Evaluate the following integrals:

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

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

$1+\frac{\cos x}{\sin x} d x=d t$

$(1+\cot ) d x=d t$

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

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

$=\ln |t|+c$

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

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