Question

Integrate the function ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$

Answer 1

Let $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$
$\therefore {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)={\sin }^{-1}\left(\sin 2\theta \right)=2\theta$
$\Rightarrow \int {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx=\int 2\theta \cdot {\sec }^2\theta d\theta =2\int \theta \cdot {\sec }^2\theta d\theta$
Integrating by parts, we obtain
$=2[\theta \cdot \int {\sec }^2\theta d\theta -\int \{\left(\frac{d}{d\theta }\theta \right)\int {\sec }^2\theta d\theta \}d\theta ]$
$=2[\theta \cdot \tan \theta -\int \tan \theta d\theta ]$
$=2[\theta \tan \theta +\log |\cos \theta |]+C$
$=2[x{\tan }^{-1}x+\log \left|\frac{1}{\sqrt{1+x^2}}\right|]+C$
$=2x{\tan }^{-1}x+2\log (1+x^2{)}^{\frac{1}{2}}+C$
$=2x{\tan }^{-1}x+2[-\frac{1}{2}\log (1+x^2\left)\right]+C$
$=2x{\tan }^{-1}x-\log (1+x^2)+C$