Evaluate the following integrals:

Assume $\log (\sin x)=t$

$d(\log (\sin x))=d t$

$\Rightarrow \frac{\cos x}{\sin x} d x=d t$

$\Rightarrow \cot x d x=d t$

Substituting the values oft and dt we get

$\Rightarrow \int \mathrm{tdt}$

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

But $t=\log (\sin x)$

$\Rightarrow \frac{\log (\sin x)^{2} x}{2}+c$