Question

Answer 1

$$\begin{gathered} \int {\sin }^{-1}\left(\frac{2\tan x}{1+{\tan }^{2}x}\right)dx \\ =\int {\sin }^{-1}\left(\sin 2x\right)dx\left[\therefore \sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}\right] \\ =2\int xdx \\ =2\left(\frac{x^2}{2}\right)+\mathrm{C} \\ =x^2+\mathrm{C} \end{gathered}$$