Class 12-NCERT Solutions-Chapter-07-Integrals-Ex 7.3

Solution:

We know the identity \( \sin^2 A = \frac{1 - \cos 2A}{2} \).

\[ \begin{aligned} \int \sin^2(2x+5) \, dx &= \int \frac{1 - \cos 2(2x+5)}{2} \, dx \\ &= \frac{1}{2} \int (1 - \cos(4x+10)) \, dx \\ &= \frac{1}{2} \left( x - \frac{\sin(4x+10)}{4} \right) + C \\ &= \frac{x}{2} - \frac{1}{8} \sin(4x+10) + C \end{aligned} \]

Solution:

Use the identity \( 2\sin A \cos B = \sin(A+B) + \sin(A-B) \).

\[ \begin{aligned} \int \sin 3x \cos 4x \, dx &= \frac{1}{2} \int 2\sin 3x \cos 4x \, dx \\ &= \frac{1}{2} \int (\sin(3x+4x) + \sin(3x-4x)) \, dx \\ &= \frac{1}{2} \int (\sin 7x + \sin(-x)) \, dx \\ &= \frac{1}{2} \int (\sin 7x - \sin x) \, dx \\ &= \frac{1}{2} \left( \frac{-\cos 7x}{7} - (-\cos x) \right) + C \\ &= -\frac{\cos 7x}{14} + \frac{\cos x}{2} + C \end{aligned} \]

Solution:

Use \( 2\cos A \cos B = \cos(A+B) + \cos(A-B) \).

\[ \begin{aligned} I &= \int \cos 2x \cos 4x \cos 6x \, dx \\ &= \frac{1}{2} \int (2\cos 2x \cos 4x) \cos 6x \, dx \\ &= \frac{1}{2} \int (\cos 6x + \cos 2x) \cos 6x \, dx \\ &= \frac{1}{2} \int (\cos^2 6x + \cos 2x \cos 6x) \, dx \\ &= \frac{1}{2} \int \left( \frac{1+\cos 12x}{2} + \frac{1}{2}(2\cos 2x \cos 6x) \right) \, dx \\ &= \frac{1}{4} \int (1 + \cos 12x + \cos 8x + \cos 4x) \, dx \\ &= \frac{1}{4} \left( x + \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + \frac{\sin 4x}{4} \right) + C \end{aligned} \]

Solution:

Use \( \sin^3 \theta = \frac{3\sin \theta - \sin 3\theta}{4} \) or substitution.

Using substitution method: \( \sin^3(2x+1) = \sin^2(2x+1) \sin(2x+1) = (1-\cos^2(2x+1))\sin(2x+1) \).

Let \( \cos(2x+1) = t \Rightarrow -2\sin(2x+1)dx = dt \Rightarrow \sin(2x+1)dx = -\frac{1}{2}dt \).

\[ \begin{aligned} I &= \int (1-t^2) \left( -\frac{1}{2} \right) dt \\ &= -\frac{1}{2} (t - \frac{t^3}{3}) + C \\ &= -\frac{1}{2} \cos(2x+1) + \frac{1}{6} \cos^3(2x+1) + C \end{aligned} \]

Solution:

\[ \begin{aligned} I &= \int \sin^3 x \cos^3 x \, dx \\ &= \int \sin^3 x \cos^2 x \cos x \, dx \\ &= \int \sin^3 x (1-\sin^2 x) \cos x \, dx \end{aligned} \]

Let \( \sin x = t \Rightarrow \cos x \, dx = dt \).

\[ \begin{aligned} I &= \int t^3 (1-t^2) \, dt \\ &= \int (t^3 - t^5) \, dt \\ &= \frac{t^4}{4} - \frac{t^6}{6} + C \\ &= \frac{1}{4}\sin^4 x - \frac{1}{6}\sin^6 x + C \end{aligned} \]

Solution:

\[ \begin{aligned} I &= \frac{1}{2} \int (2\sin x \sin 2x) \sin 3x \, dx \\ &= \frac{1}{2} \int (\cos x - \cos 3x) \sin 3x \, dx \\ &= \frac{1}{2} \int (\cos x \sin 3x - \cos 3x \sin 3x) \, dx \\ &= \frac{1}{4} \int (2\sin 3x \cos x - 2\sin 3x \cos 3x) \, dx \\ &= \frac{1}{4} \int (\sin 4x + \sin 2x - \sin 6x) \, dx \\ &= \frac{1}{4} \left( -\frac{\cos 4x}{4} - \frac{\cos 2x}{2} + \frac{\cos 6x}{6} \right) + C \end{aligned} \]

Solution:

Use \( 2\sin A \sin B = \cos(A-B) - \cos(A+B) \).

\[ \begin{aligned} I &= \frac{1}{2} \int 2\sin 4x \sin 8x \, dx \\ &= \frac{1}{2} \int (\cos(4x-8x) - \cos(4x+8x)) \, dx \\ &= \frac{1}{2} \int (\cos 4x - \cos 12x) \, dx \\ &= \frac{1}{2} \left( \frac{\sin 4x}{4} - \frac{\sin 12x}{12} \right) + C \end{aligned} \]

Solution:

Use identities \( 1-\cos x = 2\sin^2 \frac{x}{2} \) and \( 1+\cos x = 2\cos^2 \frac{x}{2} \).

\[ \begin{aligned} I &= \int \frac{2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} \, dx \\ &= \int \tan^2 \frac{x}{2} \, dx \\ &= \int (\sec^2 \frac{x}{2} - 1) \, dx \\ &= \frac{\tan \frac{x}{2}}{1/2} - x + C \\ &= 2\tan \frac{x}{2} - x + C \end{aligned} \]

Solution:

\[ \begin{aligned} I &= \int \frac{(1+\cos x) - 1}{1+\cos x} \, dx \\ &= \int \left( 1 - \frac{1}{1+\cos x} \right) \, dx \\ &= \int \left( 1 - \frac{1}{2\cos^2 \frac{x}{2}} \right) \, dx \\ &= \int \left( 1 - \frac{1}{2}\sec^2 \frac{x}{2} \right) \, dx \\ &= x - \frac{1}{2} \frac{\tan \frac{x}{2}}{1/2} + C \\ &= x - \tan \frac{x}{2} + C \end{aligned} \]

Solution:

\[ \begin{aligned} \sin^4 x &= (\sin^2 x)^2 = \left( \frac{1-\cos 2x}{2} \right)^2 \\ &= \frac{1}{4} (1 - 2\cos 2x + \cos^2 2x) \\ &= \frac{1}{4} \left( 1 - 2\cos 2x + \frac{1+\cos 4x}{2} \right) \\ &= \frac{1}{8} (2 - 4\cos 2x + 1 + \cos 4x) \\ &= \frac{1}{8} (3 - 4\cos 2x + \cos 4x) \end{aligned} \]

Integrating:

\[ I = \frac{1}{8} \left( 3x - 2\sin 2x + \frac{\sin 4x}{4} \right) + C \]

Solution:

\[ \begin{aligned} \cos^4 2x &= (\cos^2 2x)^2 = \left( \frac{1+\cos 4x}{2} \right)^2 \\ &= \frac{1}{4} (1 + 2\cos 4x + \cos^2 4x) \\ &= \frac{1}{4} \left( 1 + 2\cos 4x + \frac{1+\cos 8x}{2} \right) \\ &= \frac{1}{8} (3 + 4\cos 4x + \cos 8x) \end{aligned} \]

Integrating:

\[ I = \frac{1}{8} \left( 3x + \sin 4x + \frac{\sin 8x}{8} \right) + C \]

Solution:

\[ \begin{aligned} I &= \int \frac{1-\cos^2 x}{1+\cos x} \, dx \\ &= \int \frac{(1-\cos x)(1+\cos x)}{1+\cos x} \, dx \\ &= \int (1 - \cos x) \, dx \\ &= x - \sin x + C \end{aligned} \]

Solution:

Use \( \cos 2A = 2\cos^2 A - 1 \).

\[ \begin{aligned} I &= \int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} \, dx \\ &= \int \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha} \, dx \\ &= 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} \, dx \\ &= 2 \int (\cos x + \cos \alpha) \, dx \\ &= 2 ( \sin x + x \cos \alpha ) + C \end{aligned} \]

Solution:

We know \( 1 + \sin 2x = \cos^2 x + \sin^2 x + 2\sin x \cos x = (\cos x + \sin x)^2 \).

\[ I = \int \frac{\cos x - \sin x}{(\cos x + \sin x)^2} \, dx \]

Let \( \cos x + \sin x = t \Rightarrow (-\sin x + \cos x) \, dx = dt \).

\[ I = \int \frac{dt}{t^2} = -\frac{1}{t} + C = \frac{-1}{\cos x + \sin x} + C \]

Solution:

\[ \begin{aligned} I &= \int \tan^2 2x \cdot \sec 2x \tan 2x \, dx \\ &= \int (\sec^2 2x - 1) \cdot \sec 2x \tan 2x \, dx \end{aligned} \]

Let \( \sec 2x = t \Rightarrow 2\sec 2x \tan 2x \, dx = dt \Rightarrow \sec 2x \tan 2x \, dx = \frac{1}{2} dt \).

\[ \begin{aligned} I &= \frac{1}{2} \int (t^2 - 1) \, dt \\ &= \frac{1}{2} \left( \frac{t^3}{3} - t \right) + C \\ &= \frac{1}{6} \sec^3 2x - \frac{1}{2} \sec 2x + C \end{aligned} \]

Solution:

\[ \begin{aligned} I &= \int \tan^2 x \tan^2 x \, dx \\ &= \int \tan^2 x (\sec^2 x - 1) \, dx \\ &= \int (\tan^2 x \sec^2 x - \tan^2 x) \, dx \\ &= \int \tan^2 x \sec^2 x \, dx - \int (\sec^2 x - 1) \, dx \end{aligned} \]

For first part, let \( \tan x = t \Rightarrow \sec^2 x \, dx = dt \).

\[ I = \frac{\tan^3 x}{3} - (\tan x - x) + C = \frac{1}{3} \tan^3 x - \tan x + x + C \]

Solution:

\[ \begin{aligned} I &= \int \frac{\sin^3 x}{\sin^2 x \cos^2 x} \, dx + \int \frac{\cos^3 x}{\sin^2 x \cos^2 x} \, dx \\ &= \int \frac{\sin x}{\cos^2 x} \, dx + \int \frac{\cos x}{\sin^2 x} \, dx \\ &= \int \tan x \sec x \, dx + \int \cot x \text{cosec } x \, dx \\ &= \sec x - \text{cosec } x + C \end{aligned} \]

Solution:

We know \( \cos 2x = 1 - 2\sin^2 x \).

\[ \begin{aligned} I &= \int \frac{(1 - 2\sin^2 x) + 2\sin^2 x}{\cos^2 x} \, dx \\ &= \int \frac{1}{\cos^2 x} \, dx \\ &= \int \sec^2 x \, dx \\ &= \tan x + C \end{aligned} \]

Solution:

Rewrite numerator \( 1 = \sin^2 x + \cos^2 x \).

\[ \begin{aligned} I &= \int \frac{\sin^2 x + \cos^2 x}{\sin x \cos^3 x} \, dx \\ &= \int \frac{\sin x}{\cos^3 x} \, dx + \int \frac{1}{\sin x \cos x} \, dx \\ &= \int \tan x \sec^2 x \, dx + \int \frac{\sec^2 x}{\tan x} \, dx \end{aligned} \]

Let \( \tan x = t \Rightarrow \sec^2 x \, dx = dt \).

\[ \begin{aligned} I &= \int t \, dt + \int \frac{1}{t} \, dt \\ &= \frac{t^2}{2} + \log |t| + C \\ &= \frac{1}{2} \tan^2 x + \log |\tan x| + C \end{aligned} \]

Solution:

Use \( \cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x) \).

\[ \begin{aligned} I &= \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} \, dx \\ &= \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx \end{aligned} \]

Let \( \cos x + \sin x = t \Rightarrow (-\sin x + \cos x) \, dx = dt \).

\[ I = \int \frac{dt}{t} = \log |t| + C = \log |\cos x + \sin x| + C \]

Solution:

We know \( \cos x = \sin(\frac{\pi}{2} - x) \).

\[ \begin{aligned} I &= \int \sin^{-1} \left( \sin \left( \frac{\pi}{2} - x \right) \right) \, dx \\ &= \int \left( \frac{\pi}{2} - x \right) \, dx \\ &= \frac{\pi}{2}x - \frac{x^2}{2} + C \end{aligned} \]

Solution:

Multiply and divide by \( \sin(a-b) \).

\[ \begin{aligned} I &= \frac{1}{\sin(a-b)} \int \frac{\sin((x-b) - (x-a))}{\cos(x-a)\cos(x-b)} \, dx \\ &= \frac{1}{\sin(a-b)} \int \frac{\sin(x-b)\cos(x-a) - \cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)} \, dx \\ &= \frac{1}{\sin(a-b)} \int (\tan(x-b) - \tan(x-a)) \, dx \\ &= \frac{1}{\sin(a-b)} [-\log |\cos(x-b)| - (-\log |\cos(x-a)|)] + C \\ &= \frac{1}{\sin(a-b)} \log \left| \frac{\cos(x-a)}{\cos(x-b)} \right| + C \end{aligned} \]

Answer: (A)

Solution:

\[ \begin{aligned} I &= \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \, dx \\ &= \int (\sec^2 x - \text{cosec}^2 x) \, dx \\ &= \tan x - (-\cot x) + C \\ &= \tan x + \cot x + C \end{aligned} \]

Answer: (B)

Solution:

Let \( xe^x = t \).

Differentiating using product rule: \( (x \cdot e^x + e^x \cdot 1) \, dx = dt \Rightarrow e^x(1+x) \, dx = dt \).

\[ \begin{aligned} I &= \int \frac{dt}{\cos^2 t} \\ &= \int \sec^2 t \, dt \\ &= \tan t + C \\ &= \tan(xe^x) + C \end{aligned} \]