Question

Evaluate
${\int }_0^1{\sin }^{-1}\left(\frac{2x}{{1+x}^2}\right)dx$

Answer 1

We have,

${\int }_0^1{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$

Let

$x=\tan \theta$

$dx={\sec }^2\theta d\theta$

Change limit and we get,

$x=0$

$\tan \theta =0$

$\tan \theta =\tan 0$

$\theta =0$

And

$x=1$

$\tan \theta =1$

$\tan \theta =\tan \frac{\pi }{4}$

$\theta =\frac{\pi }{4}$

So,

${\int }_0^{\frac{\pi }{4}}{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right){\sec }^2\theta d\theta$

$\Rightarrow {\int }_0^{\frac{\pi }{4}}{\sin }^{-1}\sin 2\theta {\sec }^2\theta d\theta$

$\Rightarrow {\int }_0^{\frac{\pi }{4}}2\theta {\sec }^2\theta d\theta$

$\Rightarrow 2{\int }_0^{\frac{\pi }{4}}\theta {\sec }^2\theta d\theta$

On integrating and we get,

$2{\int }_0^{\frac{\pi }{4}}\theta {\sec }^2\theta d\theta$

$=2\left[\theta {\int }_0^{\frac{\pi }{4}}{\sec }^2\theta d\theta -{\int }_0^{\frac{\pi }{4}}\left({\int }_0^{\frac{\pi }{4}}\frac{d}{d\theta }\theta {\sec }^2\theta d\theta \right)d\theta \right]$

$=2{{\left[\theta \tan \theta \right]}_0}^{\frac{\pi }{4}}-2{{(-\log \cos \theta )}_0}^{\frac{\pi }{4}}$

$=2\left[(\frac{\pi }{4}-0)(\tan \frac{\pi }{4}-\tan 0)\right]+2[\log \cos \frac{\pi }{4}-\log \cos 0]$

$=2\left[\frac{\pi }{4}(1-0)\right]+2[\log \frac{1}{\sqrt{2}}-\log 1]$

$=2\left[\frac{\pi }{4}\right]+2[\log 1-\log \sqrt{2}-\log 1]$

$=\frac{\pi }{2}-2\log \sqrt{2}$

Hence, this is the answer.