STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

Also $I = \int_0^\pi {(\pi - x){{\sin }^3}x\,\,dx} $.....$(ii)$

Adding $(i)$ and $(ii),$ we get

$2I = \pi \int_0^\pi {{{\sin }^3}x} \,\,dx = \frac{\pi }{4}\int_0^\pi {\{ 3\sin x - \sin 3x\} dx} $

$ = \frac{\pi }{4}\left[ { - 3\cos x + \frac{{\cos 3x}}{3}} \right]_0^\pi $

$= \frac{\pi }{4}\left[ {3 - \frac{1}{3} + 3 - \frac{1}{3}} \right] = \frac{{4\pi }}{3}$

Hence, $I = \frac{{2\pi }}{3}$.