A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) 0
D) 1
Correct Answer: A
Solution :
Let \[l=\int_{0}^{\pi /2}{\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx}\] ... (i) \[l=\int_{0}^{\pi /2}{\frac{\sqrt{\sin (\pi /2-x)}}{\sqrt{\cos (\pi /2-x)}+\sqrt{\sin (\pi /2-x)}}dx}\] \[l=\int_{0}^{\pi /2}{\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}\] ... (ii) On adding Eqs. (i) and (ii), \[2l=\int_{0}^{\pi /2}{1dx\Rightarrow l=\frac{\pi }{4}}\]
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