NCERT Solutions for Class 12 Science Math Chapter 1 - Integrals

Answer:

The anti derivative of sin 2x is a function of x whose derivative is sin 2x.

It is known that,

Therefore, the anti derivative of

Answer:

The anti derivative of cos 3x is a function of x whose derivative is cos 3x.

It is known that,

Therefore, the anti derivative of .

Answer:

The anti derivative of e^2x is the function of x whose derivative is e^2x.

It is known that,

Therefore, the anti derivative of .

Answer:

The anti derivative of is the function of x whose derivative is .

It is known that,

Therefore, the anti derivative of .

Answer:

The anti derivative of is the function of x whose derivative is .

It is known that,

Therefore, the anti derivative of is .

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

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On dividing, we obtain

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

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Hence, the correct answer is C.

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It is given that,

∴Anti derivative of

∴

Also,

Hence, the correct answer is A.

Answer:

Let = t

∴2x dx = dt

Answer:

Let log |x| = t

∴

Answer:

Let 1 + log x = t

∴

Answer:

sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt

Answer:

Let

∴ 2adx = dt

Answer:

Let ax + b = t

⇒ adx = dt

Answer:

Let

∴ dx = dt

Answer:

Let 1 + 2x² = t

∴ 4xdx = dt

Answer:

Let

∴ (2x + 1)dx = dt

Answer:

Let

∴

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{I}=\int \frac{x}{\left(x+4\right)} dx \\ \mathrm{put} x+4=t \\ \Rightarrow dx=dt \\ \mathrm{Now}, \mathrm{I}=\int \frac{\left(t-4\right)}{\sqrt{t}}dt \\ =\int \left(\sqrt{t}-4t^{-1/2}\right)dt \\ =\frac{2}{3}{t}^{3/2}-4\left(2t^{1/2}\right)+\mathrm{C} \\ =\frac{2}{3}.t.t^{1/2}-8t^{1/2}+\mathrm{C} \\ =\frac{2}{3}\left(x+4\right)\sqrt{x+4}-8\sqrt{x+4}+\mathrm{C} \\ =\frac{2}{3}x\sqrt{x+4}+\frac{8}{3}\sqrt{x+4}-8\sqrt{x+4}+\mathrm{C} \\ =\frac{2}{3}x\sqrt{x+4}-\frac{16}{3}\sqrt{x+4}+\mathrm{C} \\ =\frac{2}{3}\left(\sqrt{x+4}\right)\left(x-8\right)+\mathrm{C} \end{gathered}$$

Answer:

Let

∴

Answer:

Let

∴ 9x² dx = dt

Answer:

Let log x = t

∴

Answer:

Let

∴ −8x dx = dt

Answer:

Let

∴ 2dx = dt

Answer:

Let

∴ 2xdx = dt

Answer:

Let

∴

Answer:

Dividing numerator and denominator by e^x, we obtain

Let

∴

Answer:

Let

∴

Answer:

Let 2x − 3 = t

∴ 2dx = dt

$$\begin{gathered} \Rightarrow \int {\tan }^2\left(2x-3\right)dx \\ = \int \left[{\sec }^2\left(2x-3\right) - 1\right]dx \\ =\int \left[{\sec }^{2}t- 1\right]\frac{dt}{2} \\ = \frac{1}{2}\left[\int {\sec }^{2}t\mathit{ }dt - \int 1dt\right] \\ = \frac{1}{2}\left[\tan t - t + \mathrm{C}\right] \\ = \frac{1}{2}\left[\tan \left(2x-3\right) - \left(2x-3\right) + \mathrm{C}\right] \end{gathered}$$

Answer:

Let 7 − 4x = t

∴ −4dx = dt

Answer:

Let

∴

Answer:

Let

∴

Answer:

Let

∴

Answer:

Let

∴

Answer:

Let sin 2x = t

∴

Answer:

Let

∴ cos x dx = dt

Answer:

Let log sin x = t

Answer:

Let 1 + cos x = t

∴ −sin x dx = dt

Answer:

Let 1 + cos x = t

∴ −sin x dx = dt

Answer:

Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt

Answer:

Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt

Answer:

Answer:

Let 1 + log x = t

∴

Answer:

Let

∴

Answer:

Let x⁴ = t

∴ 4x³ dx = dt

Let

∴

From (1), we obtain

Answer:

Let 

∴

Hence, the correct answer is D.

Answer:

Hence, the correct answer is B.

Answer:

Answer:

It is known that,

Answer:

It is known that,

Answer:

Let

Answer:

Answer:

It is known that,

Answer:

It is known that,

$$\begin{gathered} \sin \mathrm{A} . \sin \mathrm{B} = \frac{1}{2}\left[\cos \left(\mathrm{A}-\mathrm{B}\right)-\cos \left(\mathrm{A}+\mathrm{B}\right)\right] \\ \therefore \int \sin 4x \sin 8x dx \\ =\int \frac{1}{2}\left\{\cos \left(4x-8x\right)-\cos \left(4x+8x\right)\right\}dx \\ =\frac{1}{2}\int \left\{\cos \left(-4x\right)-\cos \left(12x\right)\right\}dx \\ =\frac{1}{2}\int \left\{\cos 4x-\cos 12x\right\}dx \\ =\frac{1}{2}\left(\frac{\sin 4x}{4}-\frac{\sin 12x}{12}\right)+\mathrm{C} \end{gathered}$$

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

From equation (1), we obtain

Answer:

Answer:

Answer:

$\frac{1}{\sin x{\cos }^{3}x}=\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x{\cos }^{3}x}=\frac{\sin x}{{\cos }^{3}x}+\frac{1}{\sin x\cos x}$
$\Rightarrow \frac{1}{\sin x{\cos }^{3}x}=\tan x{\sec }^{2}x+\frac{{\displaystyle \frac{1}{{\cos }^{2}x}}}{{\displaystyle \frac{\sin x\cos x}{{\cos }^{2}x}}}=\tan x{\sec }^{2}x+\frac{{\sec }^{2}x}{\tan x}$

Answer:

Answer:

It is known that,

Substituting in equation (1), we obtain

Answer:

Answer:

Hence, the correct answer is A.

Answer:

Let e^xx = t

Hence, the correct answer is B.

Answer:

Let x³ = t

∴ 3x² dx = dt

Answer:

Let 2x = t

∴ 2dx = dt

Answer:

Let 2 − x = t

⇒ −dx = dt

Answer:

Let 5x = t

∴ 5dx = dt

Answer:

Answer:

Let x³ = t

∴ 3x² dx = dt

Answer:

From (1), we obtain

Answer:

Let x³ = t

⇒ 3x² dx = dt

Answer:

Let tan x = t

∴ sec²x dx = dt

Answer:

Answer:

$\int \frac{1}{9x^2+6x+5}dx=\int \frac{1}{{\left(3x+1\right)}^2+2^2}dx$
$\mathrm{Let} (3x+1)=t$
∴ $3 dx=dt$
$\Rightarrow \int \frac{1}{{\left(3x+1\right)}^2+2^2}dx=\frac{1}{3}\int \frac{1}{t^2+2^2}dt$
                                 $=\frac{1}{3\times 2}{\tan }^{-1}\frac{t}{2}+C$
                                $=\frac{1}{6}{\tan }^{-1}\left(\frac{3x+1}{2}\right)+C$
 

Answer:

Answer:

Answer:

Answer:

Answer:

Equating the coefficients of x and constant term on both sides, we obtain

4A = 4 ⇒ A = 1

A + B = 1 ⇒ B = 0

Let 2x² + x − 3 = t

∴ (4x + 1) dx = dt

Answer:

Equating the coefficients of x and constant term on both sides, we obtain

From (1), we obtain

From equation (2), we obtain

Answer:

Equating the coefficient of x and constant term on both sides, we obtain

Substituting equations (2) and (3) in equation (1), we obtain

Answer:

Equating the coefficients of x and constant term, we obtain

2A = 6 ⇒ A = 3

−9A + B = 7 ⇒ B = 34

∴ 6x + 7 = 3 (2x − 9) + 34

Substituting equations (2) and (3) in (1), we obtain

Answer:

Equating the coefficients of x and constant term on both sides, we obtain

Using equations (2) and (3) in (1), we obtain

Answer:

Let x² + 2x +3 = t

⇒ (2x + 2) dx =dt

Using equations (2) and (3) in (1), we obtain

Answer:

Equating the coefficients of x and constant term on both sides, we obtain

Substituting (2) and (3) in (1), we obtain

Answer:

Equating the coefficients of x and constant term, we obtain

Using equations (2) and (3) in (1), we obtain

Answer:

Hence, the correct answer is B.

Answer:

Hence, the correct answer is B.

Answer:

Let

Equating the coefficients of x and constant term, we obtain

A + B = 1

2A + B = 0

On solving, we obtain

A = −1 and B = 2

Answer:

Let

Equating the coefficients of x and constant term, we obtain

A + B = 0

−3A + 3B = 1

On solving, we obtain

Answer:

Let

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

A = 1, B = −5, and C = 4

Answer:

Let

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

Answer:

Let

Substituting x = −1 and −2 in equation (1), we obtain

A = −2 and B = 4

Answer:

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (1 − x²) by x(1 − 2x), we obtain

Let

Substituting x = 0 and in equation (1), we obtain

A = 2 and B = 3

Substituting in equation (1), we obtain

Answer:

Let

Equating the coefficients of x², x, and constant term, we obtain

A + C = 0

−A + B = 1

−B + C = 0

On solving these equations, we obtain

From equation (1), we obtain

Answer:

Let

Substituting x = 1, we obtain

Equating the coefficients of x² and constant term, we obtain

A + C = 0

−2A + 2B + C = 0

On solving, we obtain

Answer:

Let

Substituting x = 1 in equation (1), we obtain

B = 4

Equating the coefficients of x² and x, we obtain

A + C = 0

B − 2C = 3

On solving, we obtain

Answer:

Let

Equating the coefficients of x² and x, we obtain

Answer:

Let

Substituting x = −1, −2, and 2 respectively in equation (1), we obtain

Answer:

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (x³ + x + 1) by x² − 1, we obtain

Let

Substituting x = 1 and −1 in equation (1), we obtain

Answer:

Equating the coefficient of x², x, and constant term, we obtain

A − B = 0

B − C = 0

A + C = 2

On solving these equations, we obtain

A = 1, B = 1, and C = 1

Answer:

Equating the coefficient of x and constant term, we obtain

A = 3

2A + B = −1 ⇒ B = −7

Answer:

Equating the coefficient of x³, x², x, and constant term, we obtain

On solving these equations, we obtain

Answer:

Multiplying numerator and denominator by x^{n − 1}, we obtain

Substituting t = 0, −1 in equation (1), we obtain

A = 1 and B = −1

Answer:

Substituting t = 2 and then t = 1 in equation (1), we obtain

A = 1 and B = −1

Answer:

Equating the coefficients of x³, x², x, and constant term, we obtain

A + C = 0

B + D = 4

4A + 3C = 0

4B + 3D = 10

On solving these equations, we obtain

A = 0, B = −2, C = 0, and D = 6

Answer:

Let x² = t ⇒ 2x dx = dt

Substituting t = −3 and t = −1 in equation (1), we obtain

Answer:

Multiplying numerator and denominator by x³, we obtain

Let x⁴ = t ⇒ 4x³dx = dt

Substituting t = 0 and 1 in (1), we obtain

A = −1 and B = 1

Answer:

Let e^x = t ⇒ e^x dx = dt

Substituting t = 1 and t = 0 in equation (1), we obtain

A = −1 and B = 1

Answer:

Substituting x = 1 and 2 in (1), we obtain

A = −1 and B = 2

Hence, the correct answer is B.

Answer:

Equating the coefficients of x², x, and constant term, we obtain

A + B = 0

C = 0

A = 1

On solving these equations, we obtain

A = 1, B = −1, and C = 0

Hence, the correct answer is A.

Answer:

Let I =

Taking x as first function and sin x as second function and integrating by parts, we obtain

Answer:

Let I =

Taking x as first function and sin 3x as second function and integrating by parts, we obtain

Answer:

Let

Taking x² as first function and e^x as second function and integrating by parts, we obtain

Again integrating by parts, we obtain

Answer:

Let

Taking log x as first function and x as second function and integrating by parts, we obtain

Answer:

Let

Taking log 2x as first function and x as second function and integrating by parts, we obtain

Answer:

Let

Taking log x as first function and x² as second function and integrating by parts, we obtain

Answer:

Let

Taking as first function and x as second function and integrating by parts, we obtain

Answer:

Let

Taking as first function and x as second function and integrating by parts, we obtain

Answer:

Let

Taking cos⁻¹ x as first function and x as second function and integrating by parts, we obtain

Answer:

Let

Taking as first function and 1 as second function and integrating by parts, we obtain

Answer:

Let

Taking as first function and as second function and integrating by parts, we obtain

Answer:

Let

Taking x as first function and sec²x as second function and integrating by parts, we obtain

Answer:

Let

Taking as first function and 1 as second function and integrating by parts, we obtain

Answer:

Taking as first function and x as second function and integrating by parts, we obtain

$$\begin{gathered} I={\left(\log x \right)}^2\int xdx-\int \left[\left\{\frac{d}{dx}{\left(\log x \right)}^2\right\}\int xdx\right]dx \\ =\frac{x^2}{2}{\left(\log x \right)}^2-\left[\int 2\log x .\frac{1}{x}.\frac{x^2}{2}dx\right] \\ =\frac{x^2}{2}{\left(\log x \right)}^2-\int x\log x dx \end{gathered}$$

Again integrating by parts, we obtain

$$\begin{gathered} I\hspace{0.17em}= \frac{x^2}{2}{\left(\log x \right)}^2-\left[\log x \int x dx-\int \left\{\left(\frac{d}{dx}\log x \right)\int x dx\right\}dx\right] \\ =\frac{x^2}{2}{\left(\log x \right)}^2-\left[\frac{x^2}{2}\log x -\int \frac{1}{x}.\frac{x^2}{2}dx\right] \\ =\frac{x^2}{2}{\left(\log x \right)}^2-\frac{x^2}{2}\log x +\frac{1}{2}\int x dx \\ =\frac{x^2}{2}{\left(\log x \right)}^2-\frac{x^2}{2}\log x +\frac{x^2}{4}+C \end{gathered}$$

Answer:

Let

Let I = I₁ + I₂ … (1)

Where, and

Taking log x as first function and x² as second function and integrating by parts, we obtain

Taking log x as first function and 1 as second function and integrating by parts, we obtain

Using equations (2) and (3) in (1), we obtain

Answer:

Let

Let

⇒

∴

It is known that,

Answer:

Let

Let ⇒

It is known that,

Answer:

Let ⇒

It is known that,

From equation (1), we obtain

Answer:

Also, let ⇒

It is known that,

Answer:

Let ⇒

It is known that,

Answer:

Let

Integrating by parts, we obtain

Again integrating by parts, we obtain

Answer:

Let ⇒

= 2θ

⇒

Integrating by parts, we obtain

Answer:

Let

Also, let ⇒

Hence, the correct answer is A.

Answer:

Let

Also, let ⇒

It is known that,

Hence, the correct answer is B.

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

Hence, the correct answer is A.

Answer:

Hence, the correct answer is D.

Answer:

It is known that,

Answer:

It is known that,

Answer:

It is known that,

Answer:

It is known that,

From equations (2) and (3), we obtain

Answer:

It is known that,

Answer:

It is known that,

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

Let

Equating the coefficients of x and constant term, we obtain

A = 10 and B = −25

Substituting the value of I₁ in (1), we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Answer:

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is D.

Answer:

By second fundamental theorem of calculus, we obtain

Hence, the correct answer is C.

Answer:

When x = 0, t = 1 and when x = 1, t = 2

Answer:

Also, let

Answer:

Also, let x = tanθ ⇒ dx = sec²θ dθ

When x = 0, θ = 0 and when x = 1,

Takingθas first function and sec²θ as second function and integrating by parts, we obtain

Answer:

Let x + 2 = t² ⇒ dx = 2tdt

When x = 0, and when x = 2, t = 2

Answer:

Let cos x = t ⇒ −sinx dx = dt

When x = 0, t = 1 and when

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Let ⇒ dx = dt

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Let x + 1 = t ⇒ dx = dt

When x = −1, t = 0 and when x = 1, t = 2

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Let 2x = t ⇒ 2dx = dt

When x = 1, t = 2 and when x = 2, t = 4

Answer:

Let cot θ = t  ⇒ −cosec² θ dθ= dt

Hence, the correct answer is A.

Answer:

Integrating by parts, we obtain

Hence, the correct answer is B.

Answer:

Adding (1) and (2), we obtain

Answer:

Adding (1) and (2), we obtain

Answer:

Adding (1) and (2), we obtain

Answer:

Adding (1) and (2), we obtain

Answer:

It can be seen that (x + 2) ≤ 0 on [−5, −2] and (x + 2) ≥ 0 on [−2, 5].

Answer:

It can be seen that (x − 5) ≤ 0 on [2, 5] and (x − 5) ≥ 0 on [5, 8].

Answer:

Answer:

Answer:

Answer:

Adding (1) and (2), we obtain

Answer:

As sin² (−x) = (sin (−x))² = (−sin x)² = sin²x, therefore, sin²x is an even function.

It is known that if f(x) is an even function, then

Answer:

Adding (1) and (2), we obtain

Answer:

As sin⁷ (−x) = (sin (−x))⁷ = (−sin x)⁷ = −sin⁷x, therefore, sin²x is an odd function.

It is known that, if f(x) is an odd function, then

Answer:

It is known that,

Answer:

Adding (1) and (2), we obtain

Answer:

Adding (1) and (2), we obtain

sin (π − x) = sin x

Adding (4) and (5), we obtain

Let 2x = t ⇒ 2dx = dt

When x = 0, t = 0
and when $x=\frac{\mathrm{\pi }}{2}, t=\mathrm{\pi }$
∴ $I=\frac{1}{2}{\int }_0^{\pi }\log \sin tdt-\frac{\mathrm{\pi }}{2}\log 2$
$\Rightarrow I=\frac{I}{2}-\frac{\mathrm{\pi }}{2}\log 2 [\mathrm{from} 3]$
$\Rightarrow \frac{I}{2}=-\frac{\mathrm{\pi }}{2}\log 2$
$\Rightarrow I=-\mathrm{\pi log} 2$

Answer:

It is known that,

Adding (1) and (2), we obtain

Answer:

It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4

Answer:

Adding (1) and (2), we obtain

Answer:

It is known that if f(x) is an even function, then and

if f(x) is an odd function, then

Hence, the correct answer is C.

Answer:

Adding (1) and (2), we obtain

Hence, the correct answer is C.

Answer:

Equating the coefficients of x², x, and constant term, we obtain

−A + B − C = 0

B + C = 0

A = 1

On solving these equations, we obtain

From equation (1), we obtain

Answer:

Answer:

Answer:

Answer:

On dividing, we obtain

Answer:

Equating the coefficients of x², x, and constant term, we obtain

A + B = 0

B + C = 5

9A + C = 0

On solving these equations, we obtain

From equation (1), we obtain

Answer:

Let x − a = t ⇒ dx = dt

Answer:

Answer:

Let sin x = t ⇒ cos x dx = dt

Answer:

Answer:

Answer:

Let x⁴ = t ⇒ 4x³ dx = dt

Answer:

Let e^x = t ⇒ e^x dx = dt

Answer:

Equating the coefficients of x³, x², x, and constant term, we obtain

A + C = 0

B + D = 0

4A + C = 0

4B + D = 1

On solving these equations, we obtain

From equation (1), we obtain

Answer:

= cos³ x × sin x

Let cos x = t ⇒ −sin x dx = dt

$$\begin{gathered} \Rightarrow \int {\cos }^{3}x e^{\log \sin x }dx =\int {\cos }^{3}x \sin x dx \\ =-\int t^{3}dt \\ =-\frac{t^4}{4}+ c \\ =-\frac{{\cos }^{4}x}{4}+ c \end{gathered}$$

Answer:

Answer:

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{I} =\int \frac{1}{\sqrt{{\sin }^{3}x . \sin \left(x + \alpha \right)}} dx \\ =\int \frac{1}{\sqrt{{\sin }^{3}x \left(\sin x . \cos \alpha + \cos x . \sin \alpha \right)}} dx \\ = \int \frac{1}{\sqrt{{\sin }^{4}x . \cos \alpha + {\sin }^{4}x \cot x . \sin \alpha }} dx \\ = \int \frac{1}{{\sin }^{2}x\sqrt{\cos \alpha + \sin \alpha . \cot x}} dx \\ = \int \frac{{\mathrm{cosec}}^{2}x}{\sqrt{\cos \alpha + \sin \alpha \cot x}} dx \\ \mathrm{put} \cos \alpha + \sin \alpha \cot x = t \\ \Rightarrow \left(0 - \sin \alpha {\mathrm{cosec}}^{2}x\right) dx = dt \\ \Rightarrow {\mathrm{cosec}}^{2}x dx = -\frac{1}{\sin \alpha } dt \\ \mathrm{so}, \\ \mathrm{I} = -\frac{1}{\sin \alpha } \int \frac{dt}{\sqrt{t}} \\ = -\frac{1}{\sin \alpha } \times 2\sqrt{t} + \mathrm{C} \\ = -\frac{2}{\sin \alpha } \sqrt{\cos \alpha + \sin \alpha \cot x} + \mathrm{C} \\ = -\frac{2}{\sin \alpha } \sqrt{\cos \alpha + \sin \alpha . \frac{\cos x}{\sin x}} + \mathrm{C} \\ = -\frac{2}{\sin \alpha } \sqrt{\frac{\sin x . \cos \alpha + \cos x . \sin \alpha }{\sin x}} + \mathrm{C} \\ = -\frac{2}{\sin \alpha } \sqrt{\frac{\sin \left(x + \alpha \right)}{\sin x}} + \mathrm{C} \\ \mathrm{so}, \\ \int \frac{1}{\sqrt{{\sin }^{3}x . \sin \left(x\mathit{ }+ \alpha \right)}} dx = -\frac{2}{\sin \alpha } \sqrt{\frac{\sin \left(x + \alpha \right)}{\sin x}} + \mathrm{C} \end{gathered}$$

Answer:

$\mathrm{Let} I=\int \frac{{\sin }^{-1}\sqrt{x}-{\cos }^{-1}\sqrt{x}}{{\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}}dx$
$\mathrm{It} \mathrm{is} \mathrm{known} \mathrm{that}, {\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}=\frac{\mathrm{\pi }}{2}$
$\Rightarrow I=\int \frac{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\cos }^{-1}\sqrt{x}\right)-{\cos }^{-1}\sqrt{x}}{{\displaystyle \frac{\mathrm{\pi }}{2}}}dx$
     $=\frac{2}{\mathrm{\pi }}\int \left(\frac{\mathrm{\pi }}{2}-2{\cos }^{-1}\sqrt{x}\right)dx$
     $=\frac{2}{\mathrm{\pi }}.\frac{\mathrm{\pi }}{2}\int 1.dx-\frac{4}{\mathrm{\pi }}\int {\cos }^{-1}\sqrt{x}dx$
     $=x-\frac{4}{\mathrm{\pi }}\int {\cos }^{-1}\sqrt{x}dx ...\left(1\right)$
$\mathrm{Let} I_1=\int {\cos }^{-1}\sqrt{x} dx$
$\mathrm{Also}, \mathrm{let} \sqrt{x}=t\Rightarrow dx=2 t dt$
$\Rightarrow I_1=2\int {\cos }^{-1}t.t dt$
$=2\left[{\cos }^{-1}t.\frac{t^2}{2}-\int \frac{-1}{\sqrt{1-t^2}}.\frac{t^2}{2}dt\right]$
$=t^2{\cos }^{-1}t+\int \frac{t^2}{\sqrt{1-t^2}}dt$
$=t^2{\cos }^{-1}t-\int \frac{1-t^2-1}{\sqrt{1-t^2}}dt$
$=t^2{\cos }^{-1}t-\int \sqrt{1-t^2}dt+\int \frac{1}{\sqrt{1-t^2}}dt$
$=t^2{\cos }^{-1}t-\frac{t}{2}\sqrt{1-t^2}-\frac{1}{2}{\sin }^{-1}t+{\sin }^{-1}t$
$=t^2{\cos }^{-1}t-\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t$

From equation (1), we obtain

$$\begin{gathered} I=x-\frac{4}{\mathrm{\pi }}\left[t^2{\cos }^{-1}t-\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t\right] \\ =x-\frac{4}{\mathrm{\pi }}\left[x{\cos }^{-1}\sqrt{x}-\frac{\sqrt{x}}{2}\sqrt{1-x}+\frac{1}{2}{\sin }^{-1}\sqrt{x}\right] \end{gathered}$$
  $=x-\frac{4}{\mathrm{\pi }}\left[x\left(\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\left(\sqrt{x}\right)\right)-\frac{\sqrt{x-x^2}}{2}+\frac{1}{2}{\sin }^{-1}\left(\sqrt{x}\right)\right]$
 

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Equating the coefficients of x², x,and constant term, we obtain

A + C = 1

3A + B + 2C = 1

2A + 2B + C = 1

On solving these equations, we obtain

A = −2, B = 1, and C = 3

From equation (1), we obtain

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Integrating by parts, we obtain

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When x = 0, t = 0 and

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When and when

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When and when

As , therefore, is an even function.

It is known that if f(x) is an even function, then

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From equation (1), we obtain

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Adding (1) and (2), we obtain

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From equations (1), (2), (3), and (4), we obtain

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Equating the coefficients of x², x, and constant term, we obtain

A + C = 0

A + B = 0

B = 1

On solving these equations, we obtain

A = −1, C = 1, and B = 1

Hence, the given result is proved.

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Integrating by parts, we obtain

Hence, the given result is proved.

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Therefore, f (x) is an odd function.

It is known that if f(x) is an odd function, then

Hence, the given result is proved.

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Hence, the given result is proved.

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Hence, the given result is proved.

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Integrating by parts, we obtain

Let 1 − x² = t ⇒ −2x dx = dt

Hence, the given result is proved.

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It is known that,