Answer
$\frac{1}{2} ( \ln(\sin x))^2+c $
Work Step by Step
Let $ u=\ln(\sin x)$, then $ du= \frac{\cos x}{\sin x} dx $ and hence
$$\int \cot x \ln(\sin x)dx = \int \frac{\cos x \ln(\sin x)}{\sin x}dx=\int udu\\
=\frac{1}{2} u^2+c
=\frac{1}{2} ( \ln(\sin x))^2+c.$$
