logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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

Can you answer this question?
 
 

1 Answer

0 votes
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)
 
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 $2I=\frac{\pi}{2}$
 
$=>I=\frac{\pi}{4}$

 

 

answered Mar 15, 2013 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...