$\int_0^\pi x \sin x \cos ^4 x d x=$

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Aicharya · Accepted Answer

$=\int_0^\pi(\pi-x) \sin (\pi-x)[\cos (\pi-x)]^4 d x$ 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 $\cos x=t \Rightarrow-\sin x d x=dt$ When $\mathrm{x}=0, \mathrm{t}=1$ 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) $	herefore \quad 2 I=\frac{2 \pi}{5}\left[\mathrm{t}^5ight]_0^1 \Rightarrow \mathrm{I}=\frac{\pi}{5}$

$\int_0^\pi x \sin x \cos ^4 x d x=$

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