Question

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

Answer 1

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

Simplifying the given function

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

Put $x=\tan t$

$\therefore t={\tan }^{-1}\left(x\right)$

$\therefore {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)={\sin }^{-1}\left(\frac{2\tan t}{1+{\tan }^{2}t}\right)$

$={\sin }^{-1}\left(\sin 2t\right)$

$=2t$

$2{\tan }^{-1}x$

$\therefore I=\int {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=2\int {\tan }^{-1}xdx$

$=2{\tan }^{-1}x.x-2\int \frac{x}{1+x^2}dx$

$=2x{\tan }^{-1}x-\log |1+x^2|+c$