Explanation

(d)
Let $I = \int_{0}^{\pi/2}{x\sin^{2}{xcos^{2}x\ dx}}$ ...(i)
$\Rightarrow I = \int_{0}^{\pi/2}{\left( \frac{\pi}{2} - x \right)\sin^{2}{x\cos^{2}{x\ dx}\ldots(ii)}}$
On adding Eqs. (i) and (ii), we get
$2I = \frac{\pi}{2}\int_{0}^{\pi/2}{\sin^{2}{x\cos^{2}{x\ dx}}}$
$= \frac{\pi}{8}\int_{0}^{\pi/2}{\sin^{2}{2x\ dx}}$
$= \frac{\pi}{8}\int_{0}^{\pi/2}\left( \frac{1 - \cos{4x}}{2} \right)dx$
$\Rightarrow \ 2I = \frac{\pi}{16}\left\lbrack x - \frac{\sin{4x}}{4} \right\rbrack_{0}^{\pi/2}\$
$\Rightarrow \ 2I = \frac{\pi}{16}\left\lbrack \frac{\pi}{2} \right\rbrack\ \Rightarrow \ I = \frac{\pi^{2}}{64}$