Explanation

(b)
Let $I = \int_{0}^{\pi}x\sin^{4}x\ dx$ …(i)
$I = \int_{0}^{\pi}{(\pi - x)}\sin^{4}x\ dx$ …(ii)
On adding Eqs. (i) and (ii), we get
$2I = \pi\int_{0}^{\pi}{\sin^{4}x}\ dx$
$= 2\pi\int_{0}^{\frac{\pi}{2}}{\sin^{4}x}\ dx$
$= 2\pi.\frac{4 - 1}{4}.\frac{4 - 3}{4 - 2}.\frac{\pi}{2}$
$= 2\pi.\frac{3}{4}.\frac{1}{2}.\frac{\pi}{2} = \frac{3\pi^{2}}{8}$
$\Rightarrow \ I = \frac{3\pi^{2}}{16}$