$\int_0^\pi x \sin x{\cos ^2}xdx$

Let $I = \int_0^\pi x \sin x{\cos ^2}xdx$ …….(i)

and $I = \int_0^\pi {(\pi - x)\sin (\pi - x){{\cos }^2}(\pi - x)} dx$

$\Rightarrow$ $I = \int_0^\pi {(\pi - x)} \sin x{\cos ^2}xdx$ ……(ii)

Adding Eqs. (i) and (ii), we get

$2I = \int_0^\pi \pi \sin x{\cos ^2}xdx$

Let's put $\cos x = t$

$\Rightarrow$ $- \sin xdx = dt$

As $x \to 0,$ then $t \to 1$

and $x \to \pi ,$ then $t \to - 1$

therefore,$I = - \pi \int_1^{ - 1} {{t^2}} dt \Rightarrow I = - \pi \left[ {\frac{{{t^3}}}{3}} \right]_1^{ - 1}$

$\Rightarrow$ $2I = - \frac{\pi }{3}[ - 1 - 1] \Rightarrow 2I = \frac{{2\pi }}{3}$

therefore,$I = \frac{\pi }{3}$