Answer
$$-\cos x(\ln \sin x) +\ln |\csc x-\cot x|+\cos x+C$$
Work Step by Step
Given $$\int \sin x \ln (\sin x) d x$$
Let
\begin{align*}
u&=\ln (\sin x)\ \ \ \ \ \ \ dv=\sin xdx\\
du&= \frac{\cos x}{\sin x}dx\ \ \ \ \ \ \ \ v=-\cos x
\end{align*}
Then
\begin{align*}
\int \sin x \ln (\sin x) d x&= -\cos x\ln (\sin x)+\int \frac{\cos^2 x}{\sin x}dx\\
&= -\cos x\ln (\sin x)+\int \frac{1-\sin^2 x}{\sin x}dx\\
&= -\cos x\ln (\sin x)+\int [\csc x-\sin x]dx\\
&= -\cos x\ln (\sin x) +\ln |\csc x-\cot x|+\cos x+C
\end{align*}
