Evaluate the following integrals:

Assume $\log (\operatorname{cosec} x-\cot x)=t$

$d(\log (\operatorname{cosec} x-\cot x))=d t$

(use chain rule to differentiate first differentiate $\log (\sec x+\tan x)$ then $(\sec x+\tan x)$

$\Rightarrow \frac{-\csc x \cot x+\csc ^{2} x}{\operatorname{cosec} x-\cot x} d x=d t$

$\Rightarrow \frac{\csc x(\csc x-\cot x)}{\operatorname{cosec} x-\cot x} d x=d t$

$\Rightarrow \csc x d x=d t$

Put $\mathrm{t}$ and dt in given equation we get

Substituting the values oft and dt we get

$\Rightarrow \int \mathrm{tdt}$

$\Rightarrow \frac{\mathrm{t}^{2}}{2}+\mathrm{c}$

But $t=\log (\operatorname{cosec} x-\cot x)$

$\Rightarrow \frac{\log ^{2}(\operatorname{cosec} x-\cot x)}{2}+c$