Question

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

Answer 1

$\int si{n}^{-1}\left(\right(\frac{2tanx}{1+ta{n}^{2}x}\left)\right)$

$=\int si{n}^{-1}\left(sin2x\right)dx$

$=\int 2xdx$

$=2\left(\frac{x^2}{2}\right)+C$

$=x^2+C$