# Evaluate:$\int\limits_0^{\pi/2} \frac {\sqrt {\tan x}}{1+\sqrt {\tan x}}dx$

Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• (ii) $\int \limits_a^b f(x)dx=\int \limits_a^b f(a-x) dx$
Given $\int\limits_0^{\pi/2} \frac {\sqrt {\tan x}}{1+\sqrt {\tan x}}dx \qquad But\; \tan x =\frac{\sin x}{\cos x}$

This can be written as

$I=\int \limits_0^{\pi/2} \Large\frac{\frac{\sqrt {\sin x}}{\sqrt {\cos x}}}{1+\sqrt {\frac{\sin x}{\cos x}}}$

$=\int \limits_0 ^ {\pi/2} \large\frac{\sqrt {\sin x}}{\sqrt {\cos x}+\sqrt {\sin x}}$-----(1)

By applying the property $\int \limits_a^b f(x)dx=\int \limits_a^b f(a-x) dx$

$I=\int \limits_0^{\pi/2} \large\frac{\sqrt {\sin (\pi/2-x)}}{\sqrt {\cos (\pi/2-x)}+\sqrt {\sin (\pi/2-x)}} dx$

But $\sin (\pi/2-x)=\cos x \;and\; \cos (\pi/2-x)=\sin x$

Therefore $I=\int \limits_0^{\pi/2} \large\frac{\sqrt {\cos x}}{\sqrt {\sin x}+\sqrt {\cos x}}dx$ -----(2)

$2I=\int \limits_0 ^ {\pi/2} \large\frac{\sqrt {\sin x}+\sqrt {\cos x}}{\sqrt {\sin x}+\sqrt {\cos x}}dx=\int \limits_0^{\pi/2} dx$

On integrating we get $2I=\bigg[x \bigg]_0^{\pi/2}$

On applying limits$2I=\pi/2 -0$

Therefore $2I=\frac{\pi}{2}$

$=>I=\frac{\pi}{4}$