Explanation

(d)
Let $I = \int_{0}^{1}{x^{3/2}\sqrt{1 - x}dx}$
Put $x = \sin^{2}{\theta \Rightarrow dx = 2\sin{\theta\cos{\theta d\ \theta}}}$
$\therefore I = \int_{0}^{\pi/2}{\sin^{3}{\theta.\sqrt{1 - \sin^{2}\theta}}}2\sin{\theta\cos{\theta d\theta}}$
$= 2\int_{0}^{\pi/2}{\sin^{4}{\theta\cos^{2}{\theta d\theta}}}$
$= 2\left\lbrack \frac{3.1.1}{6.4.2}.\frac{\pi}{2} \right\rbrack$ [using Walli’s formula]
$= \frac{\pi}{16}$