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# By using the properties of definite integrals,evaluate the integral$\int\limits_0^\frac{\Large \pi}{\Large 2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\;dx$

Toolbox:
• (i)$\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
• (ii)$\sin (\frac{\pi}{2}-x)=\cos x$
• (iii)$\cos(\frac{\pi}{2}-x)=\sin x$
Given $\int\limits_0^\frac{\Large \pi}{\Large 2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx\;----->(1)$
By applying the property $\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
$I=\int\limits_0^\frac{\Large \pi}{\Large 2}\frac{\sqrt{\sin (\frac{\pi}{2})-x}}{\sqrt{\sin (\frac{\pi}{2})-x}+\sqrt{\cos (\large\frac{\pi}{2})-x}}dx\;$
But $\sin (\frac{\pi}{2}-x)=\cos x \;and\;\cos(\frac{\pi}{2}-x)=\sin x$
Therefore $I=\int\limits_0^\frac{\Large \pi}{\Large 2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx\;----->(2)$
Adding equ (1) and equ (2)
$2I=\int \limits_0^{\frac{\pi}{2}} \large\frac{\sqrt {\sin x}+\sqrt {\cos x}}{\sqrt {\sin x}+\sqrt {\cos x}}$
$2I=\int \limits_0^{\frac{\pi}{2}}dx$
On integrating we get
$2I=[x]_0^{\frac{\pi}{2}}$
On Applying limits,
$2I=\frac{\pi}{2}-0$
$2I=\frac{\pi}{2}$
Therefore $I=\large\frac{\pi}{4}$
edited Jan 31, 2014 by yamini.v