Question

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

Answer 1

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

Let $x=\tan \theta$

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

$\Rightarrow \frac{2x}{1+x^2}=\frac{2\tan \theta }{1+{\tan }^2\theta }=\sin 2\theta$

When $x=0\Rightarrow \theta =0$

When $x=1\Rightarrow \theta =\frac{\pi }{4}$

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

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

Let $u=\theta \Rightarrow du=d\theta$

$dv={\sec }^2\theta d\theta \Rightarrow v=\tan \theta$

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

$=2[\frac{\pi }{4}\tan \frac{\pi }{4}-0]-2{\left[\ln \left|\sec x\right|\right]}_0^{\frac{\pi }{4}}$

$=\frac{\pi }{2}-2[\ln \left|\sec \frac{\pi }{4}\right|-\ln \left|\sec 0\right|]$

$=\frac{\pi }{2}-2[\ln \left|\sqrt{2}\right|-\ln \left|1\right|]$

$=\frac{\pi }{2}-2[\ln \left|\sqrt{2}\right|-0]$

$=\frac{\pi }{2}-2\ln \left|\sqrt{2}\right|$

$=\frac{\pi }{2}-2\ln 2^{\frac{1}{2}}$

$=\frac{\pi }{2}-\ln 2$