MCQ
$\int_0^{\pi /2} {\frac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}\,dx = } $
  • A
    $0$
  • B
    $\frac{\pi }{2}$
  • $\frac{\pi }{4}$
  • D
    None of these

Answer

Correct option: C.
$\frac{\pi }{4}$
c
(c) Let $I = \int_0^{\pi /2} {\,\,\frac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx} $ .....$(i)$

and $I = \int_0^{\pi /2} {\frac{{\sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}dx} $

$I = \int_0^{\pi /2} {\frac{{\sqrt {\sin x} }}{{\sqrt {\cos x + } \sqrt {\sin x} }}} \,dx$....$(ii)$

Adding $(i)$ and $(ii),$ we get

$2I = \int_0^{\pi /2} {(1)dx = \frac{\pi }{2} \Rightarrow I = \frac{\pi }{4}} $.

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