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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate:\[\int \limits _{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt {\tan x}}\]

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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+b-x) dx$
Given $I=\int \limits _{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt {\tan x}}$-----(1)
 
This can be written as
 
$I=\int \limits _{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt {\sin x}}=\int \limits _{\pi/6}^{\pi/3} \frac{\sqrt \cos x}{\sqrt {\cos x}+\sqrt {\sin x}}dx$
 
Applying the property $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a+b-x) dx$
 
$I=\int \limits _{\pi/6}^{\pi/3} \large\frac{\sqrt {\cos (\pi/6+\pi/3-x)}}{\sqrt {\cos (\pi/6+\pi/3-x)}+\sin \sqrt {(\pi/6+\pi/3-x)}}dx$
 
$I=\int \limits _{\pi/6}^{\pi/3} \frac{\sqrt {\cos (\pi/2-x)}}{\sqrt {\cos (\pi/2-x)}+\sqrt {(\pi/2-x)}}dx$
 
But $\cos (\pi/2-x)=\sin x$ and $ \sin (\pi/2-x)=\cos x$
 
Therefore $I=\int \limits _{\pi/6}^{\pi/3} \frac{\sqrt {\sin x}}{\sqrt {\sin x}+\sqrt {\cos x}}dx$-----(2)
 
Adding equ(1) and equ (2)
 
$I=\int \limits _{\pi/6}^{\pi/3} \frac{\sqrt {\sin x}+\sqrt {\cos x }}{\sqrt {\sin x}+\sqrt {\cos x}}dx=\int \limits _{\pi/6}^{\pi/3} dx $
 
on integrating we get,
 
$2I=\int \limits _{\pi/6}^{\pi/3} dx$
 
$=[x]_{\pi/6}^{\pi/3} $
 
Applying the limits we get
 
$2I=\bigg[\frac{\pi}{3}-\frac{\pi}{6}\bigg]$
 
$=\frac{\pi}{6}$
 
Therefore $I=\frac{\pi}{12}$

 

 

answered Mar 14, 2013 by meena.p
 
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