Evaluate:\[\int \limits _{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt {\tan x}}\]

Answer 1

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 $I=\frac{\pi}{12}$