (2 π + θ) = \sin θ] = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 - \cos 2 z}{2} d z + \int_{- \frac{π}{2}}^{\frac{π}{2}} z^{3} d z$
$= \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} d z - \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos 2 z d z + \int_{- \frac{π}{2}}^{\frac{π}{2}} z^{3} d z = \frac{1}{2} \times z_{- \frac{π}{2}}^{\frac{π}{2}} - \frac{1}{2} \times {\frac{\sin 2 z}{2}}_{- \frac{π}{2}}^{\frac{π}{2}} + {\frac{z^{4}}{4}}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{1}{2} [\frac{π}{2} - (- \frac{π}{2})] - \frac{1}{4} [sinπ - \sin (- π)] + \frac{1}{4} (\frac{π^{4}}{16} - \frac{π^{4}}{16})$
$= \frac{1}{2} \times π - \frac{1}{4} (0 + 0) + \frac{1}{4} \times 0 = \frac{π}{2}$

Page No 19.96:

Question 33:

$\int_{0}^{2} x \sqrt{2 - x} d x$

Answer:

$Let I = \int_{0}^{2} x \sqrt{2 - x} d x = \int_{0}^{2} (2 - x) \sqrt{2 - 2 + x} d x = \int_{0}^{2} (2 - x) \sqrt{x} d x = \int_{0}^{2} (2 \sqrt{x} - x \sqrt{x}) d x = \int_{0}^{2} (2 x^{\frac{1}{2}} - x^{\frac{3}{2}}) d x = {[2 \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{\frac{5}{2}}}{\frac{5}{2}}]}_{0}^{2} = {[\frac{4}{3} x^{\frac{3}{2}} - \frac{2}{5} x^{\frac{5}{2}}]}_{0}^{2} = \frac{8 \sqrt{2}}{3} - \frac{8 \sqrt{2}}{5} = \frac{16 \sqrt{2}}{15}$

Page No 19.96:

Question 34:

$\int_{0}^{1} \log (\frac{1}{x} - 1) d x$

Answer:

$Let I = \int_{0}^{1} \log (\frac{1}{x} - 1) d x . . . (i) = \int_{0}^{1} \log (\frac{1}{1 - x} - 1) d x [Using \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] I = \int_{0}^{1} \log (\frac{x}{1 - x}) d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{1} \log (\frac{1 - x}{x}) + \log (\frac{x}{1 - x}) d x = \int_{0}^{1} \log (\frac{1 - x}{x} \times \frac{x}{1 - x}) d x = \int_{0}^{1} \log 1 d x = 0 Hence I = 0$

Page No 19.96:

Question 35:

Evaluate the following integrals:

$\int_{- 1}^{1} |x cosπ x| d x$ [NCERT EXEMPLAR]

Answer:

Let I = $\int_{- 1}^{1} |x cosπ x| d x$

Consider $f (x) = |x cosπ x|$ .

$f (- x) = |(- x) cosπ (- x)| = |- x cosπ x| = |x cosπ x| = f (x)$

$∴ I = \int_{- 1}^{1} |x cosπ x| d x = 2 \int_{0}^{1} |x cosπ x| d x [\int_{- a}^{a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (- x) = f (x) \\ 0, & if f (- x) = - f (x) \end{cases}]$
Now,

$|x cosπ x| = \{\begin{cases} x cosπ x, & if 0 \leq x \leq \frac{1}{2} \\ - x cosπ x, & if \frac{1}{2} < x \leq 1 \end{cases}$

$∴ I = 2 [\int_{0}^{\frac{1}{2}} x cosπ x d x + \int_{\frac{1}{2}}^{1} (- x cosπ x) d x] = 2 [x {\frac{sinπ x}{π}}_{0}^{\frac{1}{2}} - \frac{1}{π} \int_{0}^{\frac{1}{2}} sinπ x d x] - 2 [x {\frac{sinπ x}{π}}_{\frac{1}{2}}^{1} - \frac{1}{π} \int_{\frac{1}{2}}^{1} sinπ x d x] = 2 (\frac{1}{2 π} \sin \frac{π}{2} - 0) - \frac{2}{π} \times {(- \frac{cosπx}{π})}_{0}^{\frac{1}{2}} - 2 (\frac{1}{π} sinπ - \frac{1}{2 π} \sin \frac{π}{2}) + \frac{2}{π} \times {(- \frac{cosπ x}{π})}_{\frac{1}{2}}^{1} = \frac{1}{π} + \frac{2}{π^{2}} (\cos \frac{π}{2} - \cos 0) + \frac{1}{π} - \frac{2}{π^{2}} (cosπ - \cos \frac{π}{2}) = \frac{1}{π} - \frac{2}{π^{2}} + \frac{1}{π} + \frac{2}{π^{2}} = \frac{2}{π}$

Page No 19.96:

Question 36:

Evaluate the following integrals:

$\int_{0}^{π} (\frac{x}{1 + \sin^{2} x} + \cos^{7} x) d x$

Answer:

Let I = $\int_{0}^{π} (\frac{x}{1 + \sin^{2} x} + \cos^{7} x) d x$ .....(1)
Then,

$I = \int_{0}^{π} (\frac{π - x}{1 + \sin^{2} (π - x)} + \cos^{7} (π - x)) d x = \int_{0}^{π} (\frac{π - x}{1 + \sin^{2} x} - \cos^{7} x) d x . . . . . (2)$

Adding (1) and (2), we get

$2 I = \int_{0}^{π} (\frac{x}{1 + \sin^{2} x} + \cos^{7} x + \frac{π - x}{1 + \sin^{2} x} - \cos^{7} x) d x \Rightarrow 2 I = π \int_{0}^{π} \frac{1}{1 + \sin^{2} x} d x$

Dividing the numerator and denominator by cos²x, we get

$2 I = π \int_{0}^{π} \frac{\sec^{2} x}{\sec^{2} x + \tan^{2} x} d x \Rightarrow 2 I = π \int_{0}^{π} \frac{\sec^{2} x}{1 + 2 \tan^{2} x} d x \Rightarrow 2 I = 2 π \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{1 + 2 \tan^{2} x} d x [\int_{0}^{2 a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (2 a - x) = f (x) \\ 0, & if f (2 a - x) = - f (x) \end{cases}]$
Put tanx = z

Then $\sec^{2} x d x = d z$

When $x \to 0, z \to 0$

When $x \to \frac{π}{2}, z \to \infty$

$∴ 2 I = 2 π \int_{0}^{\infty} \frac{d z}{1 + {(\sqrt{2} z)}^{2}} \Rightarrow 2 I = 2 π \times {\frac{\tan^{- 1} \sqrt{2} z}{\sqrt{2}}}_{0}^{\infty} \Rightarrow I = \frac{π}{\sqrt{2}} (\tan^{- 1} \infty - \tan^{- 1} 0) \Rightarrow I = \frac{π}{\sqrt{2}} \times (\frac{π}{2} - 0) \Rightarrow I = \frac{π^{2}}{2 \sqrt{2}}$

Page No 19.96:

Question 37:

Evaluate : $\int_{0}^{π} \frac{x}{1 + \sin α \sin x} d x$

Answer:

$Let I = \int_{0}^{π} \frac{x}{1 + \sin α \sin x} d x \Rightarrow I = \int_{0}^{π} \frac{π - x}{1 + \sin α \sin (π - x)} d x [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \Rightarrow I = \int_{0}^{π} \frac{π}{1 + \sin α \sin x} d x - \int_{0}^{π} \frac{x}{1 + \sin α \sin x} d x \Rightarrow I = \int_{0}^{π} \frac{π}{1 + \sin α \sin x} d x - I \Rightarrow 2 I = \int_{0}^{π} \frac{π}{1 + \sin α \sin x} d x \Rightarrow 2 I = π \int_{0}^{π} \frac{1}{1 + sinα sinx} d x$

$Substituting \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}, we get 2 I = π \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} + sinα \times 2 \tan \frac{x}{2}} d x I = \frac{π}{2} \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} + \sin α \times 2 \tan \frac{x}{2}} d x Let \tan \frac{x}{2} = t, d (\tan \frac{x}{2}) = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t Also, When x \to 0, t \to \tan 0 = 0 When x \to π, t \to \tan \frac{π}{2} = \infty ∴ I = \frac{π}{2} \int_{0}^{\infty} \frac{2 d t}{t^{2} + 2 t \sin α + 1} \Rightarrow I = π \int_{0}^{\infty} \frac{1}{{(t + \sin α)}^{2} + \cos^{2} α} d t \Rightarrow I = \frac{π}{\cos α} {[\tan^{- 1} (\frac{t + \sin α}{\cos α})]}_{0}^{\infty} \Rightarrow I = \frac{π}{\cos α} [\tan^{- 1} \infty - \tan^{- 1} (\tan α)] \Rightarrow I = \frac{π}{\cos α} (\frac{π}{2} - α)$

Page No 19.96:

Question 38:

Evaluate the following integrals:

$\int_{0}^{2 π} \sin^{100} x \cos^{101} x d x$

Answer:

Let I = $\int_{0}^{2 π} \sin^{100} x \cos^{101} x d x$

Suppose $f (x) = \sin^{100} x \cos^{101} x$ .

Now,

$f (2 π - x) = \sin^{100} (2 π - x) \cos^{101} (2 π - x) = {(- \sin x)}^{100} {(\cos x)}^{101} = \sin^{100} x \cos^{101} x = f (x)$

$∴ I = \int_{0}^{2 π} \sin^{100} x \cos^{101} x d x = 2 \int_{0}^{π} \sin^{100} x \cos^{101} x d x [\int_{0}^{2 a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (2 a - x) = f (x) \\ 0, & if f (2 a - x) = - f (x) \end{cases}]$
Again,

$f (π - x) = \sin^{100} (π - x) \cos^{101} (π - x) = {(\sin x)}^{100} {(- \cos x)}^{101} = - \sin^{100} x \cos^{101} x = - f (x)$

$∴ I = 2 \times 0 = 0 [\int_{0}^{2 a} f (x) d x = \{\begin{cases} 2 \int_{0}^{a} f (x) d x, & if f (2 a - x) = f (x) \\ 0, & if f (2 a - x) = - f (x) \end{cases}]$

Page No 19.96:

Question 39:

Evaluate the following integrals:

$\int_{0}^{\frac{π}{2}} \frac{a \sin x + b \sin x}{\sin x + \cos x} d x$

Answer:

Let I = $\int_{0}^{\frac{π}{2}} \frac{a \sin x + b \cos x}{\sin x + \cos x} d x$ .....(1)

Then,
$I = \int_{0}^{\frac{π}{2}} \frac{a \sin (\frac{π}{2} - x) + b \cos (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} d x [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$
$= \int_{0}^{\frac{π}{2}} \frac{a \cos x + b \sin x}{\cos x + \sin x} d x$ .....(2)

Adding (1) and (2), we get

$2 I = \int_{0}^{\frac{π}{2}} (\frac{a \sin x + b \cos x}{\cos x + \sin x} + \frac{a \cos x + b \sin x}{\sin x + \cos x}) d x \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} (\frac{a \sin x + b \cos x + a \cos x + b \sin x}{\sin x + \cos x}) d x \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \frac{(a + b) \sin x + (a + b) \cos x}{\sin x + \cos x} d x \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \frac{(a + b) (\sin x + \cos x)}{\sin x + \cos x} d x$
$\Rightarrow 2 I = \int_{0}^{\frac{π}{2}} (a + b) d x \Rightarrow 2 I = (a + b) \times x_{0}^{\frac{π}{2}} \Rightarrow 2 I = (a + b) \times (\frac{π}{2} - 0) \Rightarrow 2 I = \frac{π}{2} (a + b) \Rightarrow I = \frac{π}{4} (a + b)$

Page No 19.96:

Question 40:

$\int_{0}^{3 / 2} | x cosπ x | d x$

Answer:

Ans

Page No 19.96:

Question 41:

$\int_{0}^{1} | x \sin π x | d x$

Answer:

$For 0 < x < 1, x > 0 and sinπ x > 0 \Rightarrow x sinπ x > 0$
$∴ \int_{0}^{1} |x sinπ x| d x = \int_{0}^{1} x sinπ x d x Let I = \int x sinπ x d x = x \int sinπ x d x - \int (\frac{d}{d x} x \int sinπ x d x) d x = x (\frac{- cosπ x}{π}) - \int (\frac{- cosπ x}{π}) d x$
$= \frac{- x cosπ x}{π} + \frac{sinπ x}{π^{2}}$

Applying the limits, we get

$\int_{0}^{1} |x sinπ x| d x = {(\frac{- x cosπ x}{π} + \frac{sinπ x}{π^{2}})}_{0}^{1} = (\frac{- cosπ}{π} + \frac{sinπ}{π^{2}}) - (0 + 0)$
$= \frac{1}{π} + 0 - 0 = \frac{1}{π}$

Page No 19.96:

Question 42:

Evaluate : $\int_{0}^{3 / 2} |x \sin π x| d x$

Answer:

$For 0 < x < 1, x > 0 and sinπ x > 0 \Rightarrow x sinπ x > 0 For 1 < x < \frac{3}{2}, x > 0 and sinπ x < 0 \Rightarrow x sinπ x < 0$
$∴ \int_{0}^{\frac{3}{2}} |x sinπ x| d x = \int_{0}^{1} x sinπ x d x - \int_{1}^{\frac{3}{2}} x sinπ x d x Let I = \int x sinπ x d x = x \int sinπ x d x - \int (\frac{d}{d x} x \int sinπ x d x) d x = x (\frac{- cosπ x}{π}) - \int (\frac{- cosπ x}{π}) d x$
$= \frac{- x cosπ x}{π} + \frac{sinπ x}{π^{2}}$

Applying the limits, we get

$\int_{0}^{\frac{3}{2}} |x sinπ x| d x = {[\frac{- x cosπ x}{π} + \frac{sinπ x}{π^{2}}]}_{0}^{1} - {[\frac{- x cosπ x}{π} + \frac{sinπ x}{π^{2}}]}_{1}^{\frac{3}{2}} = [(\frac{- cosπ}{π} + \frac{sinπ}{π^{2}}) - (0 + 0)] - [(\frac{- \frac{3}{2} \cos \frac{3 π}{2}}{π} + \frac{\sin \frac{3 π}{2}}{π^{2}}) - (\frac{- cosπ}{π} + \frac{sinπ}{π^{2}})]$
$= [(\frac{1}{π} + 0)] - [(0 - \frac{1}{π^{2}}) - (\frac{1}{π} + 0)] = \frac{1}{π} + \frac{1}{π^{2}} + \frac{1}{π} = \frac{2}{π} + \frac{1}{π^{2}} = \frac{2 π + 1}{π^{2}}$

Page No 19.96:

Question 43:

If f is an integrable function such that f(2a − x) = f(x), then prove that
$\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x$

Answer:

$Let I = \int_{0}^{2 a} f (x) d x By Additive property I = \int_{0}^{a} f (x) d x + \int_{a}^{2 a} f (x) d x Consider the integral \int_{a}^{2 a} f (x) d x Let x = 2 a - t, then d x = - d t When x = a, t = a, x = 2 x, t = 0 Hence \int_{a}^{2 a} f (x) d x = - \int_{a}^{0} f (2 a - t) d t = \int_{0}^{a} f (2 a - t) d t = \int_{0}^{a} f (2 a - x) d x (Changeing the variable) Therefore, I = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (x) d x [Given \int_{0}^{a} f (x) d x = \int_{0}^{a} f (2 a - x) d x] = 2 \int_{0}^{a} f (x) d x$

Hence Proved

Page No 19.96:

Question 44:

If f(2a − x) = −f(x), prove that $\int_{0}^{2 a} f (x) d x = 0 .$

Answer:

$Let I = \int_{0}^{2 a} f (x) d x Using additive property I = \int_{0}^{a} f (x) d x + \int_{a}^{2 a} f (x) d x Consider the integral \int_{a}^{2 a} f (x) d x Let x = 2 a - t, Then dx = - dt When x = a, t = a and x = 2 a, t = 0 Therefore, \int_{a}^{2 a} f (x) d x = - \int_{a}^{0} f (2 a - t) d t = \int_{0}^{a} f (2 a - t) d t = \int_{0}^{a} f (2 a - x) d x (chang ing the variable) We have f (2 a - x) = - f (x) Therefore, I = \int_{0}^{a} f (x) d x - \int_{0}^{a} f (x) d x = 0$

Page No 19.96:

Question 45:

If f is an integrable function, show that
(i) $\int_{- a}^{a} f (x^{2}) d x = 2 \int_{0}^{a} f (x^{2}) d x$

(ii) $\int_{- a}^{a} x f (x^{2}) d x = 0$

Answer:

(i)
$I = \int_{- a}^{a} f (x^{2}) d x Here g (x) = f (x^{2}) \Rightarrow g (- x) = f {(- x)}^{2} = f (x^{2}) = g (x) i . e, g (x) is even Therefore I = 2 \int_{0}^{a} f (x^{2}) d x [Using \int_{- a}^{a} g (x) d x = 2 \int_{0}^{a} g (x) d x when g (x) is even]$

(ii)
$I = \int_{- a}^{a} x f (x^{2}) d x Let g (x) = x f (x^{2}) \Rightarrow g (- x) = (- x) f {(- x)}^{2} = - (x) f (x^{2}) = - g (x) i . e, g (x) is odd Therefore I = 0 [Using \int_{- a}^{a} g (x) d x = 0 when g (x) is odd]$

Page No 19.96:

Question 46:

If f (x) is a continuous function defined on [0, 2a]. Then, prove that
$\int_{0}^{2 a} f (x) d x = \int_{0}^{a} \{f (x) + f (2 a - x)\} d x$

Answer:

$Let I = \int_{0}^{2 a} f (x) d x By Additive property I = \int_{0}^{a} f (x) d x + \int_{a}^{2 a} f (x) d x Consider the integral \int_{a}^{2 a} f (x) d x Let x = 2 a - t, then d x = - d t When x = a . t = a, x = 2 a, t = 0 Hence, \int_{a}^{2 a} f (x) d x = - \int_{a}^{0} f (2 a - t) d t = \int_{0}^{a} f (2 a - t) d t = \int_{0}^{a} f (2 a - x) d x Therefore I = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x = \int_{0}^{a} \{f (x) + f (2 a - x)\} d x Hence, proved .$

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Question 47:

If $f (a + b - x) = f (x)$ , then prove that

$\int_{a}^{b} x f (x) d x = (\frac{a + b}{2}) \int_{a}^{b} f (x) d x$

Answer:

$\int_{a}^{b} x f (x) d x = \int_{a}^{b} (a + b - x) f (a + b - x) d x [\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] \Rightarrow \int_{a}^{b} x f (x) d x = \int_{a}^{b} (a + b - x) f (x) d x [f (a + b - x) = f (x)] \Rightarrow \int_{a}^{b} x f (x) d x = \int_{a}^{b} (a + b) f (x) d x - \int_{a}^{b} x f (x) d x \Rightarrow 2 \int_{a}^{b} x f (x) d x = (a + b) \int_{a}^{b} f (x) d x \Rightarrow \int_{a}^{b} x f (x) d x = (\frac{a + b}{2}) \int_{a}^{b} f (x) d x$

Page No 19.97:

Question 48:

If f(x) is a continuous function defined on [−a, a], then prove that
$\int_{- a}^{a} f (x) d x = \int_{0}^{a} \{f (x) + f (- x)\} d x$

Answer:

$Let I = \int_{- a}^{a} f (x) d x By Additive property I = \int_{- a}^{0} f (x) d x + \int_{0}^{a} f (x) d x Let x = - t, then d x = - d t, When x = - a, t = a, x = 0, t = 0 Hence \int_{- a}^{0} f (x) d x = - \int_{a}^{0} f (- t) d t = \int_{0}^{a} f (- t) d t = \int_{0}^{a} f (- x) d x (Changing the variable) Therefore, I = \int_{0}^{a} f (- x) d x + \int_{0}^{a} f (x) d x = \int_{0}^{a} \{f (x) + f (- x)\} dx Hence, proved .$

Page No 19.97:

Question 49:

Prove that: $\int_{0}^{π} x f (\sin x) d x = \frac{π}{2} \int_{0}^{π} f (\sin x) d x$

Answer:

$\int_{0}^{π} x f (\sin x) d x = \int_{0}^{π} (π - x) f [\sin (π - x)] d x [\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \Rightarrow \int_{0}^{π} x f (\sin x) d x = \int_{0}^{π} (π - x) f (\sin x) d x \Rightarrow \int_{0}^{π} x f (\sin x) d x = π \int_{0}^{π} f (\sin x) d x - \int_{0}^{π} x f (\sin x) d x \Rightarrow 2 \int_{0}^{π} x f (\sin x) d x = π \int_{0}^{π} f (\sin x) d x \Rightarrow \int_{0}^{π} x f (\sin x) d x = \frac{π}{2} \int_{0}^{π} f (\sin x) d x$

Page No 19.97:

Question 50:

Prove that: $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x,$ and hence evaluate $\int_{0}^{1} x^{2} {(1 - x)}^{n} d x$

Answer:

To Prove: $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$

$I = \int_{0}^{a} f (x) d x Let x = a - t \Rightarrow d x = 0 - d t \Rightarrow d x = - d t Also, if x = 0, t = a if x = a, t = 0 Thus, I = \int_{a}^{0} f (a - t) (- d t) = - \int_{a}^{0} f (a - t) d t = - (- \int_{0}^{a} f (a - t) d t) (∵ \int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x) = \int_{0}^{a} f (a - t) d t = \int_{0}^{a} f (a - x) d x (∵ \int_{a}^{b} f (t) d t = \int_{a}^{b} f (x) d x) Hence, \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$

Now,
$Let J = \int_{0}^{1} x^{2} {(1 - x)}^{n} d x = \int_{0}^{1} {(1 - x)}^{2} {(1 - (1 - x))}^{n} d x (∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x) = \int_{0}^{1} (1 + x^{2} - 2 x) {(1 - 1 + x)}^{n} d x = \int_{0}^{1} (1 + x^{2} - 2 x) {(x)}^{n} d x = \int_{0}^{1} (x^{n} + x^{n} x^{2} - 2 x^{n} x) d x = \int_{0}^{1} (x^{n} + x^{n + 2} - 2 x^{n + 1}) d x = {(\frac{x^{n + 1}}{n + 1} + \frac{x^{n + 3}}{n + 3} - 2 \frac{x^{n + 2}}{n + 2})}_{0}^{1} = \frac{1^{n + 1}}{n + 1} + \frac{1^{n + 3}}{n + 3} - 2 \frac{1^{n + 2}}{n + 2} = \frac{1}{n + 1} + \frac{1}{n + 3} - \frac{2}{n + 2}$

Hence, $\int_{0}^{1} x^{2} {(1 - x)}^{n} d x = \frac{1}{n + 1} + \frac{1}{n + 3} - \frac{2}{n + 2} .$

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