Ex 7.2 Class 12 Maths Ncert Solutions

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

Let \( \log |x|=\mathrm{t} \)
\(\Rightarrow \frac{1}{x} d x=d t\)
Now,
\(\Rightarrow \frac{t^{3}}{3}+C\)
\(\Rightarrow \frac{(\log |x|)^{3}}{3}+C\)

\(\frac{1}{x+x \log x}=\frac{1}{x(1+\log x)}\)
\(\text {let } 1+\log x=\mathrm{t}\)
\(\Rightarrow \frac{1}{x} d x=d t\)
\(\Rightarrow \int \frac{1}{x(1+\log x)} d x=\int \frac{1}{t} d t\)
\(=\log |\mathrm{t}|+\mathrm{C}\)
\(=\log |1+\log x|+\mathrm{C}\)

Let \( \cos x=\mathrm{t} \)
\(\Rightarrow-\sin x \mathrm{~d} x=\mathrm{dt}\)
\(\Rightarrow \int \sin x \cdot \sin (\cos x) d x=-\int \sin t d t\)
\(=-[-\cos t]+C\)
\(=\cos t+C\)
\(=\cos (\cos x)+C\)

Let \( \mathrm{I}=\int \sin (a x+b) \cos (a x+b) d x \)
We know that,
\( \sin 2 \mathrm{A}=2 \sin \mathrm{A} \cdot \cos \mathrm{A} \)
Therefore, \( \sin (a x+b) \cos (a x+b) \)
\(= \frac{2 \sin (a x+b) \cos (a x+b)}{2}=\frac{\sin 2(a x+b)}{2} \)
Let \( 2(a x+b)=t \)
\(\Rightarrow 2 a d x=d t\)
\(\Rightarrow \int \frac{\sin 2(a x+b)}{2} d x=\frac{1}{2} \int \frac{\sin t}{2 a} d t\)
\(=\frac{1}{4 a}[-\cos t]+C\)
\(=-\frac{1}{4 a} \cos 2(a x+b)+C\)

Let \( a x+b=t \)
\(\Rightarrow a d x=d t\)
\(\Rightarrow d x=\frac{1}{a} d t\)
\(\Rightarrow \int(a+b)^{\frac{1}{2}}=\frac{1}{a} \int t^{\frac{1}{2}} d t\)
\(=\frac{1}{a}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\)
\(=\frac{2}{3 a}(a+b)^{\frac{3}{2}}+C\)

Let \( (x+2)=t \)
\(\Rightarrow d x=d t\)
\(\Rightarrow \int x \sqrt{x+2} d x=\int(t-2) \sqrt{t} d t\)
\(= \int\left(t^{\frac{3}{2}}-2 t^{\frac{t}{2}}\right) d t\)
\(=\int t^{\frac{3}{2}}-2 \int t^{\frac{t}{2}} d t\)
\(=\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\)
\(=\frac{2}{5} t^{\frac{5}{2}}-\frac{4}{3} t^{\frac{3}{2}}+C\)
\(=\frac{2}{5}(x+2)^{\frac{5}{2}}-4(x+2)^{\frac{3}{2}}+C\)

Let \( 1+2 x^{2}=t \)
\(\Rightarrow 4 x \mathrm{~d} x=\mathrm{dt}\)
\(\Rightarrow\int x \sqrt{x+2 x^{2}} d x=\int \frac{\sqrt{t} d t}{4}\)
\(=\frac{1}{4} \int t^{\frac{1}{2}} d t\)
\(
=\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\)
\(=\frac{1}{6}\left(1+2 x^{2}\right)^{\frac{3}{2}}+C\)

Let \( x^{2}+x+1=\mathrm{t} \)
Differentiating both sides, we get,
\(\Rightarrow(2 x+1) \mathrm{d}x=\mathrm{tdt}\)
Therefore,
\(\Rightarrow \int(4 x+2) \sqrt{x^{2}+x+1} d x=\int 2 \sqrt{t} d t\)
\(=2 \int \sqrt{t} d t\)
\(=2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\)
\(=\frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+C\)

\(\frac{1}{x-\sqrt{x}}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}\)
Now, Let \( (\sqrt{x}-1)=t \)
\(\Rightarrow\frac{1}{2 \sqrt{x}} d x=d t\)
\(\Rightarrow\int \frac{1}{2 \sqrt{x}} d x=d t\)
\(\Rightarrow\int \frac{1}{\sqrt{x}(\sqrt{x}-1)} d x=\int \frac{2}{t} d t\)
\(=2 \log |t|+C\)
\(=2 \log |\sqrt{x}-1|+\mathrm{C}\)

\(\text { Let } x+4=\mathrm{t}\)
\(\Rightarrow d x=d t\)
\(\Rightarrow \int \frac{x}{\sqrt{x+4}} d x=\int \frac{(t-4)}{\sqrt{t}} d t\)
\(=\int\left(\sqrt{t}-\frac{4}{\sqrt{t}}\right) d t\)
\(=\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+C\)
\(=\frac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C\)
\(=\frac{2}{3} t \cdot t^{\frac{1}{2}}-8(t)^{\frac{1}{2}}+C\)
\(=\frac{2}{3} t^{\frac{1}{2}}(t-12)+C\)
\(=\frac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C\)
\(=\frac{2}{3} \sqrt{x+4}(x-8)+C\)

Let \( x^{3}-1=\mathrm{t} \)
\(\Rightarrow 3 x^{2} \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int\left(x^{3}-1\right)^{\frac{1}{3}} x^{5} d x=\int\left(x^{3}-1\right)^{\frac{1}{3}} x^{3} \cdot x^{2} d x\)
\(=\int t^{\frac{1}{3}}(t+1) \frac{d t}{3}\)
\(=\frac{1}{3} \int\left(t^{\frac{4}{3}}+t^{\frac{1}{3}}\right) d t\)
\(=\frac{1}{3}\left[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right]+C\)
\(=\frac{1}{3}\left[\frac{3}{7} t^{\frac{7}{3}}+\frac{3}{4} t^{\frac{4}{3}}\right]+C\)
\(=\frac{1}{7}\left(x^{3}-1\right)^{\frac{7}{3}}+\frac{1}{4}\left(x^{3}-1\right)^{\frac{4}{3}}+C\)

Let \( 2+3 x^{3}=1 \)
\(\Rightarrow 9 x^{2} \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{x^{2}}{\left(2+3 x^{3}\right)^{3}} d x=\frac{1}{9} \int \frac{d t}{t^{3}}\)
\(=\frac{1}{9}\left[\frac{t^{-2}}{-2}\right]+C\)
\(=\frac{-1}{18}\left(\frac{1}{t^{2}}\right)+C\)
\(=\frac{-1}{18\left(2+3 x^{3}\right)^{2}}+C\)

Let \( \log x=t \)
\(=\frac{1}{x} d x=d t\)
\(=\int \frac{1}{x(\log x)^{m}} d x=\int \frac{d t}{t^{m}}\)
\(=\left(\frac{t^{-m+1}}{1-m}\right)+C\)
\(=\frac{(\log x)^{1-m}}{(1-m)}+C\)

Let \( 9-4 x^{2}=\mathrm{t} \)
\(=-8 x\mathrm{d}x=\mathrm{dt}\)
\(= \int \frac{x}{9-4 x^{2}} d x\)
\(=\frac{-1}{8} \int \frac{1}{t} d t\)
\(=\frac{-1}{8} \log |t|+\mathrm{C}\)
\(=\frac{-1}{8} \log \left|9-4 x^{2}\right|+C\)

Let \( 2 x+3=\mathrm{t} \)
\(= 2 d x=d t\)
\(=\int e^{2 x+3} d x=\frac{1}{2} \int e^{t} d t\)
\(=\frac{1}{2} e^{t}+C\)
\(=\frac{1}{2} e^{2 x+3}+C\)

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

\(\text {let } \tan -1 x=\mathrm{t}\)
\( \frac{1}{1+x^{2}} d x=d t\)
\(=\int \frac{\mathrm{e}^{\tan ^{-1} }x}{1+x^{2}} d x=\int e^{t} d t\)
\(=e^{t}+C\)
\(=\mathrm{e}^{\tan ^{-1}}+C\)

We have,
\(\frac{e^{2 x}-1}{e^{2 x}+1}\)
Dividing numerator and denominator by ex, we get,
\(\frac{\frac{e^{2 x}-1}{e}}{\frac{e^{2 x}+1}{e}}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\)
Let \( \mathrm{e}^{x}+\mathrm{e}^{-x}=\mathrm{t} \)
Differentiating both sides, we get,
\(\left(e^{x}-\mathrm{e}^{-x}\right) \mathrm{d}x=\mathrm{dt}\)
Now the integral becomes,
\(=\int \frac{d t}{t}\)
\(=\log |t|+C\)
\(=\log \left|e^{x}+e^{-x}\right|+C\)

Let \( e^{2 x}+e^{-2 x}=t \)
\(\Rightarrow\left(2 e^{2 x}-2 e^{-2 x}\right) d x=d t\)
\(\Rightarrow2\left(e^{2 x}-e^{-2 x}\right) d x=d t\)
\(=\int \frac{e^{2 x} \pm}{e^{2 x}+e^{-2 x}} d x\)
\(=\int \frac{d t}{2 t}\)
\(=\frac{1}{2} \int \frac{d t}{2 t}\)
\(=\frac{1}{2} \log |t|+\mathrm{C}\)
\(=\frac{1}{2} \log \left|e^{2 x}+e^{-2 x}\right|+\mathrm{C}\)

\(\tan ^{2}(2 x-3)=\sec ^{2}(2 x-3)-1\)
Let \( 2 x-3=\mathrm{t} \)
\(\Rightarrow 2 \mathrm{d}x=\mathrm{dt}\)
\(=\int \tan ^{2}(2 x-3) d x\)
\(=\int\left[\sec ^{2}(2 x-3)-1\right] d x\)
\(=\frac{1}{2} \int\left(\sec ^{2} t\right) d t-\int 1 . d t\)
\(=\frac{1}{2} \int \sec ^{2} t d t-\int 1 . d t\)
\(=\frac{ 1 }{ 2 } \tan \mathrm{t}-\mathrm{t}+\mathrm{C} \)
\(= \frac{ 1 }{ 2 } \tan (2 x-3)-(2 x-3)+\mathrm{C}\)

Let \( 7-4 x=\mathrm{t} \)
\(\Rightarrow-4 \mathrm{d}x=\mathrm{dt}\)
\(=\int \sec ^{2}(7-4 x) d x\)
\(=\frac{-1}{4} \int \sec ^{2} t d t\)
\(=\frac{-1}{4}(\tan t)+C\)
\(=\frac{-1}{4} \tan (7-4 x)+C\)

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

\(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}=\frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}\)
\(\text {let } 3 \cos x+2 \sin x=\mathrm{t}\)
\((-3 \sin x+2 \cos x) \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x=\int \frac{d t}{2 t}\)
\(=\frac{1}{2} \int \frac{1}{t} d t\)
\(=\frac{1}{2} \log |t|+\mathrm{C}\)
\(=\frac{1}{2} \log |2 \sin x+3 \cos x|+\mathrm{C}\)

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

Let \( \sqrt{x}=t \)
\(\Rightarrow=\frac{1}{2 \sqrt{x}} d x=d t\)
\(=\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x\)
\(=2 \int \cos t d t\)
\(=2 \sin t+C\)
\(=2 \sin +C\)

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

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

Let \( \log \sin x=\mathrm{t} \)
\(\Rightarrow\frac{1}{\sin x} \cdot \cos x d x=d t\)
\(\Rightarrow \cot x d x=d t\)
\(\Rightarrow\int \cot x \log \sin x d x=\int t d t\)
\(=\frac{t^{2}}{2}+C\)
\(=\frac{1}{2}(\log \sin x)^{2}+C\)

Let \( 1+\cos x=\mathrm{t} \)
\(\Rightarrow-\sin x\mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow\int \frac{\sin x}{1+\cos x}=\int-\frac{d t}{t}\)
\(=-\log |t|+C\)
\(=-\log |1+\cos x|+\mathrm{C}\)

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

Let \( \mathrm{I}=\int \frac{1}{1+\cot x} d x \)
\(=\int \frac{1}{1+\frac{\cos x}{\sin x}} d x\)
\(=\int \frac{\sin x}{\sin x+\cos x} d x\)
\(=\frac{1}{2} \int \frac{2 \sin x}{\sin x+\cos x} d x\)
\(=\frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin x+\cos x} d x\)
\(=\frac{1}{2} x+\frac{1}{2} \int \frac{(\sin x-\cos x)}{\sin x+\cos x} d x\)
Let \( \sin x+\cos x=\mathrm{t} \)
\(\Rightarrow(\cos x-\sin x) \mathrm{d}x=\mathrm{dt}\)
Therefore, \( \mathrm{I}=\frac{x}{2}+\frac{1}{2} \int \frac{-d t}{t} \)
\(=\frac{x}{2}-\frac{1}{2} \log |t|+C\)
\(=\frac{x}{2}-\frac{1}{2} \log |\sin x+\cos x|+C\)

Let \( \mathrm{I}= \)
\(\Rightarrow \int \frac{1}{1-\tan x} d x\)
\(\Rightarrow\int \frac{1}{1-\frac{\sin x}{\cos x}} d x\)
\(\Rightarrow\int \frac{\cos x}{\cos x-\sin x} d x\)
\(\Rightarrow\frac{1}{2} \int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{\cos x-\sin x} d x\)
\(\Rightarrow\frac{1}{2} \int 1 \cdot d x+\frac{1}{2} \int \frac{(\cos x+\sin x)}{\cos x-\sin x} d x\)
\(\Rightarrow\frac{1}{2} x+\frac{1}{2} \int \frac{(\cos x+\sin x)}{\cos x-\sin x} d x\)
Let \( \cos x-\sin x=\mathrm{t} \)
\(\Rightarrow(-\sin x-\cos x) d x=d t\)
Therefore, \( \mathrm{I}=\frac{x}{2}+\frac{1}{2} \int \frac{-d t}{t} \)
\(\Rightarrow\frac{x}{2}-\frac{1}{2} \log |t|+C\)
\(\Rightarrow\frac{x}{2}-\frac{1}{2} \log |\cos x-\sin x|+C\)

Let \( \mathrm{I}=\int \frac{\sqrt{\tan x}}{\sin x \cos x} \)
\( \Rightarrow\int \frac{\sqrt{\tan x} \cdot \cos x}{\sin x \cos x \cdot \cos x} d x \)
\( \Rightarrow\int \frac{\sqrt{\tan x}}{\tan x \cos ^{2} x} d x \)
\( \Rightarrow\int \frac{\sec ^{2} x d x}{\sqrt{\tan x}} d x \)
Let \( \tan x=\mathrm{t} \)
\(\Rightarrow \sec ^{2} x \mathrm{d}x=\mathrm{dt}\)
\(\Rightarrow \mathrm{I}=\int \frac{d t}{\sqrt{t}}\)
\(=2 \sqrt{t}+C\)
\(=2 \sqrt{\tan x}+C\)

let \( 1+\log x=\mathrm{t} \)
\(\Rightarrow \frac{1}{x} d x=d t\)
\(\Rightarrow\int \frac{(1+\log x)^{2}}{x}=\int t^{2} d t\)
\(\Rightarrow\frac{t^{3}}{3}+C\)
\(\Rightarrow\int \frac{(1+\log x)^{3}}{3}+C\)

\(\frac{(x+1)(x+\log x)^{2}}{x}\)
\(=\left(\frac{x+1}{x}\right)(x+\log x)^{2}\)
\(=\left(1+\frac{1}{x}\right)(x+\log x)^{2}\)
Let \( (x+\log x)=\mathrm{t} \)
\(\Rightarrow\left(1+\frac{1}{x}\right) d x=d t\)
\(\Rightarrow\int\left(1+\frac{1}{x}\right)(x+\log x)^{2} d x=\int t^{2} d t\)
\(\Rightarrow\frac{t^{3}}{3}+C\)
\(\Rightarrow\frac{1}{3}(1+\log x)^{3}+C\)

Let \( x^{4}=\mathrm{t} \)
\( 4 x^{3} \mathrm{d}x=\mathrm{dt}\)
\(=\int \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}} d x\)
\(=\frac{1}{4} \int \frac{\sin \left(\tan ^{-1} t\right)}{1+t^{2}} d t\)
Let \( \tan -1=\mathrm{v} \)
\( \frac{1}{1+t^{2}} d t=d v\)
Thus, we get,
\(=\int \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}} d x\)
\(=\frac{1}{4} \int \sin v d v\)
\(=\frac{1}{4}(-\cos v)+C\)
\(=\frac{-1}{4} \cos \left(\tan ^{-1} t\right)+C\)
\(=\frac{-1}{4} \cos \left(\tan ^{-1} x^{4}\right)+C\)

Let \( x^{10}+10^{x}=\mathrm{t} \)
\(\left(10 x^{9}+10^{x} \log _{e} 10\right) d x=d t\)
\(=\int \frac{10 x^{9}+10^{x} \log _{e} 10 d x}{x^{10}+10^{x}} d x=\int \frac{d t}{t}\)
\(=\log \mathrm{t}+\mathrm{C}\)
\(=\log \left(x^{10}+10^{x}\right)+\mathrm{C}\)

Let \( \mathrm{I}=\int \frac{d x}{\sin ^{2} x \cos ^{2} x}=\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x \)
\(=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x\)
\(=\int \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x\)
\(=\int \sec ^{2} x d x+\int \operatorname{cosec}^{2} x d x\)
\(=\tan x+\cot x+\mathrm{C}\)