Let $\int_0^\pi x \sin ^3 x \cos ^2 x d x=I$ ...(i)
$\Rightarrow I=\int_0^\pi(\pi-x) \sin ^3(\pi-x) \cos ^2(\pi-x) d x$
$I=\int_0^\pi(\pi-x) \sin ^3(x) \cos ^2 x d x$ ...(ii)
Equation (i) + (ii)
$2 I=\pi \int_0^\pi \sin ^3 x \cos ^2 x d x$
$=\pi \int_0^\pi\left(1-\cos ^2 x\right) \cos ^2 x \sin x d x$
Let $\cos x=t \Rightarrow-\sin x d x=d t$
$\therefore 2 \mathrm{I}=\pi \int_1^{-1}\left(1-\mathrm{t}^2\right) \mathrm{t}^2(-\mathrm{dt})=\pi \int_{-1}^1\left(\mathrm{t}^2-\mathrm{t}^4\right) \mathrm{dt}$
$=\pi\left[\frac{t^3}{3}-\frac{t^5}{5}\right]_{-1}^1$
$\Rightarrow 2 I=\frac{4 \pi}{15} \Rightarrow I=\frac{2 \pi}{15}$