Class 12-NCERT Solutions-Chapter-07-Integrals-Ex 7.6

Solution:

Using integration by parts: Let \( u = x \) and \( dv = \sin x \, dx \).

Then \( du = dx \) and \( v = \int \sin x \, dx = -\cos x \).

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

Solution:

Let \( u = x \) and \( dv = \sin 3x \, dx \).

Then \( du = dx \) and \( v = \frac{-\cos 3x}{3} \).

\[ \begin{aligned} \int x \sin 3x \, dx &= x \left( \frac{-\cos 3x}{3} \right) - \int \frac{-\cos 3x}{3} \, dx \\ &= -\frac{x \cos 3x}{3} + \frac{1}{3} \int \cos 3x \, dx \\ &= -\frac{x \cos 3x}{3} + \frac{1}{9} \sin 3x + C \end{aligned} \]

Solution:

Let \( u = x^2 \) and \( dv = e^x \, dx \). Then \( du = 2x \, dx \) and \( v = e^x \).

\[ \int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx \]

Apply integration by parts again for \( \int x e^x \, dx \): Let \( u = x, dv = e^x \, dx \).

\[ \begin{aligned} &= x^2 e^x - 2 \left( x e^x - \int e^x \, dx \right) \\ &= x^2 e^x - 2x e^x + 2e^x + C \\ &= e^x (x^2 - 2x + 2) + C \end{aligned} \]

Solution:

Using ILATE, \( u = \log x \) and \( dv = x \, dx \).

Then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).

\[ \begin{aligned} \int x \log x \, dx &= \frac{x^2}{2} \log x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \\ &= \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx \\ &= \frac{x^2}{2} \log x - \frac{x^2}{4} + C \end{aligned} \]

Solution:

Let \( u = \log 2x \) and \( dv = x \, dx \).

Then \( du = \frac{1}{2x} \cdot 2 \, dx = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).

\[ \begin{aligned} \int x \log 2x \, dx &= \frac{x^2}{2} \log 2x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \\ &= \frac{x^2}{2} \log 2x - \frac{1}{2} \int x \, dx \\ &= \frac{x^2}{2} \log 2x - \frac{x^2}{4} + C \end{aligned} \]

Solution:

Let \( u = \log x \) and \( dv = x^2 \, dx \).

Then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^3}{3} \).

\[ \begin{aligned} \int x^2 \log x \, dx &= \frac{x^3}{3} \log x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx \\ &= \frac{x^3}{3} \log x - \frac{1}{3} \int x^2 \, dx \\ &= \frac{x^3}{3} \log x - \frac{x^3}{9} + C \end{aligned} \]

Solution:

Let \( u = \sin^{-1} x \) and \( dv = x \, dx \).

Then \( du = \frac{1}{\sqrt{1-x^2}} \, dx \) and \( v = \frac{x^2}{2} \).

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

Solution:

Let \( u = \tan^{-1} x \) and \( dv = x \, dx \). Then \( du = \frac{1}{1+x^2} \, dx \) and \( v = \frac{x^2}{2} \).

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

Solution:

Similar to Q7. Result will be analogous.

\[ \int x \cos^{-1} x \, dx = \frac{(2x^2-1)}{4} \cos^{-1} x - \frac{x}{4}\sqrt{1-x^2} + C \]

Solution:

Let \( \sin^{-1} x = \theta \Rightarrow x = \sin \theta \Rightarrow dx = \cos \theta \, d\theta \).

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

Solution:

Let \( \cos^{-1} x = t \Rightarrow x = \cos t \). Then \( \frac{-1}{\sqrt{1-x^2}} dx = dt \Rightarrow \frac{dx}{\sqrt{1-x^2}} = -dt \).

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

Solution:

Let \( u = x \) and \( dv = \sec^2 x \, dx \).

Then \( du = dx \) and \( v = \tan x \).

\[ \begin{aligned} \int x \sec^2 x \, dx &= x \tan x - \int \tan x \, dx \\ &= x \tan x - \log|\sec x| + C \end{aligned} \]

Solution:

Let \( u = \tan^{-1} x \) and \( dv = 1 \, dx \).

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

Solution:

Let \( u = (\log x)^2 \) and \( dv = x \, dx \).

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

Solution:

Let \( u = \log x \) and \( dv = (x^2+1) \, dx \).

Then \( v = \frac{x^3}{3} + x \).

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

Solution:

This is of the form \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \).

Here \( f(x) = \sin x \) and \( f'(x) = \cos x \).

Therefore, integral is \( e^x \sin x + C \).

Solution:

Rewrite integrand: \( e^x \left[ \frac{x+1-1}{(1+x)^2} \right] = e^x \left[ \frac{1}{1+x} + \frac{-1}{(1+x)^2} \right] \).

Here \( f(x) = \frac{1}{1+x} \) and \( f'(x) = \frac{-1}{(1+x)^2} \).

Result: \( \frac{e^x}{1+x} + C \).

Solution:

Simplify fraction:

\[ \frac{1+2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{1}{2}\sec^2(x/2) + \tan(x/2) \]

Let \( f(x) = \tan(x/2) \), then \( f'(x) = \frac{1}{2}\sec^2(x/2) \).

Result: \( e^x \tan \frac{x}{2} + C \).

Solution:

\( f(x) = \frac{1}{x} \), \( f'(x) = -\frac{1}{x^2} \).

Result: \( \frac{e^x}{x} + C \).

Solution:

Rewrite: \( e^x \left[ \frac{(x-1)-2}{(x-1)^3} \right] = e^x \left[ \frac{1}{(x-1)^2} + \frac{-2}{(x-1)^3} \right] \).

\( f(x) = \frac{1}{(x-1)^2} \), \( f'(x) = -2(x-1)^{-3} \).

Result: \( \frac{e^x}{(x-1)^2} + C \).

Solution:

Let \( I = \int e^{2x} \sin x \, dx \).

Parts: \( u = \sin x, dv = e^{2x} dx \Rightarrow v = e^{2x}/2 \).

\( I = \frac{1}{2} e^{2x} \sin x - \int \frac{1}{2} e^{2x} \cos x \, dx \).

Again parts on integral: \( u = \cos x, v = e^{2x}/2 \).

\( I = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \left[ \frac{1}{2} e^{2x} \cos x - \int \frac{1}{2} e^{2x} (-\sin x) dx \right] \).

\( I = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x - \frac{1}{4} I \).

\( \frac{5}{4} I = \frac{e^{2x}}{4} (2\sin x - \cos x) \).

Result: \( \frac{e^{2x}}{5} (2\sin x - \cos x) + C \).

Solution:

Put \( x = \tan \theta \). Then \( \frac{2x}{1+x^2} = \sin 2\theta \).

\( \sin^{-1}(\sin 2\theta) = 2\theta = 2 \tan^{-1} x \).

Integral becomes \( \int 2 \tan^{-1} x \, dx \).

From Q13, \( \int \tan^{-1} x \, dx = x \tan^{-1} x - \frac{1}{2} \log(1+x^2) \).

Result: \( 2x \tan^{-1} x - \log(1+x^2) + C \).

Answer: (A)

Solution:

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

\( \int e^t \frac{dt}{3} = \frac{1}{3} e^t + C = \frac{1}{3} e^{x^3} + C \).

Answer: (B)

Solution:

\( \int e^x (\sec x + \sec x \tan x) \, dx \).

\( f(x) = \sec x \), \( f'(x) = \sec x \tan x \).

Result: \( e^x \sec x + C \).