Explanation

(c)
Using Walli’s formula, we get
$\int_{0}^{\frac{\pi}{2}}{\sin^{n}x}\ dx = \frac{n - 1}{n}.\frac{n - 3}{n - 2}.\frac{n - 5}{n - 4}\ldots..\frac{2}{3}\ if\ n\ is\ odd$
$\Rightarrow \int_{0}^{\frac{\pi}{2}}{\sin^{5}x}\ dx = \frac{5 - 1}{5}.\frac{5 - 3}{5 - 2}$
$= \frac{4}{5}.\frac{2}{3}$
$= \frac{8}{15}$