STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

$ = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}\;dx} $ ..$(i)$

$I = \int_{\pi /6}^{\pi /3} {\frac{{\sqrt {\sin x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}\;} $ ..$(ii)$

(Since $\int_a^b {f(x)dx} = \int_a^b {f(a + b - x)\,dx} $)

Adding $(i)$ and $(ii),$ we get, 

$2I = \int_{\pi /6}^{\pi /3} {\;dx} $

==> $I = \frac{1}{2}\left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) = \frac{\pi }{{12}}$.