Question

Prove that

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

Answer 1

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

Put $x=\tan \theta ⟹dx={\sec }^2\theta d\theta$

When, $x=0$, then $0=\tan \theta ⟹\theta =0$

When, $x=1$, then $1=\tan \theta ⟹\theta =\frac{\pi }{4}$

Therefore,

$I={{\int }_o}^{\pi /4}{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right).{\sec }^2\theta d\theta$

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

$I={{\int }_0}^{\pi /4}2\theta .{\sec }^2\theta d\theta$

Integrating by parts, we get,

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

$I={[2\theta \tan \theta {]}_0}^{\pi /4}-{{\int }_0}^{\pi /4}2\tan \theta d\theta$

$I=[2\times \frac{\pi }{4}\tan \frac{\pi }{4}-0]-2{[\log \sec \theta {]}_0}^{\pi /4}$

$I=\frac{\pi }{2}-2[\log \sec \frac{\pi }{4}-\log \sec 0]$

$I=\frac{\pi }{2}-2[\log \sqrt{2}-\log 1]$

$I=\frac{\pi }{2}-2.\frac{1}{2}\log 2=\frac{\pi }{2}-\log 2$