Answer
$$
\int \sin ^{2} x \cos x \ln (\sin x) d x=\frac{1}{9} \sin ^{3} x[3 \ln (\sin x)-1]+C
$$
Work Step by Step
$$
\int \sin ^{2} x \cos x \ln (\sin x) d x
$$
If we make the substitution
$$
u=\sin x, \quad d u=\cos x d x
$$
and we look at the Table of Integrals , we see that the closest entry is number $101 $ with $n=2$ so we have:
$$
\begin{aligned}
\int \sin ^{2} x \cos x \ln (\sin x) d x &=\int u^{2} \ln u d u\\
& \stackrel{101}{=} \frac{u^{2+1}}{(2+1)^{2}}[(2+1) \ln u-1]+C\\
&=\frac{1}{9} u^{3}(3 \ln u-1)+C \\
&=\frac{1}{9} \sin ^{3} x[3 \ln (\sin x)-1]+C
\end{aligned}
$$
