EXERCISE 7.6(Integration)

Integrate the functions in Exercises 1 to 22.(Ex 7.6 integration ncert maths solution class 12)

Question 1: $\int x \sin xdx$

Solution: Let $I=\int x \sin x d x$

$I =x \int \sin x d x-\int\left\{\left(\frac{d}{d x}(x)\right) \int \sin x d x\right\} d x$

$=x(-\cos x)-\int 1 \cdot(-\cos x) d x$

$=-x \cos x+\sin x+C$

Question 2: $\int x \sin 3 xdx$

Solution: Let $I=\int x \sin 3 x d x$

$I =x \int \sin 3 x d x-\int\left\{\left(\frac{d}{d x} x\right) \int \sin 3 x d x\right\} d x$

$=x\left(\frac{-\cos 3 x}{3}\right)-\int 1 \cdot\left(\frac{-\cos 3 x}{3}\right) d x$

$=\frac{-x \cos 3 x}{3}+\frac{1}{3} \int \cos 3 x d x$

$=\frac{-x \cos 3 x}{3}+\frac{1}{9} \sin 3 x+C$

Question 3: $\int x^2 e^xdx$

Solution: Let $I=\int x^2 e^x d x$

$I =x^2 \int e^x d x-\int\left\{\left(\frac{d}{d x} x^2\right) \int e^x d x\right\} d x$

$=x^2 e^x-\int 2 x \cdot e^x d x$

$=x^2 e^x-2 \int x \cdot e^x d x$

$=x^2 e^x-2\left[x \int e^x d x-\int\left\{\left(\frac{d}{d x} x\right) \int e^x d x\right\} d x\right]$

$=x^2 e^x-2\left[x e^x-\int e^x d x\right]$

$=x^2 e^x-2\left[x e^x-e^x\right]$

$=x^2 e^x-2 x e^x+2 e^x+C$

$=e^x\left(x^2-2 x+2\right)+C$

Question 4: $\int x \log xdx$

Solution: Let $I=\int x \log x d x$

$I =\log x \int x d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x d x\right\} d x$

$=\log x \cdot \frac{x^2}{2}-\int \frac{1}{x} \cdot \frac{x^2}{2} d x$

$=\frac{x^2 \log x}{2}-\int \frac{x}{2} d x$

$=\frac{x^2 \log x}{2}-\frac{x^2}{4}+C$

Question 5: $\int x \log 2 xdx$

Solution: Let $I=\int x \log 2 x d x$

$I =\log 2 x \int x d x-\int\left\{\left(\frac{d}{d x} \log 2 x\right) \int x d x\right\} d x$

$=\log 2 x \cdot \frac{x^2}{2}-\int \frac{2}{2 x} \cdot \frac{x^2}{2} d x$

$=\frac{x^2 \log 2 x}{2}-\int \frac{x}{2} d x$

$=\frac{x^2 \log 2 x}{2}-\frac{x^2}{4}+C$

Question 6: $\int x^2 \log xdx$

Solution: Let $I=\int x^2 \log x d x$

$I =\log x \int x^2 d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x^2 d x\right\} d x$

$=\log x \cdot\left(\frac{x^3}{3}\right)-\int \frac{1}{x} \cdot \frac{x^3}{3} d x$

$=\frac{x^3 \log x}{3}-\int \frac{x^2}{3} d x$

Question 7: $\int x \sin ^{-1} x$

Solution: Let $I=\int x \sin ^{-1} x d x$

$I=\sin ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \sin ^{-1} x\right) \int x d x\right\} d x$

$=\sin ^{-1} x\left(\frac{x^2}{2}\right)-\int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} d x$

$=\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2} \int \frac{-x^2}{\sqrt{1-x^2}} d x$

$=\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\right\} d x$

$=\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}}\right\} d x$

$=\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2}\left\{\int \sqrt{1-x^2} d x-\int \frac{1}{\sqrt{1-x^2}} d x\right\}$

$=\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2}\left\{\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x-\sin ^{-1} x\right\}+C$

$=\frac{x^2 \sin ^{-1} x}{2}+\frac{x}{4} \sqrt{1-x^2}+\frac{1}{4} \sin ^{-1} x-\frac{1}{2} \sin ^{-1} x+C$

$=\frac{1}{4}\left(2 x^2-1\right) \sin ^{-1} x+\frac{x}{4} \sqrt{1-x^2}+C$

Question 8: $\int x \tan ^{-1} xdx$

Solution: Let $I=\int x \tan ^{-1} x d x$

$I=\tan ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \tan ^{-1} x\right) \int x d x\right\} d x$

$=\tan ^{-1} x\left(\frac{x^2}{2}\right)-\int \frac{1}{1+x^2} \cdot \frac{x^2}{2} d x$

$=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^2}{1+x^2} d x$

$=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2} \int\left\{\frac{x^2+1}{1+x^2}-\frac{1}{1+x^2}\right\} d x$

$=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2} \int\left(1-\frac{1}{1+x^2}\right) d x$

$=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+C$

$=\frac{x^2}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x+C$

Question 9: $\int x \cos ^{-1} xdx$

Solution: Let $I=\int x \cos ^{-1} x d x$

$I=\cos ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int x d x\right\} d x$

$=\cos ^{-1} x\left(\frac{x^2}{2}\right)-\int \frac{-1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} d x$

$=\frac{x^2 \cos ^{-1} x}{2}-\frac{1}{2} \int \frac{1-x^2-1}{\sqrt{1-x^2}} d x$

$=\frac{x^2 \cos ^{-1} x}{2}-\frac{1}{2} \int\left\{\sqrt{1-x^2}+\left(\frac{-1}{\sqrt{1-x^2}}\right)\right\} d x$

$=\frac{x^2 \cos ^{-1} x}{2}-\frac{1}{2} \int \sqrt{1-x^2} d x-\frac{1}{2} \int\left(\frac{-1}{\sqrt{1-x^2}}\right) d x$

$=\frac{x^2 \cos ^{-1} x}{2}-\frac{1}{2} I_1-\frac{1}{2} \cos ^{-1} x--(i)$

Where, $I_1=\int \sqrt{1-x^2} d x$

$\Rightarrow I_1=\sqrt{1-x^2} \int 1 d x-\int \frac{d}{d x} \sqrt{1-x^2} \int 1 d x$

$\Rightarrow I_1=x \sqrt{1-x^2}-\int \frac{-2 x}{2 \sqrt{1-x^2}} x d x$

$\Rightarrow I_1=x \sqrt{1-x^2}-\int \frac{-x^2}{\sqrt{1-x^2}} d x$

$\Rightarrow I_1=x \sqrt{1-x^2}-\int \frac{1-x^2-1}{\sqrt{1-x^2}} d x$

$\Rightarrow I_1=x \sqrt{1-x^2}-\left\{\int \sqrt{1-x^2} d x+\int \frac{-d x}{\sqrt{1-x^2}}\right\}$

$\Rightarrow I_1=x \sqrt{1-x^2}-\left\{I_1+\cos ^{-1} x\right\}$

$\Rightarrow 2 I_1=x \sqrt{1-x^2}-\cos ^{-1} x$

$\therefore I_1=\frac{x}{2} \sqrt{1-x^2}-\frac{1}{2} \cos ^{-1} x$

Substituting in (1),

$I=\frac{x^2 \cos ^{-1} x}{2}-\frac{1}{2}\left(\frac{x}{2} \sqrt{1-x^2}-\frac{1}{2} \cos ^{-1} x\right)-\frac{1}{2} \cos ^{-1} x$

$=\frac{\left(2 x^2-1\right)}{4} \cos ^{-1} x-\frac{x}{4} \sqrt{1-x^2}+C$

Question 10: $\int\left(\sin ^{-1} x\right)^2dx$

Solution: Let $I=\int\left(\sin ^{-1} x\right)^2 \cdot 1 d x$

$I=\left(\sin ^{-1} x\right)^2 \cdot \int 1 d x-\int\left\{\frac{d}{d x}\left(\sin ^{-1} x\right)^2 \cdot \int 1 d x\right\} d x$

$=\left(\sin ^{-1} x\right)^2 \cdot x-\int \frac{2 \sin ^{-1} x}{\sqrt{1-x^2}} \cdot x d x$

$=x\left(\sin ^{-1} x\right)^2+\int \sin ^{-1} x\left(\frac{-2 x}{\sqrt{1-x^2}}\right) d x$

$=x\left(\sin ^{-1} x\right)^2+\left[\sin ^{-1} x \int \frac{-2 x}{\sqrt{1-x^2}} d x-\int\left\{\left(\frac{d}{d x} \sin ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^2}} d x\right\} d x\right]$

$=x\left(\sin ^{-1} x\right)^2+\left[\sin ^{-1} x \cdot 2 \sqrt{1-x^2}-\int \frac{1}{\sqrt{1-x^2}} \cdot 2 \sqrt{1-x^2} d x\right]$

$=x\left(\sin ^{-1} x\right)^2+2 \sqrt{1-x^2} \sin ^{-1} x-\int 2 d x$

$=x\left(\sin ^{-1} x\right)^2+2 \sqrt{1-x^2} \sin ^{-1} x-2 x+C$

Question 11: $\int\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}dx$

Solution: Let $I=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^2}} d x$

$I=\frac{-1}{2} \int \frac{-2 x}{\sqrt{1-x^2}} \cdot \cos ^{-1} x d x$

$I=\frac{-1}{2}\left[\cos ^{-1} x \int \frac{-2 x}{\sqrt{1-x^2}} d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^2}} d x\right\} d x\right]$

$=\frac{-1}{2}\left[\cos ^{-1} x \cdot 2 \sqrt{1-x^2}-\int \frac{-1}{\sqrt{1-x^2}} \cdot 2 \sqrt{1-x^2} d x\right]$

$=\frac{-1}{2}\left[2 \sqrt{1-x^2} \cos ^{-1} x+\int 2 d x\right]$

$=\frac{-1}{2}\left[2 \sqrt{1-x^2} \cos ^{-1} x+2 x\right]+C$

$=-\left[\sqrt{1-x^2} \cos ^{-1} x+x\right]+C$

Question 12: $\int x \sec ^2 xdx$

Solution: Let $I=\int x \sec ^2 x d x$

$I=x \int \sec ^2 x d x-\int\left\{\left(\frac{d}{d x} x\right) \int \sec ^2 x d x\right\} d x$

$=x \tan x-\int 1 \cdot \tan x d x$

$=x \tan x+\log |\cos x|+C$

Question 13: $\int \tan ^{-1} xdx$

Solution: Let $I=\int 1 \cdot \tan ^{-1} x d x$

$I=\tan ^{-1} x \int 1 d x-\int\left\{\left(\frac{d}{d x} \tan ^{-1} x\right) \int 1 \cdot d x\right\} d x$

$=\tan ^{-1} x \cdot x-\int \frac{1}{1+x^2} x d x$

$=x \tan ^{-1} x-\frac{1}{2} \int \frac{2}{1+x^2} d x$

$=x \tan ^{-1} x-\frac{1}{2} \log \left|1+x^2\right|+C$

$=x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^2\right)+C$

Question 14: $\int x(\log x)^2dx$

Solution: Let $I=\int x(\log x)^2 d x$

$I=(\log x)^2 \int x d x-\int\left[\left\{\frac{d}{d x}(\log x)^2\right\} \int x d x\right] d x$

$=\frac{x^2}{2}(\log x)^2-\left[\int 2 \log x \cdot \frac{1}{x} \cdot \frac{x^2}{2} d x\right]$

$=\frac{x^2}{2}(\log x)^2-\int x \log x d x$

Again, using integration by parts, we get

$I =\dfrac{x^2}{2}(\log x)^2-\left[\log x\int xdx-\int\left(\frac{d}{dx}\log x\int xdx\right)dx\right]$

$=\dfrac{x^2}{2}(\log x)^2-\left[\log x\frac{x^2}{2}-\int\left(\frac{1}{x}\frac{x^2}{2}\right)dx\right]$

$=\dfrac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{1}{2}\int xdx+C$

$=\dfrac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{1}{2}\frac{x^2}{2}+C$

$=\dfrac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{x^2}{4}+C$

Question 15: $\int (x^2+1)\log x$

Solution: Let $I=\int (x^2+1)\log x$

$=\log x\int(x^2+1)dx-\int\left(\frac{d}{dx}\log x\int(x^2+1)dx\right)dx$

$=\log x(\frac{x^3}{3}+x)-\int\frac{1}{x}\left(\frac{x^3}{3}+x\right)dx$

$=(\frac{x^3}{3}+x)\log x-\int\frac{1}{x}x\left(\frac{x^2}{3}+1\right)dx$

$=(\frac{x^3}{3}+x)\log x-\int\left(\frac{x^2}{3}+1\right)dx+C$

$=(\frac{x^3}{3}+x)\log x-\left(\frac{1}{3}\frac{x^3}{3}+x\right)+C$

$=(\frac{x^3}{3}+x)\log x-\left(\frac{x^3}{9}+x\right)+C$

Question 16: $e^x(\sin x+\cos x)$

Solution: Let $I=\int e^x(\sin x+\cos x) d x$

Let $f(x)=\sin x$

$f'(x)=\cos x$

$I=\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$

Since, $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C$

$\therefore I=e^x \sin x+C$

Question 17: $\int \frac{x e^x}{(1+x)^2}dx$

Solution: Let, $I=\int \frac{x e^x}{(1+x)^2} d x=\int e^x\left\{\frac{x}{(1+x)^2}\right\} d x$

$=\int e^x\left\{\frac{1+x-1}{(1+x)^2}\right\} d x=\int e^x\left\{\frac{1}{1+x}-\frac{1}{(1+x)^2}\right\} d x$

Let $f(x)=\frac{1}{1+x} \Rightarrow f^{\prime}(x)=\frac{-1}{(1+x)^2}$

$I= \int \frac{x e^x}{(1+x)^2} d x=\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$

Since, $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C$

$\therefore \int \frac{x e^x}{(1+x)^2} d x=e^x \frac{1}{1+x}+C$

Question 18: $\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)dx$

Solution: $e^x\left(\frac{1+\sin x}{1+\cos x}\right)=e^x\left(\frac{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right)$

$=\frac{e^x\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}{2 \cos ^2 \frac{x}{2}}=\frac{1}{2} e^x\left(\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}}\right)^2$

$=\frac{1}{2} e^x\left[\tan \frac{x}{2}+1\right]^2$

$=\frac{1}{2} e^x\left[1+\tan \frac{x}{2}\right]^2$

$=\frac{1}{2} e^x\left[1+\tan ^2 \frac{x}{2}+2 \tan \frac{x}{2}\right]$

$=\frac{1}{2} e^x\left[\sec ^2 \frac{x}{2}+2 \tan \frac{x}{2}\right]$

$\frac{e^x(1+\sin x) d x}{(1+\cos x)}=e^x\left[\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right] .$

It is known that, $\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C$

$\int \frac{e^x(1+\sin x)}{(1+\cos x)} d x=e^x \tan \frac{x}{2}+C$

Question 19: $\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)dx$

Solution: Let $I=\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$

Here, $f(x)=\frac{1}{x} \Rightarrow f'(x)=\frac{-1}{x^2}$

It is known that,

$\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x$

$=e^x f(x)+C$

$\therefore I=\frac{e^x}{x}+C$

Question 20: $\int \frac{(x-3) e^x}{(x-1)^3}dx$

Solution: $\int e^x\left\{\frac{x-3}{(x-1)^3}\right\} d x=\int e^x\left\{\frac{x-1-2}{(x-1)^3}\right\} d x$

$=\int e^x\left\{\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\right\} d x$

Let $f(x)=\frac{1}{(x-1)^2} \quad f^{\prime}(x)=\frac{-2}{(x-1)^3}$

It is known that,

$\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C$

$\therefore \int e^x\left\{\frac{(x-3)}{(x-1)^2}\right\} d x=\frac{e^x}{(x-1)^2}+C$

Question 21: $\int e^{2 x} \sin xdx$

Solution: Let $I=e^{2 x} \sin x d x$ .

$I=\sin x \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \sin x\right) \int e^{2 x} d x\right\} d x$

$\Rightarrow I=\sin x \cdot \frac{e^{2 x}}{2}-\int \cos x \cdot \frac{e^{2 x}}{2} d x$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2} \int e^{2 x} \cos x d x$

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

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \frac{e^{2 x}}{2}-\int(-\sin x) \frac{e^{2 x}}{2} d x\right]$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\frac{e^{2 x} \cos x}{2}+\frac{1}{2} \int e^{2 x} \sin x d x\right]$

$\Rightarrow I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} I$

$\Rightarrow I+\frac{1}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}$

$\Rightarrow \frac{5}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}$

$\Rightarrow I=\frac{4}{5}\left[\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}\right]+C$

$\Rightarrow I=\frac{e^{2 x}}{5}[2 \sin x-\cos x]+C$

Question 22: $\sin { }^{-1}\left(\frac{2 x}{1+x^2}\right)$

Solution: Let $x=\tan \theta \quad d x=\sec ^2 \theta d \theta$

$\therefore \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$

$=\sin ^{-1}(\sin 2 \theta)=2 \theta$

$\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=\int 2 \theta \cdot \sec ^2 \theta d \theta$

$=2 \int \theta \cdot \sec ^2 \theta d \theta$

Using integration by parts, we get

$I=2\left[\theta \cdot \int \sec ^2 \theta d \theta-\int\left\{\left(\frac{d}{d \theta} \theta\right) \int \sec ^2 \theta d \theta\right\} d \theta\right]$