Explanation

(a)
$I_{1} = \int_{0}^{\pi/2}x\sin{x\ dx}$
$= - \left\lbrack x\cos x \right\rbrack_{0}^{\pi/2} - \int_{0}^{\pi/2}{( - \cos{x)dx}}$
$= \lbrack 0 - 0\rbrack + \left\lbrack \sin x\ \right\rbrack\ _{0}^{\pi/2} = 1\ \ \ \$...(i)
And $I_{2} = \int_{0}^{\pi/2}{x\cos\ x\ dx =}\left\lbrack x\sin x\ \right\rbrack\ _{0}^{\pi/2} - \int_{0}^{\pi/2}{\sin{x\ dx}}$
$= \left\lbrack \frac{\pi}{2} - 0 \right\rbrack + \left\lbrack \cos x\ \right\rbrack\ _{0}^{\pi/2}$
$= \frac{\pi}{2} - 1$ …(ii)
From Eqs. (i)and (ii), we get
$I_{1} + I_{2} = \frac{\pi}{2}$