Question

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

Answer 1

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

put $x=\tan \theta$

Put $x=\tan \theta$ $\theta ={\theta }^{-1}x$

$=\int {\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)dx$

$=\int {\sin }^{-1}\left(\frac{2\tan \theta }{{\sec }^2\theta }\right)dx$

$=\int {\sin }^{-1}\left(\sin 2\theta \right)dx$

$=\int 2\theta dx$

$\int 2{\tan }^{-1}xdx$

$=2x{\tan }^{-1}x-ln(x^2+1)+c$