Evaluate the following integrals:

Assume $1+\cot x=t$

$\mathrm{d}(1+\cot x)=\mathrm{dt}$

$\Rightarrow \operatorname{cosec}^{2} x=d t$

Put $t$ and $d t$ in given equation we get

$\Rightarrow \int \frac{d t}{t}$

$=\ln |t|+c$

But $t=1+\cot x$

$=\ln |1+\cot x|+c$