Q83 / 150 · MHT CET 2021 (24 Sep Shift 2) · Mathematics
$\int_0^\pi x \sin x \cos ^4 x d x=$
- A$\frac{\pi}{10}$
- B$\frac{2 \pi}{5}$
- Correct answerC$\frac{\pi}{5}$
- D$\frac{\pi}{8}$
Solution
$=\int_0^\pi(\pi-x) \sin (\pi-x)[\cos (\pi-x)]^4 d x$
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) $\therefore \quad 2 I=\frac{2 \pi}{5}\left[\mathrm{t}^5\right]_0^1 \Rightarrow \mathrm{I}=\frac{\pi}{5}$
Hence, the correct answer is option: 'C'.