Answer
$$\sin x(\ln(\sin x)-1)+c$$
Work Step by Step
Let $u=\sin x$; then $du=\cos xdx$. Hence, we have
$$\int \cos x \ln (\sin x) d x
= \int \ln u \ du.$$
Now we use integration by parts as follows
$$\int \ln u \ du=u \ln -\int du=u(\ln u -1).$$
Hence
$$\int \cos x \ln (\sin x) d x=\sin x(\ln(\sin x)-1)+c$$
