Question

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

Answer 1

Let $I={\int }_0^1{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$
Also, let $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$
When $x=0,\theta =0$ and when $x=1,\theta =\frac{\pi }{4}$
$\therefore I={\int }_0^{\frac{\pi }{4}}{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right){\sec }^2\theta d\theta$
$={\int }_0^{\frac{\pi }{4}}{\sin }^{-1}\left(\sin 2\theta \right){\sec }^2\theta d\theta$
$={\int }_0^{\frac{\pi }{4}}2\theta \cdot {\sec }^2\theta d\theta$
$=2{\int }_0^{\frac{\pi }{4}}\theta \cdot {\sec }^2\theta d\theta$
Taking $\theta$ as first function and ${\sec }^2\theta$ as second function and integrating by parts, we obtain
$I=2{[\theta \int {\sec }^2\theta d\theta -\int \{\left(\frac{d}{dx}\theta \right)\int {\sec }^2\theta d\theta \}d\theta ]}_0^{\frac{\pi }{4}}$
$=2[\theta \tan \theta -\int \tan \theta d\theta {]}_0^{\frac{\pi }{4}}$
$=2[\theta \tan \theta +\log |\cos \theta |{]}_0^{\frac{\pi }{4}}$
$=2[\frac{\pi }{4}\tan \frac{\pi }{4}+\log \left|\cos \frac{\pi }{4}\right|-\log |\cos 0|]$
$=2[\frac{\pi }{4}+\log \left(\frac{1}{\sqrt{2}}\right)-\log 1]$
$=2[\frac{\pi }{4}-\frac{1}{2}\log 2]$
$=\frac{\pi }{2}-\log 2$