Explanation

(b)
Given $I_{10} = \int_{0}^{\pi/2}x^{10}\sin xdx$
$= \lbrack - x^{10}{\cos{x\rbrack}}_{0}^{\pi/2} + \int_{0}^{\pi/2}{10x^{9}.\cos{xdx}}$
$= 0 + \left\lbrack 10x^{9}\sin x \right\rbrack_{0}^{\pi/2} - \int_{0}^{\pi/2}{90x^{8}\sin{x\ dx}}$
$= 10\left( \frac{\pi}{2} \right)^{9} - 90I_{8}$
$\Rightarrow I_{10} + 90I_{8} = 10\left( \frac{\pi}{2} \right)^{9}$