$=\int_0^\pi(\pi-x) \sin (\pi-x)[\cos (\pi-x)]^4 d x$

Eq. (1) $+(2)$ gives,
$2 I=\int_0^\pi \pi \sin x \cos ^4 x d x$
Eq. (1) $+(2)$ gives,
$2 I=\int_0^\pi \pi \sin x \cos ^4 x d x$
Put $\cos x=t \Rightarrow-\sin x d x=$
When $\mathrm{x}=0, \mathrm{t}=$ and when $\mathrm{x}=\pi, \mathrm{t}=-1$
$2 \mathrm{I}=\pi \int_1^{-1}(\mathrm{t})^4(-\mathrm{dt})$
... ( $\mathrm{t}^4$ is an even function)
$\therefore \quad 2 I=\frac{2 \pi}{5}\left[\mathrm{t}^5\right]_0^1 \Rightarrow \mathrm{I}=\frac{\pi}{5}$