Explanation

(d)
Let $f(\theta) = \left\lbrack \int_{0}^{\sin^{2}\theta}{\sin^{- 1}\sqrt{\phi}}\ d\phi + \int_{0}^{\cos^{2}\theta}{\cos^{- 1}\sqrt{\phi}}\ d\phi \right\rbrack$
$f^{'}(\theta) = \left( \frac{d}{d\theta}\sin^{2}\theta \right)\left\lbrack \sin^{- 1}\sqrt{\sin^{2}\theta} \right\rbrack + \left( \frac{d}{d\theta}\cos^{2}\theta \right)\left\lbrack \cos^{- 1}\sqrt{\cos^{2}\theta} \right\rbrack$
$\left( 2\sin\theta\cos{\theta)\theta - (2}\sin\theta\cos\theta \right)\theta = 0$
$\therefore\ f(\theta) = constant = a$ [say]
$\therefore\ f\left( \frac{\pi}{4} \right) = a$
$\Rightarrow \ \int_{0}^{\frac{1}{2}}{(\sin^{- 1}\sqrt{\phi}} + \cos^{- 1}\sqrt{\phi})d\phi = a$
$\Rightarrow \ \frac{\pi}{2}\lbrack\phi\rbrack_{0}^{\frac{1}{2}} = a\ \Rightarrow \ \frac{\pi}{4} = a$