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 xSol :
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 xSol :
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 xSol :
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 xSol :
=\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 xSol :
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 xSol :
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 xSol :
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 xSol :
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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