Class 12-NCERT Solutions-Chapter-07-Integrals-Ex 7.2

Solution:

Let \( 1 + x^2 = t \).

Differentiating both sides with respect to \( x \):

\( 2x \, dx = dt \)

Now, substitute into the integral:

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

Solution:

Let \( \log x = t \).

Differentiating with respect to \( x \):

\( \frac{1}{x} \, dx = dt \)

Substitute into the integral:

\[ \begin{aligned} \int \frac{(\log x)^2}{x} \, dx &= \int t^2 \, dt \\ &= \frac{t^3}{3} + C \\ &= \frac{(\log x)^3}{3} + C \end{aligned} \]

Solution:

Rewrite the expression:

\( \frac{1}{x(1 + \log x)} \)

Let \( 1 + \log x = t \).

\( \frac{1}{x} \, dx = dt \)

Substitute into the integral:

\[ \begin{aligned} \int \frac{1}{x(1 + \log x)} \, dx &= \int \frac{1}{t} \, dt \\ &= \log |t| + C \\ &= \log |1 + \log x| + C \end{aligned} \]

Solution:

Let \( \cos x = t \).

\( -\sin x \, dx = dt \Rightarrow \sin x \, dx = -dt \)

Substitute into the integral:

\[ \begin{aligned} \int \sin(\cos x) \cdot \sin x \, dx &= \int \sin t \cdot (-dt) \\ &= -\int \sin t \, dt \\ &= -(-\cos t) + C \\ &= \cos t + C \\ &= \cos(\cos x) + C \end{aligned} \]

Solution:

Method 1: Substitution

Let \( \sin(ax+b) = t \).

\( a \cos(ax+b) \, dx = dt \Rightarrow \cos(ax+b) \, dx = \frac{1}{a} dt \)

\[ \begin{aligned} \int t \cdot \frac{1}{a} \, dt &= \frac{1}{a} \frac{t^2}{2} + C \\ &= \frac{1}{2a} \sin^2(ax+b) + C \end{aligned} \]

Method 2: Trigonometric Identity

\[ \begin{aligned} \int \sin(ax+b)\cos(ax+b) \, dx &= \frac{1}{2} \int 2\sin(ax+b)\cos(ax+b) \, dx \\ &= \frac{1}{2} \int \sin(2(ax+b)) \, dx \\ &= \frac{1}{2} \cdot \frac{-\cos(2(ax+b))}{2a} + C \\ &= -\frac{1}{4a} \cos(2ax+2b) + C \end{aligned} \]

Solution:

Let \( ax + b = t \).

\( a \, dx = dt \Rightarrow dx = \frac{1}{a} dt \)

\[ \begin{aligned} \int \sqrt{t} \cdot \frac{1}{a} \, dt &= \frac{1}{a} \int t^{1/2} \, dt \\ &= \frac{1}{a} \left( \frac{t^{3/2}}{3/2} \right) + C \\ &= \frac{2}{3a} (ax+b)^{3/2} + C \end{aligned} \]

Solution:

Let \( x + 2 = t \Rightarrow x = t - 2 \).

\( dx = dt \)

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

Solution:

Let \( 1 + 2x^2 = t \).

\( 4x \, dx = dt \Rightarrow x \, dx = \frac{1}{4} dt \)

\[ \begin{aligned} \int \sqrt{t} \cdot \frac{1}{4} \, dt &= \frac{1}{4} \int t^{1/2} \, dt \\ &= \frac{1}{4} \left( \frac{t^{3/2}}{3/2} \right) + C \\ &= \frac{1}{4} \cdot \frac{2}{3} t^{3/2} + C \\ &= \frac{1}{6} (1+2x^2)^{3/2} + C \end{aligned} \]

Solution:

Factor out 2: \( 2(2x+1)\sqrt{x^2+x+1} \)

Let \( x^2 + x + 1 = t \).

\( (2x + 1) \, dx = dt \)

\[ \begin{aligned} \int 2\sqrt{t} \, dt &= 2 \int t^{1/2} \, dt \\ &= 2 \cdot \frac{2}{3} t^{3/2} + C \\ &= \frac{4}{3} (x^2+x+1)^{3/2} + C \end{aligned} \]

Solution:

Factor out \( \sqrt{x} \): \( \frac{1}{\sqrt{x}(\sqrt{x} - 1)} \)

Let \( \sqrt{x} - 1 = t \).

\( \frac{1}{2\sqrt{x}} \, dx = dt \Rightarrow \frac{1}{\sqrt{x}} \, dx = 2dt \)

\[ \begin{aligned} \int \frac{1}{t} \cdot 2 \, dt &= 2 \int \frac{1}{t} \, dt \\ &= 2 \log |t| + C \\ &= 2 \log |\sqrt{x} - 1| + C \end{aligned} \]

Solution:

Let \( x + 4 = t \Rightarrow x = t - 4 \).

\( dx = dt \)

\[ \begin{aligned} \int \frac{t-4}{\sqrt{t}} \, dt &= \int \left( \frac{t}{t^{1/2}} - \frac{4}{t^{1/2}} \right) \, dt \\ &= \int (t^{1/2} - 4t^{-1/2}) \, dt \\ &= \frac{t^{3/2}}{3/2} - 4\frac{t^{1/2}}{1/2} + C \\ &= \frac{2}{3}(x+4)^{3/2} - 8\sqrt{x+4} + C \end{aligned} \]

Solution:

Let \( x^3 - 1 = t \Rightarrow x^3 = t + 1 \).

\( 3x^2 \, dx = dt \Rightarrow x^2 \, dx = \frac{1}{3} dt \)

Rewrite integrand: \( (x^3-1)^{1/3} \cdot x^3 \cdot x^2 \)

\[ \begin{aligned} \int t^{1/3} (t+1) \cdot \frac{1}{3} \, dt &= \frac{1}{3} \int (t^{4/3} + t^{1/3}) \, dt \\ &= \frac{1}{3} \left[ \frac{t^{7/3}}{7/3} + \frac{t^{4/3}}{4/3} \right] + C \\ &= \frac{1}{3} \left[ \frac{3}{7}t^{7/3} + \frac{3}{4}t^{4/3} \right] + C \\ &= \frac{1}{7}(x^3-1)^{7/3} + \frac{1}{4}(x^3-1)^{4/3} + C \end{aligned} \]

Solution:

Let \( 2 + 3x^3 = t \).

\( 9x^2 \, dx = dt \Rightarrow x^2 \, dx = \frac{1}{9} dt \)

\[ \begin{aligned} \int \frac{1}{t^3} \cdot \frac{1}{9} \, dt &= \frac{1}{9} \int t^{-3} \, dt \\ &= \frac{1}{9} \left( \frac{t^{-2}}{-2} \right) + C \\ &= -\frac{1}{18} \frac{1}{t^2} + C \\ &= -\frac{1}{18(2+3x^3)^2} + C \end{aligned} \]

Solution:

Let \( \log x = t \).

\( \frac{1}{x} \, dx = dt \)

\[ \begin{aligned} \int \frac{1}{t^m} \, dt &= \int t^{-m} \, dt \\ &= \frac{t^{-m+1}}{-m+1} + C \\ &= \frac{(\log x)^{1-m}}{1-m} + C \end{aligned} \]

Solution:

Let \( 9 - 4x^2 = t \).

\( -8x \, dx = dt \Rightarrow x \, dx = -\frac{1}{8} dt \)

\[ \begin{aligned} \int \frac{1}{t} \cdot \left( -\frac{1}{8} \right) \, dt &= -\frac{1}{8} \log |t| + C \\ &= -\frac{1}{8} \log |9-4x^2| + C \end{aligned} \]

Solution:

Let \( 2x + 3 = t \Rightarrow 2 \, dx = dt \Rightarrow dx = \frac{1}{2} dt \).

\[ \begin{aligned} \int e^t \cdot \frac{1}{2} \, dt &= \frac{1}{2} e^t + C \\ &= \frac{1}{2} e^{2x+3} + C \end{aligned} \]

Solution:

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

\[ \begin{aligned} \int \frac{1}{e^t} \cdot \frac{1}{2} \, dt &= \frac{1}{2} \int e^{-t} \, dt \\ &= \frac{1}{2} (-e^{-t}) + C \\ &= -\frac{1}{2e^{x^2}} + C \end{aligned} \]

Solution:

Let \( \tan^{-1}x = t \).

\( \frac{1}{1+x^2} \, dx = dt \)

\[ \begin{aligned} \int e^t \, dt &= e^t + C \\ &= e^{\tan^{-1}x} + C \end{aligned} \]

Solution:

Divide numerator and denominator by \( e^x \):

\( \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

Let \( e^x + e^{-x} = t \).

\( (e^x - e^{-x}) \, dx = dt \)

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

Solution:

Let \( e^{2x} + e^{-2x} = t \).

Differentiating with respect to \( x \):

\( (2e^{2x} - 2e^{-2x}) \, dx = dt \)

\( 2(e^{2x} - e^{-2x}) \, dx = dt \Rightarrow (e^{2x} - e^{-2x}) \, dx = \frac{1}{2} dt \)

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

Solution:

Use the identity \( \tan^2 A = \sec^2 A - 1 \).

\[ \int \tan^2(2x-3) \, dx = \int (\sec^2(2x-3) - 1) \, dx \]

Let \( 2x - 3 = t \Rightarrow 2 \, dx = dt \Rightarrow dx = \frac{1}{2} dt \).

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

Solution:

Let \( 7 - 4x = t \).

\( -4 \, dx = dt \Rightarrow dx = -\frac{1}{4} dt \).

\[ \begin{aligned} \int \sec^2(7-4x) \, dx &= \int \sec^2 t \cdot \left(-\frac{1}{4}\right) \, dt \\ &= -\frac{1}{4} \tan t + C \\ &= -\frac{1}{4} \tan(7-4x) + C \end{aligned} \]

Solution:

Let \( \sin^{-1} x = t \).

\( \frac{1}{\sqrt{1-x^2}} \, dx = dt \).

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

Solution:

Rewrite denominator: \( 2(3\cos x + 2\sin x) \).

Let \( 3\cos x + 2\sin x = t \).

\( (-3\sin x + 2\cos x) \, dx = dt \Rightarrow (2\cos x - 3\sin x) \, dx = dt \).

\[ \begin{aligned} \int \frac{1}{2t} \, dt &= \frac{1}{2} \log |t| + C \\ &= \frac{1}{2} \log |6\cos x + 4\sin x| + C \end{aligned} \]

Solution:

Rewrite as \( \int \frac{\sec^2 x}{(1-\tan x)^2} \, dx \).

Let \( 1 - \tan x = t \).

\( -\sec^2 x \, dx = dt \Rightarrow \sec^2 x \, dx = -dt \).

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

Solution:

Let \( \sqrt{x} = t \).

\( \frac{1}{2\sqrt{x}} \, dx = dt \Rightarrow \frac{1}{\sqrt{x}} \, dx = 2 dt \).

\[ \begin{aligned} \int \cos t \cdot 2 \, dt &= 2 \sin t + C \\ &= 2 \sin \sqrt{x} + C \end{aligned} \]

Solution:

Let \( \sin 2x = t \).

\( 2\cos 2x \, dx = dt \Rightarrow \cos 2x \, dx = \frac{1}{2} dt \).

\[ \begin{aligned} \int \sqrt{t} \cdot \frac{1}{2} \, dt &= \frac{1}{2} \cdot \frac{t^{3/2}}{3/2} + C \\ &= \frac{1}{2} \cdot \frac{2}{3} t^{3/2} + C \\ &= \frac{1}{3} (\sin 2x)^{3/2} + C \end{aligned} \]

Solution:

Let \( 1 + \sin x = t \).

\( \cos x \, dx = dt \).

\[ \begin{aligned} \int \frac{1}{\sqrt{t}} \, dt &= \int t^{-1/2} \, dt \\ &= 2t^{1/2} + C \\ &= 2\sqrt{1+\sin x} + C \end{aligned} \]

Solution:

Let \( \log(\sin x) = t \).

\( \frac{1}{\sin x} \cdot \cos x \, dx = dt \Rightarrow \cot x \, dx = dt \).

\[ \begin{aligned} \int t \, dt &= \frac{t^2}{2} + C \\ &= \frac{1}{2} (\log(\sin x))^2 + C \end{aligned} \]

Solution:

Let \( 1 + \cos x = t \).

\( -\sin x \, dx = dt \Rightarrow \sin x \, dx = -dt \).

\[ \begin{aligned} \int -\frac{1}{t} \, dt &= -\log |t| + C \\ &= -\log |1+\cos x| + C \end{aligned} \]

Solution:

Let \( 1 + \cos x = t \).

\( -\sin x \, dx = dt \Rightarrow \sin x \, dx = -dt \).

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

Solution:

Rewrite: \( \frac{1}{1 + \frac{\cos x}{\sin x}} = \frac{\sin x}{\sin x + \cos x} \).

Divide and multiply by 2: \( \frac{1}{2} \int \frac{2\sin x}{\sin x + \cos x} \, dx \).

Write \( 2\sin x \) as \( (\sin x + \cos x) + (\sin x - \cos x) \).

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

(Note: Let \( \sin x + \cos x = t \), then \( (\cos x - \sin x)dx = dt \)).

Solution:

Rewrite: \( \frac{1}{1 - \frac{\sin x}{\cos x}} = \frac{\cos x}{\cos x - \sin x} \).

Divide and multiply by 2: \( \frac{1}{2} \int \frac{2\cos x}{\cos x - \sin x} \, dx \).

Write \( 2\cos x \) as \( (\cos x - \sin x) + (\cos x + \sin x) \).

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

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

Solution:

Divide numerator and denominator by \( \cos^2 x \) (inside the implicit fraction steps):

Actually, multiply and divide denominator by \( \cos x \):

\( \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x} \cdot \cos^2 x} = \frac{\sqrt{\tan x}}{\tan x \cdot \cos^2 x} = \frac{1}{\sqrt{\tan x}} \sec^2 x \).

Let \( \tan x = t \).

\( \sec^2 x \, dx = dt \).

\[ \begin{aligned} \int t^{-1/2} \, dt &= 2t^{1/2} + C \\ &= 2\sqrt{\tan x} + C \end{aligned} \]

Solution:

Let \( 1 + \log x = t \).

\( \frac{1}{x} \, dx = dt \).

\[ \begin{aligned} \int t^2 \, dt &= \frac{t^3}{3} + C \\ &= \frac{(1+\log x)^3}{3} + C \end{aligned} \]

Solution:

Rewrite as \( \left( 1 + \frac{1}{x} \right) (x + \log x)^2 \).

Let \( x + \log x = t \).

\( (1 + \frac{1}{x}) \, dx = dt \).

\[ \begin{aligned} \int t^2 \, dt &= \frac{t^3}{3} + C \\ &= \frac{(x+\log x)^3}{3} + C \end{aligned} \]

Solution:

Let \( \tan^{-1} x^4 = t \).

Differentiating: \( \frac{1}{1+(x^4)^2} \cdot 4x^3 \, dx = dt \Rightarrow \frac{4x^3}{1+x^8} \, dx = dt \).

\( \frac{x^3}{1+x^8} \, dx = \frac{1}{4} dt \).

\[ \begin{aligned} \int \sin t \cdot \frac{1}{4} \, dt &= -\frac{1}{4} \cos t + C \\ &= -\frac{1}{4} \cos(\tan^{-1} x^4) + C \end{aligned} \]

Answer: (D)

Solution:

Let \( x^{10} + 10^x = t \).

Differentiating: \( (10x^9 + 10^x \log_e 10) \, dx = dt \).

\[ \begin{aligned} \int \frac{1}{t} \, dt &= \log |t| + C \\ &= \log |x^{10} + 10^x| + C \end{aligned} \]

Answer: (B)

Solution:

Replace \( 1 \) in numerator with \( \sin^2 x + \cos^2 x \).

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