Question

$Evaluate:{\int }_0^{\frac{1}{2}}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$

Answer 1

Let $I={\int }_0^{\frac{1}{2}}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$
Substituting $x=\sin \theta$, where $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$
$\Rightarrow dx=\cos \theta d\theta$
The limits of the integration become
$x=0\Rightarrow \sin \theta =0\Rightarrow \theta =0$
$x=\frac{1}{2}\Rightarrow \sin \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{6}$
$I={\int }_0^{\frac{\pi }{6}}\frac{\cos \theta d\theta }{(1+{\sin }^2\theta )\sqrt{1-{\sin }^2\theta }}$
$\text{Solution:}$
$\Rightarrow I={\int }_0^{\frac{\pi }{6}}\frac{\cos \theta d\theta }{(1+{\sin }^2\theta )\cos \theta }[\because \theta \in [-\frac{\pi }{2},\frac{\pi }{2}]]$
$\Rightarrow I={\int }_0^{\frac{\pi }{6}}\frac{d\theta }{1+{\sin }^2\theta }$
$\Rightarrow I={\int }_0^{\frac{\pi }{6}}\frac{{\sec }^2\theta d\theta }{{\sec }^2\theta +{\sin }^2\theta {\sec }^2\theta }$
$\Rightarrow I={\int }_0^{\frac{\pi }{6}}\frac{{\sec }^2\theta d\theta }{1+{\tan }^2\theta +{\tan }^2\theta }$
$\Rightarrow I={\int }_0^{\frac{\pi }{6}}\frac{{\sec }^2\theta d\theta }{1+2{\tan }^2\theta }$
$\Rightarrow I=\frac{1}{2}{\int }_0^{\frac{\pi }{6}}\frac{{\sec }^2\theta d\theta }{{\tan }^2\theta +\frac{1}{2}}$
$\Rightarrow I=\frac{1}{2}{\int }_0^{\frac{\pi }{6}}\frac{{\sec }^2\theta d\theta }{{\tan }^2\theta +{\left(\frac{1}{\sqrt{2}}\right)}^2}$
Substituting $\tan \theta =t\Rightarrow dt={\sec }^2\theta d\theta$
The limits of the integration become
$\theta =0\Rightarrow t=\tan 0=0$
$\theta =\frac{\pi }{6}\Rightarrow t=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$
$\Rightarrow I=\frac{1}{2}{\int }_0^{\frac{1}{\sqrt{3}}}\frac{dt}{t^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}$
$\Rightarrow I=\frac{1}{2}{\left[\frac{1}{\frac{1}{\sqrt{2}}}{\tan }^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)\right]}_0^{\frac{1}{\sqrt{3}}}$
$\Rightarrow I=\frac{1}{\sqrt{2}}[{\tan }^{-1}\left(\frac{\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{2}}}\right)-{\tan }^{-1}\left(\frac{0}{\frac{1}{\sqrt{2}}}\right)]$
$\Rightarrow I=\frac{1}{\sqrt{2}}{\tan }^{-1}\sqrt{\frac{2}{3}}$
Hence, the value of required integration is $\frac{1}{\sqrt{2}}{\tan }^{-1}\sqrt{\frac{2}{3}}$