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

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  • (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)
Adding equ(1) and equ (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 $2I=\frac{\pi}{2}$



answered Mar 15, 2013 by meena.p
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