\(\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\)

\(\frac{\pi}{4}\)

The correct answer is Option (4) → \(\frac{\pi}{4}\)

$I=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$   ....(1)

$I=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\cos(\frac{\pi}{2}-x)}+\sqrt{\sin(\frac{\pi}{2}-x)}}$

$=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$   ....(2)

Add (1) and (2),

$2I=\int\limits_{0}^{\frac{\pi}{2}}1\,dx$

$=\frac{\pi}{2}$

$⇒I=\frac{\pi}{4}$