Answer
$=\displaystyle \frac{1}{9}\sin^{3}x[3\ln(\sin x)-1]+C$
Work Step by Step
First, use the fact that $(\sin x)'=\cos x$ and substitute $\left[\begin{array}{l}
u=\sin x\\
du=\cos xdx
\end{array}\right]$
$\displaystyle \int\sin^{2}x\cos x\ln(\sin x)dx=\int u^{2}\ln udu=$
Table of integrals:
$\color{blue}{101.\quad \displaystyle \int u^{n}\ln udu=\frac{u^{n+1}}{(n+1)^{2}}[(n+1)\ln u-1]+C }$
$=\displaystyle \frac{u^{2+1}}{(2+1)^{2}}[(2+1)\ln u-1]+C$
$=\displaystyle \frac{1}{9}u^{3}(3\ln u-1)+C$
...bring x back ($u=\sin x$)...
$=\displaystyle \frac{1}{9}\sin^{3}x[3\ln(\sin x)-1]+C$
