Let $\mathrm{I}=\int_0^\pi x \sin x \cos ^2 \mathrm{x} d \mathrm{x}$
$$
\begin{aligned}
&\therefore \mathrm{I}=\int_0^\pi(\pi-x) \sin (\pi-x) \cos ^2(\pi-x) d x \\
&\Rightarrow \mathrm{I}=\int_0^\pi(\pi-\mathrm{x}) \sin \mathrm{x} \cos ^2 \mathrm{xdx}
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
2 I=\int_0^\pi \pi \sin x \cos ^2 x d x
$$
Put $\quad \cos x=t$
$\Rightarrow-\sin x d x=d t$
As $x \rightarrow 0$, then $t \rightarrow 1$
and $\mathrm{x} \rightarrow \pi$, then $\mathrm{t} \rightarrow-1$
$$
\therefore \mathrm{I}=-\pi \int_1^{-1} \mathrm{t}^2 \mathrm{dt}
$$
$\Rightarrow \mathrm{I}=-\pi\left[\frac{\mathrm{t}^3}{3}\right]_1^{-1}$
$\Rightarrow 2 \mathrm{I}=-\frac{\pi}{3}[-1-1] \Rightarrow 2 \mathrm{I}=\frac{2 \pi}{3}$
$\therefore \mathrm{I}=\frac{\pi}{2}$