Evaluate the following integrals:

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

We know $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

Therefore, we can write the integral as

$I=\int \sin ^{-1}(\sin 2 x) d x$

$\Rightarrow I=\int 2 x d x$

$\Rightarrow I=2 \int x d x$

Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow \mathrm{I}=2 \times \frac{\mathrm{x}^{1+1}}{1+1}+\mathrm{c}$

$\Rightarrow \mathrm{I}=2 \times \frac{\mathrm{x}^{2}}{2}+\mathrm{c}$

$\therefore \mathrm{I}=\mathrm{x}^{2}+\mathrm{c}$

Thus, $\int \sin ^{-1}\left(\frac{2 \tan x}{1+\tan ^{2} x}\right) d x=x^{2}+c$