Ex 7.3 Class 12 Maths Ncert Solutions

\(\sin ^{2}(2 x+5)=\frac{1-\cos 2(2 x+5)}{2}=\frac{1-\cos (4 x+10)}{2}\)
\(\Rightarrow\int \sin ^{2}(2 x+5) \mathrm{d} x=\int \frac{1-\cos (4 x+10)}{2} d x\)
\(\Rightarrow\frac{1}{2} \int 1 \cdot d x-\frac{1}{2} \int \cos (4 x+10) d x\)
\(\Rightarrow\frac{1}{2} x-\frac{1}{2} \int \cos (4 x+10) d x\)
\(\Rightarrow\frac{1}{2} x-\frac{1}{2}\left(\frac{\sin (4 x+10)}{4}\right)+C\)
\(\Rightarrow\frac{1}{2} x-\frac{1}{8} \sin (4 x+10)+C\)

\(\int \sin 3 x \cos 4 x d x\)\(=\frac{1}{2} \int\{\sin (3 x+4 x)+\sin (3 x-4 x)\} d x\)
\(\Rightarrow\frac{1}{2} \int\{\sin 7 x+\sin (-x)\} d x\)
\(\Rightarrow\frac{1}{2} \int\{\sin 7 x-\sin x\} d x\)
\(\Rightarrow\frac{1}{2} \int \sin 7 x d x-\frac{1}{2} \int \sin x d x\)
\(\Rightarrow\frac{1}{2}\left(\frac{-\cos 7 x}{14}\right)-\frac{1}{2}(-\cos x)+C\)
\(\Rightarrow\frac{-\cos 7 x}{14}+\frac{\cos x}{2}+C\)

\(\int \cos 2 x \cos 4 x \cos 6 x d x\)\(=\int \cos 2 x\left[\frac{1}{2}\{\cos (4 x+6 x)+\cos (4 x+6 x)\}\right] d x\)
\(\Rightarrow \frac{1}{2} \int\{\cos 2 x \cos 10 x+\cos 2 x \cos (-2 x)\} d x\)
\(\Rightarrow\frac{1}{2} \int\left\{\cos 2 x \cos 10 x+\cos ^{2} 2 x\right\} d x\)
\(\Rightarrow\frac{1}{2} \int\left[\left\{\frac{1}{2} \cos (2 x+10 x)+\cos (2 x-10 x)\right\}\right.\) \(\left.+\left(\frac{1+\cos 4 x}{2}\right)\right] d x\)
\(\Rightarrow\frac{1}{4} \int(\cos 12 x+\cos 8 x+\cos 4 x) d x\)
\(\Rightarrow\frac{1}{4}\left[\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+x+\frac{\sin 4 x}{4}\right]+C\)

Let \( \mathrm{I}=\sin ^{3}(2 x+1) \)
\(\Rightarrow\int \sin ^{3}(2 x+1) d x=\int \sin ^{2}(2 x+1) \cdot \sin (2 x+1) d x\)
\(\Rightarrow\int\left(1-\cos ^{2}(2 x+1) )\sin (2 x+1) d x\right.\)
Let \( \cos (2 x+1)=\mathrm{t} \)
\(\Rightarrow-2 \sin (2 x+1) d x=d t\)
\(\Rightarrow \sin (2 x+1) \mathrm{d}x=\frac{-d t}{2}\)
\(\Rightarrow\mathrm{I}=\frac{-1}{2} \int\left(1-t^{2}\right) d t\)
\(\Rightarrow\frac{-1}{2}\left\{t-\frac{t^{3}}{3}\right\}\)
\(\Rightarrow\frac{-1}{2}\left\{\cos (2 x+1)-\frac{\cos ^{3}(2 x+1)}{3}\right\}\)
\(\Rightarrow\frac{-\cos (2 x+1)}{2}+\frac{\cos ^{3}(2 x+1)}{3}+C\)

\( \text { Let } \mathrm{I}=\int \sin ^{3} x \cos ^{3} x \cdot d x\)
\(\Rightarrow\int \cos ^{3} x \cdot \sin ^{3} x \cdot \sin x \cdot d x\)
\(\Rightarrow\int \cos ^{3} x\left(1-\cos ^{2} x\right) \sin x \cdot d x \)
\( \text {Let } \cos x=\mathrm{t}\)
\(\Rightarrow-\sin x \cdot \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \mathrm{I}=-\int t^{3}\left(1-t^{2}\right) d t\)
\(\Rightarrow-\int\left(t^{3}-t^{5}\right) d t\)
\(\Rightarrow-\left\{\frac{t^{4}}{4}-\frac{t^{6}}{6}\right\}+C\)
\(\Rightarrow-\left\{\frac{\cos ^{4} x}{6}-\frac{\cos ^{6} x}{4}\right\}+C\)
\(\Rightarrow\frac{\cos ^{6} x}{6}-\frac{\cos ^{4} x}{4}+C \)

\(\int \sin x \sin 2 x \sin 3 x d x=\int \sin x \cdot \frac{1}{2}[\{\cos (2 x-3 x)-\cos (2 x+3 x)\}] d x\)
\(\Rightarrow\frac{1}{2} \int\{\sin x \cos (-x)-\sin x \cos 5 x\} d x\)
\(\Rightarrow\frac{1}{2} \int\{\sin x \cos x-\sin x \cos 5 x\} d x\)
\(\Rightarrow\frac{1}{2} \int \frac{\sin 2 x}{2} d x-\frac{1}{2} \int \sin x \cos 5 x d x\)
\(\Rightarrow\frac{1}{4}\left[\frac{-\cos 2 x}{2}\right]-\frac{1}{2} \int\left\{\frac{1}{2} \sin (x+5 x)+\sin (x-5 x)\right\} d x\)
\(\Rightarrow\frac{-\cos 2 x}{2}-\frac{1}{4} \int(\sin 6 x+\sin (-4 x)) d x\)
\(\Rightarrow\frac{-\cos 2 x}{2}-\frac{1}{4}\left[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{4}\right]+C\)
\(\Rightarrow\frac{-\cos 2 x}{2}-\frac{1}{8}\left[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{2}\right]+C\)
\(\Rightarrow\frac{1}{8}\left[\frac{-\cos 6 x}{3}-\frac{\cos 4 x}{2}-\cos 2 x\right]+C\)

\(\int \sin 4 x \sin 8 x d x=\int\left\{\frac{1}{2} \cos (4 x-8 x)-\cos (4 x+8 x)\right\} d x\)
\(\Rightarrow\frac{1}{2} \int\{\cos (-4 x)-\cos 12 x \}d x\)
\(\Rightarrow\frac{1}{2} \int\{\cos 4 x-\cos 12 x\} d x\)
\(\Rightarrow\frac{1}{2}\left[\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right]+C\)

\(\frac{1-\cos x}{1+\cos x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=2 \tan ^{2} \frac{x}{2}=\left(\sec ^{2} \frac{x}{2}-1\right)\)
\( \therefore \int \frac{1-\cos x}{1+\cos x} d x=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x\)
\(\Rightarrow\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x\right]+C\)
\(\Rightarrow2 \tan \frac{x}{2}-x+C\)

\(\frac{\cos x}{1+\cos x}=\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=\frac{1}{2}\left[1-\tan ^{2} \frac{x}{2}\right]\)
\(\Rightarrow \int \frac{\cos x}{1+\cos x} d x=\int \frac{1}{2}\left[1-\tan ^{2} \frac{x}{2}\right] d x\)
\(\Rightarrow\frac{1}{2} \int\left[1-\sec ^{2} \frac{x}{2}\right] d x\)
\(\Rightarrow\frac{1}{2} \int\left[2-\sec ^{2} \frac{x}{2}\right] d x\)
\(
\Rightarrow\frac{1}{2}\left[2 x-\frac{\tan \frac{x}{2}}{\frac{1}{2}}\right]+C\)
\(\Rightarrow x-\tan \frac{x}{2}+C\)

\(\sin ^{4} x=\sin ^{2} x \sin ^{2} x\)
\(\Rightarrow\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1-\cos 2 x}{2}\right)\)
\(\Rightarrow\frac{1}{4}(1-\cos 2 x)^{2}\)
\(\Rightarrow\frac{1}{4}\left[1+\cos ^{2} 2 x-2 \cos 2 x\right]\)
\(\Rightarrow\frac{1}{4}\left[1+\left(\frac{1+\cos 4 x}{2}\right)-2 \cos 2 x\right]\)
\(\Rightarrow\frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]\)
\(\Rightarrow\frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]\)
Now, \( \int \sin ^{4} x d x=\frac{1}{4} \int\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right] d x \)
\(\Rightarrow\frac{1}{4}\left[\frac{3}{2} x+\frac{1}{2}\left(\frac{\sin 4 x}{4}\right)-\frac{2 \sin 2 x}{2}\right]+C\)
\(\Rightarrow\frac{1}{8}\left[3 x+\left(\frac{\sin 4 x}{4}\right)-2 \sin 2 x\right]+C\)
\(\Rightarrow\frac{3 x}{8}-\frac{1}{4} \sin 2 x+\frac{1}{32} \sin 4 x+C\)

\(\cos ^{4} 2 x=\left(\cos ^{2} 2\right)^{2}\)
\(\Rightarrow\left(\frac{1+\cos 4 x}{2}\right)^{2}\)
\(\Rightarrow\frac{1}{4}\left[1+\cos ^{2} 4 x-2 \cos 4 x\right]\)
\(\Rightarrow\frac{1}{4}\left[1+\left(\frac{1+\cos 8 x}{2}\right)+2 \cos 4 x\right]\)
\(\Rightarrow\frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2} \cos 8 x+2 \cos 4 x\right]\)
\(\Rightarrow\frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 8 x+2 \cos 4 x\right]\)
Now, \( \int \cos ^{4} 2 x d x=\int\left[\frac{3}{8}+\frac{1}{8} \cos 8 x+\frac{1}{2} \cos 4 x\right] d x \)
\(\Rightarrow\frac{3 x}{8}+\frac{1}{64} \sin 8 x+\frac{1}{8} \sin 4 x+C\)

\(\frac{\sin ^{2} x}{1+\cos x}=\frac{\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)^{2}}{2 \cos ^{2} \frac{x}{2}}\)
\(\Rightarrow\frac{4 \sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\)
\(\Rightarrow2 \sin ^{2} \frac{x}{2}\)
\(=1-\cos x\)
\(\Rightarrow \int \frac{\sin ^{2} x}{1+\cos x} d x=\int(1-\cos x) d x\)
\(=x-\sin x+C\)

Using the identity \( \cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} \), we have
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}=\frac{-2 \sin \frac{2 x+2 \alpha}{2} \sin \frac{2 x-2 \alpha}{2}}{-2 \sin \frac{x+\alpha}{2} \sin \frac{x-\alpha}{2}}\)
\(=\frac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \left(\frac{x+\alpha}{2}\right) \sin \left(\frac{x-\alpha}{2}\right)}\)
Now, using the identity \( \sin 2 x=2 \sin x\cos x \), we have
\(=\frac{\left[2 \sin \left(\frac{x+\alpha}{2}\right) \cos \left(\frac{x+\alpha}{2}\right)\right]\left[2 \sin \left(\frac{x-\alpha}{2}\right) \cos \left(\frac{x-\alpha}{2}\right)\right]}{\sin \left(\frac{x+\alpha}{2}\right) \sin \left(\frac{x-\alpha}{2}\right)}\)
\(=4 \cos \left(\frac{x+\alpha}{2}\right) \cos \left(\frac{x-\alpha}{2}\right)\)
Using identity \( 2 \cos \mathrm{A} \cos \mathrm{B}=\cos (\mathrm{A}+\mathrm{B})+\cos (\mathrm{A}-\mathrm{B}) \), we have
\(=2\left[\cos \left(\frac{x+\alpha}{2}+\frac{x-\alpha}{2}\right)+\cos \frac{x+\alpha}{2}+\frac{x-\alpha}{2}\right]\)
\(=2[\cos (x)+\cos \alpha]\)
\(=2 \cos x+2 \cos \alpha\)
\(\Rightarrow \int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x=\int(2 \cos x+2 \cos \alpha) d x\)
\(=2[\sin x+x \cos \alpha]+C\)

\(\frac{\cos x-\sin x}{1+\sin 2 x}\)
\(\Rightarrow\frac{\cos x-\sin x}{\left(\sin ^{2} x+\cos ^{2} x\right)+2 \sin x \cos x}\)
\(\Rightarrow\frac{\cos x-\sin x}{(\sin x+\cos x)^{2}}\)
Let \( \sin x+\cos x=\mathrm{t} \)
\(\Rightarrow(\cos x-\sin x) \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow\int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x\)
\(\Rightarrow\int \frac{d t}{t^{2}}\)
\(=-\mathrm{t}^{-1}+\mathrm{C}\)
\(\Rightarrow-\frac{1}{t}+C\)
\(\Rightarrow\frac{-1}{\sin x+\cos x}+C\)

\(\tan ^{3} 2 x \sec 2 x=\tan ^{2} 2 x \tan 2 \mathrm{xsec} 2 x\)
\(=\left(\sec ^{2} 2 x-1\right) \tan 2 x\mathrm{sec} 2 x\)
\(=\sec ^{2} 2 x\cdot \tan 2x \mathrm{sec} 2 x-\tan 2 x\mathrm{sec} 2x\)
\(\Rightarrow \int \tan ^{3} 2 x \sec 2 x d x=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\int \tan 2 x \sec 2 x d x\)
\(\Rightarrow \int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\frac{\sec 2 x}{2}+C\)
Now, Let \( \sec 2 x=\mathrm{t} \)
\(\Rightarrow 2 \sec 2 x \tan 2 x d x=d t\)
Thus, \( \int \tan ^{3} 2 x \sec 2 x d x=\frac{1}{2} \int t^{2} d t-\frac{\sec 2 x}{2}+C \)
\(\Rightarrow\frac{t^{3}}{6}-\frac{\sec 2 x}{2}+C\)
\(\Rightarrow\frac{(\sec 2 x)^{3}}{6}-\frac{\sec 2 x}{2}+C\)

\(\tan ^{4} x=\tan ^{2} x \cdot \tan ^{2} x\)
\(=\left(\sec ^{2} x-1\right) \tan ^{2} x\)
\(=\sec ^{2} x \tan ^{2} x-\tan ^{2} x\)
\(=\sec ^{2} x \tan ^{2} x-\left(\sec ^{2} x-1\right)\)
\(=\sec ^{2} x \tan ^{2} x-\sec ^{2} x+1\)
Now, \( \int \tan ^{4} x d x=\int \sec ^{2} x \tan ^{2} x d x-\int \sec ^{2} x d x-\int 1 . d x \)
\(\Rightarrow\int \sec ^{2} x \tan ^{2} x d x-\tan x+x+C\)
Now, let \( \tan x=\mathrm{t} \)
\(\Rightarrow \sec ^{2} xdx=\mathrm{dt}\)
\(\Rightarrow \int \sec ^{2} x \tan ^{2} x d x=\int t^{2} d t=\frac{t^{3}}{3}=\frac{\tan ^{3} x}{3}\)
\(\Rightarrow \int \tan ^{4} x d x=\frac{1}{3} \tan ^{3} x-\tan x+x+C\)

\(\Rightarrow \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}=\frac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x}+\frac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}\)
\(\Rightarrow\frac{\sin x}{\cos ^{2} x}+\frac{\cos x}{\sin ^{2} x}\)
\(=\tan x \sec x+\cot x \operatorname{cosec} x\)
\( \text {Now, } \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x=\int \tan x \sec x+\cot x \operatorname{cosec} x\)
\(=\sec x-\operatorname{cosec} x+C\)

\(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}=\frac{\cos 2 x+(1+\cos 2 x)}{\cos ^{2} x}\)
\(=\frac{1}{\cos ^{2} x}\)
\(=\sec ^{2} x\)
Now, \( \int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x=\int \sec ^{2} x d x \)
\(=\tan x+C\)

\(\frac{1}{\sin x \cos ^{3} x}=\frac{\sin x}{\cos ^{3} x}+\frac{1}{\sin x \cos x}\)
\(\Rightarrow\tan x \sec ^{2} x+\frac{\frac{1}{\cos ^{2} x}}{\frac{\sin x \cos x}{\cos ^{2} x}}\)
\(\Rightarrow\tan x \sec ^{2} x+\frac{\sec ^{2} x}{\tan x}\)
Now, \( \int \frac{1}{\sin x \cos ^{3} x} d x=\int \tan x \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x\)
let \( \tan x=\mathrm{t} \)
\(\Rightarrow \sec ^{2} x \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{1}{\sin x \cos ^{3} x} d x=\int t d t+\int \frac{1}{t} d t\)
\(\Rightarrow \frac{t^{2}}{2}+\log |t|+C\)
\(\Rightarrow\frac{1}{2} \tan ^{2} x+\log |\tan x|+\mathrm{C}\)

\(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}=\frac{\cos 2 x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x}=\frac{\cos 2 x}{1+\sin 2 x}\)
Now, \( \int \frac{\cos 2 x}{(\cos x+\sin x)^{2}} d x=\int \frac{\cos 2 x}{1+\sin 2 x} d x \)
Let \( 1+\sin 2 x=\mathrm{t} \)
\(\Rightarrow 2 \cos 2 xdx=\mathrm{dt}\)
Thus, \( \int \frac{\cos 2 x}{(\cos x+\sin x)^{2}} d x=\frac{1}{2} \int \frac{1}{t} d t \)
\(\Rightarrow \frac{1}{2} \log |t|+C\)
\(\Rightarrow \frac{1}{2} \log |1+\sin 2 x|+C\)
\(\Rightarrow \frac{1}{2} \log \left|(\cos x+\sin x)^{2}\right|+C\)
\(=\log |\sin x+\cos x|+C\)

\(\sin ^{-1(\cos x)}\)
Let \( \cos x=\mathrm{t} \quad\ldots \text{(1)}\)
Then, \( \sin x=\sqrt{1-t^{2}} \)
Differentiating both sides of (1), we get,
\(\Rightarrow(-\sin x) dx=\mathrm{dt}\)
\(\Rightarrow dx=\frac{-d t}{\sin x}\)
\(\Rightarrow dx=\frac{-d t}{\sqrt{1-t^{2}}}\)
\(\Rightarrow \text { Now, } \int \sin ^{-1}(\cos x) d x=\int \sin ^{-1} t\left(\frac{-d t}{\sqrt{1-t^{2}}}\right)\)
\(\Rightarrow\int \frac{\sin ^{-1} t}{\sqrt{1-t^{2}}} d t\)
Let \( \sin ^{-1} \mathrm{t}=v \)
\(\Rightarrow \frac{-d t}{\sqrt{1-t^{2}}}=d v\)
\(\Rightarrow \int \sin ^{-1}(\cos x) dx=-\int v \mathrm{d}v\)
\(\Rightarrow-\frac{v^{2}}{2}+C\)
\(\Rightarrow-\frac{\left(\sin ^{-1} t\right)^{2}}{2}+C\)
\(=-\frac{\left(\sin ^{-1}(\cos x)\right)^{2}}{2}+C \quad\ldots \text{(2)}\)
We know that,
\(\sin -1 x+\cos -1 x=\frac{\pi}{2}\)
\(\Rightarrow \sin ^{-1}(\cos x)=\frac{\pi}{2}-\cos ^{-1}(\cos x)=\left(\frac{\pi}{2}-x\right)\)
Now, substituting in eq (2), we get,
\(\Rightarrow \int \sin ^{-1}(\cos x) d x=\frac{-\left[\frac{\pi}{2}-x\right]^{2}}{2}+C\)
\(\Rightarrow-\frac{1}{2}\left(\frac{\pi^{2}}{2}-x^{2}-\pi x\right)+C\)
\(\Rightarrow-\frac{\pi^{2}}{8}-\frac{x^{2}}{2}+\frac{1}{2} \pi x+C\)
\(\Rightarrow\frac{1}{2} \pi x-\frac{x^{2}}{2}+\left(C-\frac{\pi^{2}}{8}\right)\)
\(\Rightarrow\frac{1}{2} \pi x-\frac{x^{2}}{2}+C_{1}\)

\(\frac{1}{\cos (x-a) \cos (x-b)}=\frac{1}{\sin (a-b)}\left[\frac{\sin (a-b)}{\cos (x-a) \cos (x-b)}\right]\)
\(\Rightarrow\frac{1}{\sin (a-b)}\left[\frac{\sin [(x-b)-(x-a)]}{\cos (x-a) \cos (x-b)}\right]\)
\(\Rightarrow\frac{1}{\sin (a-b)}\left[\frac{\sin (x-b) \cos (x-a)-\cos (x-b) \sin (x-a)}{\cos (x-a) \cos (x-b)}\right]\)
\(\Rightarrow\frac{1}{\sin (a-b)}[\tan (x-b)-\tan (x-a)]\)
Now, \( \int \frac{1}{\cos (x-a) \cos (x-b)} d x=\frac{1}{\sin (a-b)} \int[\tan (x-b)-\tan (x-a)] d x \)
\(\Rightarrow\frac{1}{\sin (a-b)}[-\log |\cos (x-b)|+|\log \cos (x-a)|]\)
\(\Rightarrow\frac{1}{\sin (a-b)}\left[\log \left|\frac{\cos (x-a)}{\cos (x-b)}\right|\right]+C\)

\(\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x=\int\left(\frac{\sin x}{\sin ^{2} x \cos ^{2} x}-\frac{\cos x}{\sin ^{2} x \cos ^{2} x}\right) d x\)
\(=\int\left(\sec ^{2} x-\operatorname{cosec}^{2} x\right) d x\)
We know that,
\(\int \sec ^{2} x d x=\tan x+c\)
\(\int \operatorname{cosec}^{2} x d x=-\cot x+c\)
\(=\tan x+\cot x+C \cdot A \text { Answer }\)

Let \( x . e^{x}=\mathrm{t} \)
Differentiating both sides we get,
\(\Rightarrow\left(e^{x} \cdot x+e^{x} \cdot 1\right) \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow e^{x}(x+1)=\mathrm{dt}\)
\(\text {Now, } \int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} d x=\int \frac{d t}{\cos ^{2} t}\)
\(=\int \sec ^{2} t \cdot d t\)
\(=\tan \mathrm{t}+\mathrm{C}\)
\(=\tan \left(e^{x} \cdot x\right)+\mathrm{C}\)