\(\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\)
Answer & explanation
Correct answer: option 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),
$2I=\int\limits_{0}^{\frac{\pi}{2}}1\,dx$
$=\frac{\pi}{2}$
$⇒I=\frac{\pi}{4}$