Explanation

(a)
Let $I = \frac{2}{3}\int_{0}^{\pi/2}{\frac{\sqrt{\cos\theta}}{(\sqrt{\sin\theta} + \sqrt{\cos\theta})}d\theta}$ ...(i)
And $I = \frac{2}{3}\int_{0}^{\pi/2}{\frac{\sqrt{\cos(\pi/2 - \theta)}}{\sqrt{sin(\pi/2 - \theta)} + \sqrt{\cos(\pi/2 - \theta)}}d\theta}$
$\Rightarrow I = \frac{2}{3}\int_{0}^{\pi/2}{\frac{\sqrt{\sin\theta}}{\sqrt{\sin\theta +}\sqrt{\cos\theta}}d\theta\ \ \ }$…(ii)
On adding Eqs.(i) and (ii), we get
$2I = \frac{2}{3}f_{0}^{\pi/2}1d\theta = \frac{2}{3}\lbrack\theta\rbrack_{0}^{\pi/2} \Rightarrow I = \frac{1}{3}.\frac{\pi}{2} = \frac{\pi}{6}$