Explanation

(b)
Let $I = \int_{0}^{\pi}x\sin^{3}{dx}$ ...(i)
Also, $I = \int_{0}^{\pi}{(\pi - x)\sin^{3}{xdx}}$ ...(ii)
On adding Eqs. (i) and (ii), we get
$2I = \pi\int_{0}^{\pi}{\sin^{3}{x\ dx}}$
$= \frac{\pi}{4}\int_{0}^{\pi}{(3\sin{x - \sin{3x)dx}}}$
$= \frac{\pi}{4}\left\lbrack - 3\cos{x + \frac{\cos{3x}}{3}} \right\rbrack_{0}^{\pi} = \frac{4\pi}{3}$
Hence, $I = \frac{2\pi}{3}$