MCQ
$\int_0^\pi x \sin ^3 x d x=$
  • A
    $\frac{4 \pi}{3}$
  • $\frac{2 \pi}{3}$
  • C
    $0$
  • D
    $\frac{\pi}{4}$

Answer

Correct option: B.
$\frac{2 \pi}{3}$
(B)
Let $I =\int_0^\pi x \sin ^3 x d x$ ...(i)
$=\int_0^\pi(\pi-x) \sin ^3 x d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I=\pi \int_0^\pi \sin ^3 x d x=\frac{\pi}{4} \int_0^\pi(3 \sin x-\sin 3 x) d x$
$\begin{array}{l}=\frac{\pi}{4}\left[-3 \cos x+\frac{\cos 3 x}{3}\right]_0^\pi \\ =\frac{\pi}{4}\left[3-\frac{1}{3}+3-\frac{1}{3}\right]=\frac{4 \pi}{3}\end{array}$
$\therefore \quad I=\frac{2 \pi}{3}$

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