Exercise 19.9
Question 1
$\int \sqrt{1-x^{2}} d x$
Sol :
putting x=sinθ then dx=cosθdθ
Now , $\int \sqrt{1-x^{2}} d x$
$=\int \sqrt{1-\sin ^{2} \theta} \cos \theta d \theta$
$=\int \cos \theta \cdot \cos \theta d \theta$
$=\int \cos ^{2} \theta d \theta$
$=\int \frac{1+\cos 2 \theta}{2} d \theta$
$=\int \frac{1}{2} d \theta+\frac{1}{2} \int \cos 2 \theta d \theta$
$=\frac{1}{2} \theta+\frac{1}{2}\times \frac{1}{2} \sin 2 \theta+c$
$=\frac{1}{2}\left[0+\frac{1}{2} \sin 2 \theta\right]+c$
$=\frac{1}{2}\left[\sin ^{2} x+\frac{1}{2} \cdot 2 \sin \theta \cdot \cos \theta\right]+c$
$=\frac{1}{2}\left[\sin^{-1} x+x \cdot \sqrt{1-x^{2}}\right]+c$
Question 2
$\int \frac{d x}{\sqrt{1-x^{2}}}$
Sol :
putting x=sinθ then dz=cosθdθ
also , $\theta=\sin ^{-1} x$
Now , $\int \frac{d x}{\sqrt{1-x^{2}}}=\int \frac{\cos \theta d \theta}{\sqrt{1-\sin ^{2} \theta}}$
$=\int \frac{cos \theta d \theta}{\cos \theta}=\int d \theta=\theta+c$
$=\sin ^{-1} x+c$
Question 3
$\int \frac{d x}{x^{2} \sqrt{a^{2}-x^{2}}}$
Sol :
putting x=asinθ then dx=a.cosθdθ
x2=a2sin2θ
Now , $\int \frac{d x}{x^{2} \sqrt{a^{2}-x^{2}}}$
$\int \frac{x+1}{\sqrt{1+x^{2}}} d x$
Sol :
putting x=tanθ then $d x=\sec ^{2} \theta d \theta$
$\therefore \sec \theta=\sqrt{1+\tan ^{2} \theta}=\sqrt{1+x^{2}}$
Now , $\int \frac{x+1}{\sqrt{1+x^{2}}} d x=\int \frac{\tan \theta+1}{\sqrt{1+\tan ^{2} \theta}} \cdot \sec ^{2} \theta d \theta$
$=\int \frac{(\tan \theta+1) \cdot \sec ^{2} \theta}{\sec \theta} d \theta$
$=\int(\tan \theta+1) \sec \theta d \theta$
$=\int \sec \theta \cdot \tan \theta d \theta+\int \sec \theta d \theta$
$=\sec \theta+\log (\sec \theta+\tan \theta)+c$
$\sqrt{1+x^{2}}+\log (x+\sqrt{1+x^{2}})+c$
Sol :
putting $x^{2}=\tan \theta$ then $2 x d x=\sec ^{2} \theta d \theta$
$\Rightarrow x d x=\frac{\sec ^{2} \theta}{2} d \theta$
Now , $\int \frac{x}{1+x^{4}} d x=\int \frac{x d x}{1+\left(x^{2}\right)^{2}}$
$=\int \frac{\sec ^{2} \theta d \theta}{2\left(1+\tan ^{2} \theta\right)}$
$=\frac{1}{2} \int \frac{\sec ^{2} \theta}{\sec ^{2} \theta} d \theta$
$=\frac{1}{2} \int d \theta=\frac{1}{2} \theta+c$
$=\frac{1}{2} \tan ^{-1} x^{2}+c$
Sol :
putting $x^{6}=\tan \theta$ then $6 \cdot x^{5} d x=\sec ^{2} \theta d \theta$
$x^{5} d x=\frac{\sec ^{2} \theta}{6} d \theta$
Now ,$\int \frac{x^{5}}{1+x^{12}} d x=\int \frac{x^{5} d x}{1+x^{6} \cdot x^{6}}$
$=\int \frac{\sec^2 \theta}{6\left(1+\tan ^{2} \theta\right)}$
$=\frac{1}{6} \int \frac{\sec ^{2} \theta}{\sec ^{2} \theta} d \theta$
$=\frac{1}{6} \int d \theta=\frac{1}{6} \theta+c$
$=\frac{1}{6} \tan ^{-1}\left(x^{6}\right)+c$
Sol :
putting x=tanθ then $d x=\sec ^{2} \theta d \theta$
Now , $\int \frac{d x}{x^{2} \sqrt{1+x^{2}}}=\int \frac{\sec ^{2} \theta}{\tan ^{2} \theta \cdot \sqrt{1+\tan ^{2} \theta}}$
$=\int \frac{\sec ^{2} \theta}{\tan ^{2} \theta \cdot \sec \theta} d \theta$
$=\int \frac{\sec \theta}{\tan ^{2} \theta} d \theta$
$=\int \frac{1}{\cos \theta \cdot \frac{\sin^2 \theta}{\cos ^{2} \theta}} d \theta$
$=\int \frac{\cos \theta}{\sin^2 \theta} d \theta$
Let z=sinθ then dz=cosθdθ
Now , $\int \frac{\cos \theta}{\sin ^2 \theta} d \theta=\int \frac{d z}{z^{2}}$
$=\int z^{-2} d z=\frac{z^{-2+1}}{-2+1}+0$
$=-z^{-1}+c$
$=\frac{-1}{z}+c=-\frac{1}{\sin \theta}+c$
$=-\frac{1}{\frac{x}{\sqrt{1+x^{2}}}}+c$
$=\frac{-\sqrt{1+ x^{2}}}{x}+c$
$=-\frac{\sqrt{1+x^{2}}}{x}+c$
Sol :
putting $e^{x}=\tan \theta$ then
$e^{x} d x=\sec ^{2} \theta d \theta$
$d x=\frac{\sec ^{2} \theta d \theta}{\tan \theta}$
Now , $\int \frac{d x}{1+e^{2 x}}=\int \frac{\sec ^{2} \theta d \theta}{\tan \theta \cdot\left(1+\tan ^{2} \theta\right)}$
$=\int \frac{\sec ^{2} \theta}{\tan \theta \cdot \sec ^{2} \theta} d \theta$
$=\int \frac{1}{\tan \theta} d \theta=\int \frac{\cos \theta}{\sin \theta} d \theta$
=log(sin θ)+c
$\because \tan \theta=e^{x}$ then $\sin \theta=\frac{e^{x}}{\sqrt{1+e^{b x}}}$
Now , log(sin θ)+c
$=\log \left(\frac{e^{x}}{\sqrt{1+e^{2 x}}}\right)+c$
$\because \log \left(\frac{m}{n}\right)=\log m-log n$
$=\log e^{x}-\log (\sqrt{1+e^{2x}})+c$
$=x-\log \left(1+e^{2x}\right)^{\frac{1}{2}}+c$
$=x-\frac{1}{2} \log \left(1+e^{2 x}\right)+c$
$\because \log m^{n}=n\log m$
$x=\frac{1}{2} \log \left(1+e^{2 x}\right)$
Sol :
putting $e^{2x}=\tan \theta$ then $2 e^{2 x} d x=\sec ^{2} \theta d \theta$
$e^{2x} d x=\frac{\sec ^{2} \theta}{2} d \theta$
Now , $\int \frac{e^{2x} d x}{1+e^{4 x}}=\int \frac{e^{2 x}}{1+e^{2 x} \cdot e^{2 x}} d x$
$=\int \frac{sec^{2} \theta}{2\left(1+\tan ^{2} \theta\right)} d \theta$
$=\frac{1}{2} \int \frac{\sec ^{2} \theta}{\sec ^{2} \theta} d \theta$
$=\frac{1}{2} \int d \theta=\frac{\theta}{2}+c$
$=\frac{1}{2} \tan ^{-1}\left(e^{2 x}\right)+c$
Sol :
$=\int \frac{e^{\left(\frac{x}{2}\right)^{2}}-1}{e^{(\frac{x}{2})^{2}}-1} d x$
putting $e^{\frac{x}{2}}=\tan \theta$ then
$\frac{1}{2} e^{\frac{x}{2}} d x=\sec ^{2} \theta d \theta$
$d x=\frac{2 \sec ^{2} \theta d \theta}{\tan \theta}$
Now , $\int \frac{\tan ^{2} \theta-1}{\tan ^{2} \theta+1} \cdot \frac{2 \sec ^{2} \theta}{\tan \theta} d \theta$
$=2 \int \frac{\tan ^{2} \theta-1}{\tan \theta} d \theta$
$=2 \int \frac{\tan ^{2} \theta}{\tan \theta}-2 \int \frac{1}{\tan \theta} d \theta$
$=2 \int \tan \theta d \theta-2 \int \frac{\cos \theta}{\sin \theta} d \theta$
$=2 \log (\sec \theta)-2 \log (\sin \theta)+c$
$=2 \log \left(\frac{\sec \theta}{\sin \theta}\right)+c$
∴$\tan \theta=e^{\frac{\pi}{2}}$ then
$\sec \theta=\sqrt{1+\tan ^{2} \theta}=\sqrt{1+e^{x}}$
also , $\sin \theta=\frac{e^{x / 2}}{\sqrt{1+e^{x}}}$
Now , $2 \log \left(\frac{\sqrt{1+e^{x}}}{\frac{e^{x / 2}}{\sqrt{1+e^{x}}}}\right)+c$
$=2 \log \left(\frac{1+e^{x}}{e^{x / 2}}\right)+c$
=$2 \log \left(1+e^{x}\right)-2 \log \left(e^{x / 2}\right)+c$
=$2 \log \left(1+e^{x}\right)-2 \times \frac{x}{2}+c$
$=2 \log \left(1+e^{x}\right)-x+c$
Sol :
putting x=secθ then dx=secθ.tanθ.dθ
Now , $\int \frac{d x}{x \cdot \sqrt{x^{2}-1}}=\int \frac{\sec \theta \cdot \tan \theta d \theta}{\sec \theta \cdot \sqrt{\sec ^{2} \theta-1}}$
$=\int \frac{\sec \theta.\tan \theta}{\sec \theta \cdot \tan \theta} d \theta$
$=\int d \theta$
$=\theta+c=\sec ^{-1} x+c$
Sol :
putting x=asecθ then dx=asecθ.tanθdθ
Now , $\int \frac{x^{3}}{\left(x^{2}-a^{2}\right)^{3 / 2}} d x$
$=\int \frac{a^{3} \sec ^{3} \theta \cdot a \cdot \sec \theta \cdot \tan \theta}{\left(a^{2} \sec ^{2} \theta-a^{2}\right)^{3 / 2}} d \theta$
$\Rightarrow \int \frac{a^{4} \cdot \sec ^{4} \theta \cdot \tan \theta d \theta}{\left\{a^{2}\left(\sec ^{2} \theta-1\right)^{3/ 2}\right.}$
$=\int \frac{a^{4} \cdot \sec ^{4} \theta \cdot \tan \theta d \theta}{a^{2 \times \frac{3}{2}}+\tan \theta^{2\times \frac{3}{2}}}$
$=\int \frac{a^{4} \sec ^{4} \theta \cdot \tan \theta d \theta}{a^{3} \cdot \tan ^{3} \theta}$
$=\int \frac{a \cdot \sec ^{4} \theta}{\tan ^{2} \theta} d \theta$
$=\int \frac{a}{\cos ^{4} \theta \cdot \frac{\sin ^{2} \theta}{\cos ^{2} \theta}}d\theta$
$=\int \frac{a}{\sin ^{2} \theta \cdot \cos ^{2} \theta} d \theta$
$=\int \frac{4 a}{4 \sin ^{2} \theta \cdot \cos ^{2} \theta} d \theta$
$=\int \frac{4 a}{(2 \sin \theta \cdot \cos \theta)^{2}}$
$=4 a \int \frac{d \theta}{(\sin 2 \theta)^{2}}$
$=4 a \int \frac{d \theta}{\sin ^{2} 2 \theta}$
$=4 a \int \operatorname{cosec}^{2} 2 \theta d \theta$
$=\frac{4 a(-cot2\theta)}{2}+c$
$=\frac{-4 a}{2} \cot 2 \theta+c$
$=\frac{-4 a}{2 \tan 2\theta}+c$
$=\frac{-4a}{2.\frac{2 \tan \theta}{1-\tan ^{2} \theta}} d \theta+c$
$=\frac{-a\left(1-\tan ^{2} \theta\right)}{\tan \theta}+c$
$\because \sec \theta=\frac{x}{a}$
then $\tan \theta=\sqrt{\sec ^{2} \theta-1}$
$=\sqrt{\frac{x^{2}}{a^{2}}-1}$
$=\frac{\sqrt{x^{2}-a^{2}}}{a}$
Now , $\frac{-a\left(1-\tan ^{2} \theta\right)}{\tan \theta}+c$
$=\frac{-a\left(1-\frac{x^{2}-a^{2}}{a^{2}}\right)^{2}}{\frac{\sqrt{x^{2}-a^{2}}}{a}}+c$
$\frac{-a\left(a^{2}-x^{2}+a^{2}\right)}{\frac{a^{2}}{\frac{\sqrt{x^{2}+a^{2}}}{a}}}+c$
$=\frac{-\left(2a^{2}-x^{2}\right)}{\sqrt{x^{2}-a^{2}}}+c$
$\frac{x^{2}-2a^{2}}{\sqrt{x^{2}-a^2}}+c$
Sol :
putting $e^{x}=3 \sec \theta$ then $e^{x} d x=3 \sec \theta \cdot \tan \theta d \theta$
Now , $\int \frac{e^{x}}{\sqrt{e^{2 x}-9}} d x=\int \frac{3 \sec \theta \cdot \tan \theta d \theta}{\sqrt{9 \sec ^{2} \theta-9}}$
$=\int \frac{3 \sec \theta \cdot \tan \theta d \theta}{\sqrt{9\left(\sec ^{2} \theta-1\right)}}$
$=\int \frac{3 \sec \theta \cdot \tan \theta d \theta}{3 \tan \theta}$
$=\int \sec \theta d \theta$
$=\log |\sec \theta+\tan \theta|+c$
$=\log \left(\frac{e^{x}}{3}+\frac{\sqrt{e^{2 x}-9}}{3}\right)$
$=\log \left(\frac{e^{x}+\sqrt{e^{2 x}-9}}{3}\right)+c$
$=\log \left(e^{x}+\sqrt{e^{2 x}-9}\right)+\log 3+c$
$\begin{aligned}\left(\because \log \frac{m}{n}\right.&=\log m +\log n) \end{aligned}$
$=\log \left(e^{x}+\sqrt{e^{2 x}-9}\right)+c$
Sol :
putting $e^{x}=\sec \theta$ then $e^{x} d x=\sec \theta \cdot \tan \theta d \theta$
$d x=\frac{\sec \theta \cdot \tan \theta}{e^{x}} d \theta$
Now , $\int \frac{d x}{\sqrt{e^{2 x}-1}}=\int \frac{\sec \theta \cdot \tan \theta}{e^{x} \sqrt{\sec ^{2} \theta-1}} d \theta$
$=\int \frac{\sec \theta \cdot \tan \theta}{\sec \theta \cdot \tan \theta} d \theta$
$=\int d \theta=\theta+c$
$=\sec ^{-1}\left(e^{x}\right)+c$
Sol :
putting $x=a \cos 2 \theta$ then
$d x=2 a \cdot(-\sin 2 \theta)$
$d x=-2 a \sin 2 \theta d \theta$
Now , $\int \sqrt{\frac{a-a \cos 2\theta}{a+a \cos 2 \theta}}(-2 a \sin 2 \theta d \theta)$
$=\int \sqrt{\frac{a(1-\cos 2 \theta)}{a(1+\cos 2 \theta)}} \times(-2 a \sin 2 \theta)d\theta$
$=\int \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}}(-2 a \cdot 2 \sin \theta \cdot \cos \theta) d \theta$
$=-4 a \int \frac{\sin \theta}{\cos \theta} \sin \theta \cdot \cos \theta d \theta$
$=-4\theta\int \sin ^{2} \theta d \theta$
$=-2 a \int 2 \sin ^{2} \theta d \theta$
$=-2 a \int(1-\cos 2 \theta) d \theta$
$=-2 a \int d \theta+2 a \int \cos 2 \theta d \theta$
$=-2 a \theta+2 a \frac{\sin 2 \theta}{2}+c$
$=-2 a \theta+a \sin 2 \theta+c$
$-2 a \times \frac{1}{2} \cos ^{-1} \frac{x}{a}+a \sin 2\theta+c$
$=-a \cos^{-1} \frac{x}{a}+a \cdot \frac{\sqrt{a^{2}-x^{2}}}{a}+c$
$=-a\left[\cos^{-1} \frac{x}{a}-\frac{\sqrt{a^{2}-x^{2}}}{a}\right]+c$
Sol :
putting $x=\cos 2 \theta$ then $d x=-\sin 2 \theta \cdot 2 d \theta$
$d x=-2 \sin 2\theta d \theta$
Now , $\int \sqrt{\frac{1+\cos 20}{1-\cos 2 \theta}}(-2 \sin 2 \theta) d \theta$
$=\displaystyle\int\sqrt{\frac{2 \cos ^{2} \theta}{2 \sin ^{2} \theta}}(-2 \cdot 2 \sin \theta \cdot \cos \theta)d\theta$
$=\int \frac{\cos \theta}{\sin \theta} \times(-4 \sin \theta \cdot \cos \theta) d \theta$
$=-2 \int 2 \cos ^{2} \theta d \theta$
$=-2 \int(1+\cos 2 \theta) d \theta$
$=-2 \int d \theta-2 \int \cos 2 \theta$
$=-2 \theta-2 \frac{\sin 2 \theta}{2}+c$
$=-2 \times \frac{1}{2} \cos ^{-1} x-2 \frac{\sqrt{1-x^{2}}}{2}+c$
$=-\cos ^{-1} x-\sqrt{1-x^{2}}+c$
$=\int \frac{a \cos \theta d \theta}{a^{2} \sin ^{2} \theta \cdot \sqrt{a^{2}-a^{2} \sin ^{2} \theta}}$
$=\int \frac{a \cos \theta}{a^{2} \sin ^{2} \theta \sqrt{a^{2}\left(1-\sin ^{2} \theta\right)}}$
$=\int \frac{a \cos \theta d \theta}{a^{2} \sin ^{2} \theta \cdot a \cdot \cos \theta}$
$=\frac{1}{a^{2}} \int \operatorname{cosec}^{2} \theta d \theta$
$=\frac{1}{a^{2}}(-\cot \theta)+c$
$=-\frac{1}{a^{2}} \cot \theta+c$
$=-\frac{1}{a^{2}} \times \frac{\cos \theta}{\sin \theta}+c$..(i)
$\because \quad \sin \theta=\frac{x}{a}$
$\therefore \cos \theta=\sqrt{1-\sin ^{2} \theta}$
$=\sqrt{1-\frac{x^{2}}{a^{2}}}=\sqrt{\frac{a^{2}-x^{2}}{a^{2}}}$
$=\frac{\sqrt{a^{2}-x^{2}}}{a}$
Now , equation (i) becomes
$-\frac{1}{a^{2}} \frac{\sqrt{a^{2}-x^{2}}}{a.\frac{x}{a}}+c$
$=\frac{-1}{a^{2}} \frac{\sqrt{a^{2}-x^{2}}}{x}+c$
Question 4
$\int \frac{1+x^{2}}{\sqrt{1-x^{2}}} d x$
Sol :
putting x=sinθ then dx=cosθdθ , $\theta=\sin ^{-1} x$
Now , $\int \frac{1+x^{2}}{\sqrt{1-x^{2}}} d x$
$=\int \frac{1+\sin ^{2} \theta}{\sqrt{1-\sin ^{2} \theta}} \cdot \cos \theta d \theta$
$=\int \frac{1+\sin ^{2} \theta}{\cos \theta} \cdot \cos \theta d \theta$
$=\int 1+\sin ^{2} \theta d \theta$
$=\int\left(1+\frac{1-\cos 2 \theta}{2}\right) d \theta$
$=\int\left(1+\frac{1}{2}-\frac{\cos 2 \theta}{2}\right) d \theta$
$=\int\left(\frac{3}{2}-\frac{\cos 2 \theta}{2}\right) d \theta$
$=\frac{3}{2} \int d \theta-\frac{1}{2} \int \cos 2 \theta d \theta$
$=\frac{3}{2} \theta-\frac{1}{2}\times \frac{1}{2} \sin 2 \theta+c$
$=\frac{3}{2} \cdot \theta-\frac{1}{2} \times \frac{1}{2} \cdot 2 \sin \theta \cdot \cos \theta+c$
$=\frac{3}{2} \theta-\frac{1}{2} \sin \theta \cdot \cos \theta+c$
$=\frac{3}{2} \sin ^{-1} x-\frac{1}{2} x \sqrt{1-x^{2}}+c$
$=\frac{3}{2} \sin ^{-1} x-\frac{x}{2} \sqrt{1-x^{2}}+c$
Question 5
$\int \frac{2 x}{\sqrt{1-x^{4}}} d x$
Sol :
putting $x^{2}=\sin \theta$ then 2xdx=cosθdθ
also , $\theta=\sin ^{-1} x^{2}$
Now , $\int \frac{2 x d x}{\sqrt{1-x^{2}}}=\int \frac{\cos \theta d \theta}{\sqrt{1-\left(x^{2}\right)^{2}}}$
$=\int \frac{\cos \theta d \theta}{\sqrt{1-\sin ^{2} \theta}}=\int \frac{\cos \theta d \theta}{\cos \theta}$
$=\int d \theta=\theta+c=\sin ^{-1}(x)^2+c$
Question 6
$\int \frac{x^{2}}{\sqrt{1-x^{6}}} d x$
Sol :
putting $x^{3}=\sin \theta$
also , $\theta=\sin ^{-1} x^{3}$
then $3 x^{2} d x=\cos \theta d \theta$
$x^{2} d x=\frac{\cos \theta}{3} d \theta$
Now ,
$\int \frac{x^{2} d x}{\sqrt{1-x^{2}}}=\int \frac{x^{2} d x}{\sqrt{1-\left(x^{3}\right)^{2}}}$
$=\int \frac{\cos \theta}{3 \cdot \sqrt{1-\sin ^{2} \theta}} d \theta=\int \frac{\cos \theta d \theta}{3 \cdot \cos \theta}$
$=\frac{1}{3} \int d \theta=\frac{1}{3} \theta+c$
$=\frac{1}{3} \sin ^{-1} x^{3}+c$
Question 7
(i) $\int \frac{d x}{\sqrt{4-x^{2}}}$
Sol :
putting x=2sinθ
also, $\theta=\sin ^{-1} \frac{x}{2}$
then dx=2cosθdθ
Now , $\int \frac{d x}{\sqrt{4-x^{2}}}=\int \frac{2 \cos \theta d \theta}{\sqrt{4-4 \sin ^{2} \theta}}$
$=\int \frac{2 \cos \theta d \theta}{\sqrt{4\left(1-\sin ^{2} \theta\right)}}$
$=\int \frac{2 \cos \theta}{2 \cos \theta} d \theta$
$=\int d \theta+c=\theta+c$
$=\sin ^{-1} \frac{x}{2}+c$
(ii) $\int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x$
Sol :
$\int \frac{\cos x d x}{\sqrt{4-\sin^2x}}=\int \frac{\cos x d x}{\sqrt{4-\frac{4}{4} \sin ^{2} x}}$
$=\int \frac{\cos x d x}{\sqrt{4\left(1-\frac{\sin ^{2} x}{4}\right)}}$
$=\int \frac{\cos x d x}{8 \cdot \sqrt{1-\left(\frac{\sin x}{2}\right)^{2}}}$
Let $z=\frac{\sin x}{2}$
then $dz=\frac{1}{2} \cos x d x$
Now , $\int \frac{\cos x d x}{2 \cdot \sqrt{1-\left(\frac{\sin x}{2}\right)^{2}}}$
$=\int \frac{2 d z}{2 \cdot \sqrt{1-z^{2}}}$
$=\int \frac{d z}{\sqrt{1-z^{2}}}+c$
$=\sin ^{-1} z+c$
$=\sin ^{-1}\left(\frac{\sin x}{2}\right)+c$
(iii) $\int \frac{d x}{x \cdot \sqrt{a x-x^{2}}}$
Sol :
putting $x=a \sin ^{2} \theta$ then
$d x=a \cdot 2 \sin \theta \cdot \cos \theta d \theta$
Now , $\int \frac{d x}{x \cdot \sqrt{a x-x^{2}}}$
$=\int \frac{2 a \sin \theta \cdot \cos \theta d \theta}{a \sin ^{2} \theta \cdot \sqrt{a \cdot a \sin ^{2} \theta-a^{2} \sin ^{2} \theta}}$
$=\int \frac{2 \cos \theta d \theta}{\sin \theta \sqrt{a^{2} \sin ^{2} \theta\left(1-\sin ^{2} \theta\right)}}$
$=\int \frac{2 \cos \theta d \theta}{\sin \theta \cdot \operatorname{asin} \theta \cdot \cos \theta}$
$=\frac{2}{a} \int \frac{d \theta}{\sin ^{2} \theta}$
$=\frac{2}{a} \int \operatorname{cosec}^{2} \theta d \theta$
$=\frac{2}{a}(-\cot \theta)+c$
$=-\frac{2}{a} \cot \theta+c$
$\because x=a \sin ^{2} \theta$
$\Rightarrow \sin ^{2} \theta=\frac{x}{a}$
$\therefore \sin \theta=\sqrt{\frac{x}{a}}$
also , $\cos \theta=\sqrt{1-\sin ^{2} \theta}$
$=\sqrt{1-\frac{x}{a}}$
$=\sqrt{\frac{a-x}{a}}$
$\therefore \quad \cot \theta=\frac{\cos \theta}{\sin \theta}$
$=\frac{\sqrt{a-x}}{\sqrt{a} \sqrt{x}}\times \sqrt{a}$
$=\sqrt{\frac{a-x}{x}}$
$\therefore \cot \theta=\sqrt{\frac{a-x}{x}}$
Now , $-\frac{2}{a} \cot \theta+c$
$=\frac{-2}{a} \cdot \sqrt{\frac{a-x}{x}}+c$
Question 8
$\int \frac{d x}{9+x^{2}}$
Sol :
putting x=3tanθ then $d x=3 \sec ^{2} \theta d \theta$
Now , $\int \frac{d x}{9+x^{2}}$
$=\int \frac{3 \sec ^{2} \theta d \theta}{9+9 \tan ^{2} \theta}$
$=\int \frac{3 \sec ^{2} \theta d \theta}{9\left(1+\tan ^{2} \theta\right)}$
$=\int \frac{3 \sec ^{2} \theta d \theta}{9 \sec ^{2} \theta}$
$=\frac{1}{3} \int d \theta+c=\frac{1}{3} \theta+c$
$=\frac{1}{3} \tan^{-1} \frac{x}{3}+c$
Sol :
putting x=3tanθ then $d x=3 \sec ^{2} \theta d \theta$
Now , $\int \frac{d x}{9+x^{2}}$
$=\int \frac{3 \sec ^{2} \theta d \theta}{9+9 \tan ^{2} \theta}$
$=\int \frac{3 \sec ^{2} \theta d \theta}{9\left(1+\tan ^{2} \theta\right)}$
$=\int \frac{3 \sec ^{2} \theta d \theta}{9 \sec ^{2} \theta}$
$=\frac{1}{3} \int d \theta+c=\frac{1}{3} \theta+c$
$=\frac{1}{3} \tan^{-1} \frac{x}{3}+c$
Question 9
Sol :
putting x=tanθ then $d x=\sec ^{2} \theta d \theta$
$\therefore \sec \theta=\sqrt{1+\tan ^{2} \theta}=\sqrt{1+x^{2}}$
Now , $\int \frac{x+1}{\sqrt{1+x^{2}}} d x=\int \frac{\tan \theta+1}{\sqrt{1+\tan ^{2} \theta}} \cdot \sec ^{2} \theta d \theta$
$=\int \frac{(\tan \theta+1) \cdot \sec ^{2} \theta}{\sec \theta} d \theta$
$=\int(\tan \theta+1) \sec \theta d \theta$
$=\int \sec \theta \cdot \tan \theta d \theta+\int \sec \theta d \theta$
$=\sec \theta+\log (\sec \theta+\tan \theta)+c$
$\sqrt{1+x^{2}}+\log (x+\sqrt{1+x^{2}})+c$
Question 10
$\int \frac{x}{1+x^{4}} d x$Sol :
putting $x^{2}=\tan \theta$ then $2 x d x=\sec ^{2} \theta d \theta$
$\Rightarrow x d x=\frac{\sec ^{2} \theta}{2} d \theta$
Now , $\int \frac{x}{1+x^{4}} d x=\int \frac{x d x}{1+\left(x^{2}\right)^{2}}$
$=\int \frac{\sec ^{2} \theta d \theta}{2\left(1+\tan ^{2} \theta\right)}$
$=\frac{1}{2} \int \frac{\sec ^{2} \theta}{\sec ^{2} \theta} d \theta$
$=\frac{1}{2} \int d \theta=\frac{1}{2} \theta+c$
$=\frac{1}{2} \tan ^{-1} x^{2}+c$
Question 11
$\int \frac{x^{5}}{1+x^{12}} d x$Sol :
putting $x^{6}=\tan \theta$ then $6 \cdot x^{5} d x=\sec ^{2} \theta d \theta$
$x^{5} d x=\frac{\sec ^{2} \theta}{6} d \theta$
Now ,$\int \frac{x^{5}}{1+x^{12}} d x=\int \frac{x^{5} d x}{1+x^{6} \cdot x^{6}}$
$=\int \frac{\sec^2 \theta}{6\left(1+\tan ^{2} \theta\right)}$
$=\frac{1}{6} \int \frac{\sec ^{2} \theta}{\sec ^{2} \theta} d \theta$
$=\frac{1}{6} \int d \theta=\frac{1}{6} \theta+c$
$=\frac{1}{6} \tan ^{-1}\left(x^{6}\right)+c$
Question 12
$\int \frac{d x}{x^{2} \sqrt{1+x^{2}}}$Sol :
putting x=tanθ then $d x=\sec ^{2} \theta d \theta$
Now , $\int \frac{d x}{x^{2} \sqrt{1+x^{2}}}=\int \frac{\sec ^{2} \theta}{\tan ^{2} \theta \cdot \sqrt{1+\tan ^{2} \theta}}$
$=\int \frac{\sec ^{2} \theta}{\tan ^{2} \theta \cdot \sec \theta} d \theta$
$=\int \frac{\sec \theta}{\tan ^{2} \theta} d \theta$
$=\int \frac{1}{\cos \theta \cdot \frac{\sin^2 \theta}{\cos ^{2} \theta}} d \theta$
$=\int \frac{\cos \theta}{\sin^2 \theta} d \theta$
Let z=sinθ then dz=cosθdθ
Now , $\int \frac{\cos \theta}{\sin ^2 \theta} d \theta=\int \frac{d z}{z^{2}}$
$=\int z^{-2} d z=\frac{z^{-2+1}}{-2+1}+0$
$=-z^{-1}+c$
$=\frac{-1}{z}+c=-\frac{1}{\sin \theta}+c$
$=-\frac{1}{\frac{x}{\sqrt{1+x^{2}}}}+c$
$=\frac{-\sqrt{1+ x^{2}}}{x}+c$
$=-\frac{\sqrt{1+x^{2}}}{x}+c$
Question 13
$\int \frac{d x}{1+e^{2 x}}$Sol :
putting $e^{x}=\tan \theta$ then
$e^{x} d x=\sec ^{2} \theta d \theta$
$d x=\frac{\sec ^{2} \theta d \theta}{\tan \theta}$
Now , $\int \frac{d x}{1+e^{2 x}}=\int \frac{\sec ^{2} \theta d \theta}{\tan \theta \cdot\left(1+\tan ^{2} \theta\right)}$
$=\int \frac{\sec ^{2} \theta}{\tan \theta \cdot \sec ^{2} \theta} d \theta$
$=\int \frac{1}{\tan \theta} d \theta=\int \frac{\cos \theta}{\sin \theta} d \theta$
=log(sin θ)+c
$\because \tan \theta=e^{x}$ then $\sin \theta=\frac{e^{x}}{\sqrt{1+e^{b x}}}$
Now , log(sin θ)+c
$=\log \left(\frac{e^{x}}{\sqrt{1+e^{2 x}}}\right)+c$
$\because \log \left(\frac{m}{n}\right)=\log m-log n$
$=\log e^{x}-\log (\sqrt{1+e^{2x}})+c$
$=x-\log \left(1+e^{2x}\right)^{\frac{1}{2}}+c$
$=x-\frac{1}{2} \log \left(1+e^{2 x}\right)+c$
$\because \log m^{n}=n\log m$
$x=\frac{1}{2} \log \left(1+e^{2 x}\right)$
Question 14
$\int \frac{e^{2x}}{1+e^{4x}} d x$Sol :
putting $e^{2x}=\tan \theta$ then $2 e^{2 x} d x=\sec ^{2} \theta d \theta$
$e^{2x} d x=\frac{\sec ^{2} \theta}{2} d \theta$
Now , $\int \frac{e^{2x} d x}{1+e^{4 x}}=\int \frac{e^{2 x}}{1+e^{2 x} \cdot e^{2 x}} d x$
$=\int \frac{sec^{2} \theta}{2\left(1+\tan ^{2} \theta\right)} d \theta$
$=\frac{1}{2} \int \frac{\sec ^{2} \theta}{\sec ^{2} \theta} d \theta$
$=\frac{1}{2} \int d \theta=\frac{\theta}{2}+c$
$=\frac{1}{2} \tan ^{-1}\left(e^{2 x}\right)+c$
Question 15
$\int \frac{e^{x}-1}{e^{x}+1} d x$Sol :
$=\int \frac{e^{\left(\frac{x}{2}\right)^{2}}-1}{e^{(\frac{x}{2})^{2}}-1} d x$
putting $e^{\frac{x}{2}}=\tan \theta$ then
$\frac{1}{2} e^{\frac{x}{2}} d x=\sec ^{2} \theta d \theta$
$d x=\frac{2 \sec ^{2} \theta d \theta}{\tan \theta}$
Now , $\int \frac{\tan ^{2} \theta-1}{\tan ^{2} \theta+1} \cdot \frac{2 \sec ^{2} \theta}{\tan \theta} d \theta$
$=2 \int \frac{\tan ^{2} \theta-1}{\tan \theta} d \theta$
$=2 \int \frac{\tan ^{2} \theta}{\tan \theta}-2 \int \frac{1}{\tan \theta} d \theta$
$=2 \int \tan \theta d \theta-2 \int \frac{\cos \theta}{\sin \theta} d \theta$
$=2 \log (\sec \theta)-2 \log (\sin \theta)+c$
$=2 \log \left(\frac{\sec \theta}{\sin \theta}\right)+c$
∴$\tan \theta=e^{\frac{\pi}{2}}$ then
$\sec \theta=\sqrt{1+\tan ^{2} \theta}=\sqrt{1+e^{x}}$
also , $\sin \theta=\frac{e^{x / 2}}{\sqrt{1+e^{x}}}$
Now , $2 \log \left(\frac{\sqrt{1+e^{x}}}{\frac{e^{x / 2}}{\sqrt{1+e^{x}}}}\right)+c$
$=2 \log \left(\frac{1+e^{x}}{e^{x / 2}}\right)+c$
=$2 \log \left(1+e^{x}\right)-2 \log \left(e^{x / 2}\right)+c$
=$2 \log \left(1+e^{x}\right)-2 \times \frac{x}{2}+c$
$=2 \log \left(1+e^{x}\right)-x+c$
Question 16
$\int \frac{d x}{x \sqrt{x^{2}-1}}$Sol :
putting x=secθ then dx=secθ.tanθ.dθ
Now , $\int \frac{d x}{x \cdot \sqrt{x^{2}-1}}=\int \frac{\sec \theta \cdot \tan \theta d \theta}{\sec \theta \cdot \sqrt{\sec ^{2} \theta-1}}$
$=\int \frac{\sec \theta.\tan \theta}{\sec \theta \cdot \tan \theta} d \theta$
$=\int d \theta$
$=\theta+c=\sec ^{-1} x+c$
Question 17
$\int \frac{x^{3}}{\left(x^{2}-a^{2}\right)^{3 / 2}} d x$Sol :
putting x=asecθ then dx=asecθ.tanθdθ
Now , $\int \frac{x^{3}}{\left(x^{2}-a^{2}\right)^{3 / 2}} d x$
$=\int \frac{a^{3} \sec ^{3} \theta \cdot a \cdot \sec \theta \cdot \tan \theta}{\left(a^{2} \sec ^{2} \theta-a^{2}\right)^{3 / 2}} d \theta$
$\Rightarrow \int \frac{a^{4} \cdot \sec ^{4} \theta \cdot \tan \theta d \theta}{\left\{a^{2}\left(\sec ^{2} \theta-1\right)^{3/ 2}\right.}$
$=\int \frac{a^{4} \cdot \sec ^{4} \theta \cdot \tan \theta d \theta}{a^{2 \times \frac{3}{2}}+\tan \theta^{2\times \frac{3}{2}}}$
$=\int \frac{a^{4} \sec ^{4} \theta \cdot \tan \theta d \theta}{a^{3} \cdot \tan ^{3} \theta}$
$=\int \frac{a \cdot \sec ^{4} \theta}{\tan ^{2} \theta} d \theta$
$=\int \frac{a}{\cos ^{4} \theta \cdot \frac{\sin ^{2} \theta}{\cos ^{2} \theta}}d\theta$
$=\int \frac{a}{\sin ^{2} \theta \cdot \cos ^{2} \theta} d \theta$
$=\int \frac{4 a}{4 \sin ^{2} \theta \cdot \cos ^{2} \theta} d \theta$
$=\int \frac{4 a}{(2 \sin \theta \cdot \cos \theta)^{2}}$
$=4 a \int \frac{d \theta}{(\sin 2 \theta)^{2}}$
$=4 a \int \frac{d \theta}{\sin ^{2} 2 \theta}$
$=4 a \int \operatorname{cosec}^{2} 2 \theta d \theta$
$=\frac{4 a(-cot2\theta)}{2}+c$
$=\frac{-4 a}{2} \cot 2 \theta+c$
$=\frac{-4 a}{2 \tan 2\theta}+c$
$=\frac{-4a}{2.\frac{2 \tan \theta}{1-\tan ^{2} \theta}} d \theta+c$
$=\frac{-a\left(1-\tan ^{2} \theta\right)}{\tan \theta}+c$
$\because \sec \theta=\frac{x}{a}$
then $\tan \theta=\sqrt{\sec ^{2} \theta-1}$
$=\sqrt{\frac{x^{2}}{a^{2}}-1}$
$=\frac{\sqrt{x^{2}-a^{2}}}{a}$
Now , $\frac{-a\left(1-\tan ^{2} \theta\right)}{\tan \theta}+c$
$=\frac{-a\left(1-\frac{x^{2}-a^{2}}{a^{2}}\right)^{2}}{\frac{\sqrt{x^{2}-a^{2}}}{a}}+c$
$\frac{-a\left(a^{2}-x^{2}+a^{2}\right)}{\frac{a^{2}}{\frac{\sqrt{x^{2}+a^{2}}}{a}}}+c$
$=\frac{-\left(2a^{2}-x^{2}\right)}{\sqrt{x^{2}-a^{2}}}+c$
$\frac{x^{2}-2a^{2}}{\sqrt{x^{2}-a^2}}+c$
Question 18
$\int \frac{e^{x}}{\sqrt{e^{2 x}-9}} d x$Sol :
putting $e^{x}=3 \sec \theta$ then $e^{x} d x=3 \sec \theta \cdot \tan \theta d \theta$
Now , $\int \frac{e^{x}}{\sqrt{e^{2 x}-9}} d x=\int \frac{3 \sec \theta \cdot \tan \theta d \theta}{\sqrt{9 \sec ^{2} \theta-9}}$
$=\int \frac{3 \sec \theta \cdot \tan \theta d \theta}{\sqrt{9\left(\sec ^{2} \theta-1\right)}}$
$=\int \frac{3 \sec \theta \cdot \tan \theta d \theta}{3 \tan \theta}$
$=\int \sec \theta d \theta$
$=\log |\sec \theta+\tan \theta|+c$
$=\log \left(\frac{e^{x}}{3}+\frac{\sqrt{e^{2 x}-9}}{3}\right)$
$=\log \left(\frac{e^{x}+\sqrt{e^{2 x}-9}}{3}\right)+c$
$=\log \left(e^{x}+\sqrt{e^{2 x}-9}\right)+\log 3+c$
$\begin{aligned}\left(\because \log \frac{m}{n}\right.&=\log m +\log n) \end{aligned}$
$=\log \left(e^{x}+\sqrt{e^{2 x}-9}\right)+c$
Question 19
$\int \frac{d x}{\sqrt{e^{2 x}-1}}$Sol :
putting $e^{x}=\sec \theta$ then $e^{x} d x=\sec \theta \cdot \tan \theta d \theta$
$d x=\frac{\sec \theta \cdot \tan \theta}{e^{x}} d \theta$
Now , $\int \frac{d x}{\sqrt{e^{2 x}-1}}=\int \frac{\sec \theta \cdot \tan \theta}{e^{x} \sqrt{\sec ^{2} \theta-1}} d \theta$
$=\int \frac{\sec \theta \cdot \tan \theta}{\sec \theta \cdot \tan \theta} d \theta$
$=\int d \theta=\theta+c$
$=\sec ^{-1}\left(e^{x}\right)+c$
Question 20
$\int \sqrt{\frac{a-x}{a+x}} d x$Sol :
putting $x=a \cos 2 \theta$ then
$d x=2 a \cdot(-\sin 2 \theta)$
$d x=-2 a \sin 2 \theta d \theta$
Now , $\int \sqrt{\frac{a-a \cos 2\theta}{a+a \cos 2 \theta}}(-2 a \sin 2 \theta d \theta)$
$=\int \sqrt{\frac{a(1-\cos 2 \theta)}{a(1+\cos 2 \theta)}} \times(-2 a \sin 2 \theta)d\theta$
$=\int \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}}(-2 a \cdot 2 \sin \theta \cdot \cos \theta) d \theta$
$=-4 a \int \frac{\sin \theta}{\cos \theta} \sin \theta \cdot \cos \theta d \theta$
$=-4\theta\int \sin ^{2} \theta d \theta$
$=-2 a \int 2 \sin ^{2} \theta d \theta$
$=-2 a \int(1-\cos 2 \theta) d \theta$
$=-2 a \int d \theta+2 a \int \cos 2 \theta d \theta$
$=-2 a \theta+2 a \frac{\sin 2 \theta}{2}+c$
$=-2 a \theta+a \sin 2 \theta+c$
$-2 a \times \frac{1}{2} \cos ^{-1} \frac{x}{a}+a \sin 2\theta+c$
$=-a \cos^{-1} \frac{x}{a}+a \cdot \frac{\sqrt{a^{2}-x^{2}}}{a}+c$
$=-a\left[\cos^{-1} \frac{x}{a}-\frac{\sqrt{a^{2}-x^{2}}}{a}\right]+c$
Question 21
$\int \sqrt{\frac{1+x}{1-x}} d x$Sol :
putting $x=\cos 2 \theta$ then $d x=-\sin 2 \theta \cdot 2 d \theta$
$d x=-2 \sin 2\theta d \theta$
Now , $\int \sqrt{\frac{1+\cos 20}{1-\cos 2 \theta}}(-2 \sin 2 \theta) d \theta$
$=\displaystyle\int\sqrt{\frac{2 \cos ^{2} \theta}{2 \sin ^{2} \theta}}(-2 \cdot 2 \sin \theta \cdot \cos \theta)d\theta$
$=\int \frac{\cos \theta}{\sin \theta} \times(-4 \sin \theta \cdot \cos \theta) d \theta$
$=-2 \int 2 \cos ^{2} \theta d \theta$
$=-2 \int(1+\cos 2 \theta) d \theta$
$=-2 \int d \theta-2 \int \cos 2 \theta$
$=-2 \theta-2 \frac{\sin 2 \theta}{2}+c$
$=-2 \times \frac{1}{2} \cos ^{-1} x-2 \frac{\sqrt{1-x^{2}}}{2}+c$
$=-\cos ^{-1} x-\sqrt{1-x^{2}}+c$
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