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:\[\int\limits_0^1 \sqrt{\frac{1-x}{1+x}}dx\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii) $ 1-\cos 2x =2 \sin ^2 x$
  • (iii) $ 1+\cos 2x=2 \cos ^2x$
Given $\int\limits_0^1 \sqrt{\frac{1-x}{1+x}}dx$
 
Put $x=\cos 2\theta$ on differentiating w.r.t.x we get,
 
$dx=-2 \sin 2 \theta d\theta$
 
The limits also change when we substitude $ \theta$
 
when $x=0, \cos 2 \theta\;=>2 \theta =\frac{\pi}{2}=>\frac{\pi}{4}$
 
when $x=1, \cos \theta =1=> \theta =0$
 
Therefore $I=\int \limits_{\pi/4}^0 \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin 2 \theta d \theta)$
 
we know that $1-\cos \theta=2 \sin ^2 \theta\;and\; 1+\cos \theta =2\cos ^2 \theta$
 
Therefore $I=-2 \int \limits_{\pi/4}^0 \sqrt {\frac{\sin ^2 \theta}{\cos ^2 \theta}}(\sin 2 \theta d\theta)$
 
$\sin 2 \theta=2 \sin \theta \cos \theta$
 
The negative symbol can be removed by changing the limits,
 
(ie) $I=2 \int \limits_0^{\pi/4} \frac{\sin \theta}{\cos \theta} 2 \sin \theta \cos \theta d\theta$
 
$=2 \int \limits_0^{\pi/4} 2 \sin ^2 \theta d \theta \qquad But \;2\sin^2\theta =1-\cos 2 \theta$
 
$=2 \int \limits_0^{\pi/4}(1-\cos 2 \theta) d \theta$
 
on integrating we get , $I=2 \bigg[\int \limits_0^{\pi/4} d \theta- \int \limits_0^{\pi/4} \cos 2 \theta d\theta\bigg]$
 
$=2\bigg\{[\theta]_0^{\pi/4}-\frac{1}{2}[\sin 2 \theta]_0^{\pi/4} \bigg\}$
 
On applying limits,
 
$2\{[\pi/4-0]-\frac{1}{2}[\sin 2.\pi/4-\sin 0]\} \qquad\; But\; \sin \pi/2=1$
 
Therefore $I=2.\frac{\pi}{4}-1=\frac{\pi}{2}-1$

 

 

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
...