${\displaystyle \int \frac{\cot x}{1+{\sin }^{2}x}dx=}$

Put $\sin x=t$

$\cos xdx=dt$

$={\displaystyle \int \frac{dt}{t\left(1+t^2\right)}=\frac{1}{2}{\displaystyle \int \frac{2tdt}{t^2\left(1+t^2\right)}}}$

Put $t^2=p$

$2tdt=dp$

$=\frac{1}{2}{\displaystyle \int \frac{dp}{p\left(1+p\right)}=\frac{1}{2}\left[{\displaystyle \int \left|\frac{1}{p}-\frac{1}{1+p}\right|dp}\right]}$

$=\frac{1}{2}\log \left|\frac{p}{1+p}\right|+c=\frac{1}{2}\log \left|\frac{t^2}{1+t^2}\right|+c$

$=\frac{1}{2}\log \left|\frac{{\sin }^{2}x}{1+{\sin }^{2}x}\right|+c$

$=\log \left|\frac{\sin x}{\sqrt{1+{\sin }^{2}x}}\right|+c$