$\int\limits_0^{\pi /2} {\cfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x + \sqrt {\cos x} } }}} dx$

: Let$I = \int\limits_0^{\pi /2} {\cfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x + \sqrt {\cos x} } }}} dx$

….(i)

$\Rightarrow$ $I = \int\limits_0^{\pi /2} {\cfrac{{\sqrt {\sin \left( {\cfrac{\pi }{2} - x} \right)} }}{{\sqrt {\sin \left( {\cfrac{\pi }{2} - x} \right)} + \sqrt {\cos \left( {\cfrac{\pi }{2} - x} \right)} }}} dx$

$= \int\limits_0^{\pi /2} {\cfrac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}} dx$

….(ii)
Adding (i) and (ii),

we have
$2I = \int\limits_0^{\pi /2} {\left[ {\cfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} + \cfrac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}} \right]} dx$

$= \int\limits_0^{\pi /2} {\cfrac{{\sqrt {\cos x} + \sqrt {\sin x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}} dx$

$= \int\limits_0^{\pi /2} {dx = \left[ x \right]_0^{\pi /2} = \cfrac{\pi }{2} - 0} = \cfrac{\pi }{2}$ $\Rightarrow$ $I = \cfrac{\pi }{4}$