Evaluate the following integrals:

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

$\mathrm{d}(\log (\sin x))=\mathrm{d} t$

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

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

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

$\Rightarrow \int \frac{\mathrm{dt}}{\mathrm{t}}$

$=\ln |\mathrm{t}|+c$

But $\mathrm{t}=\log (\sin \mathrm{x})$

$=\ln |\log (\sin x)|+c$