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 the definite integral $\int\limits_0^\pi(\sin^2\frac{x}{2}-\cos^2\frac{x}{2})dx$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int \cos x dx=\sin x+c$
  • (iii)$\cos x=\cos ^2 \frac{x}{2}-\sin^2\frac{x}{2}$
Given:$I=\int\limits_0^\pi(\sin^2\frac{x}{2}-\cos^2\frac{x}{2})dx$
 
$\sin ^2 \frac{x}{2}-\cos ^2\frac{x}{2}=-(\cos^2 \frac{x}{2}-\sin ^2\frac{x}{2})$
 
$=-\cos x$
 
Therefore $I=\int \limits_0^\pi -\cos x dx$
 
On integrating we get,
 
$-[\sin x]_0^{\pi}$
 
On applying limits,
 
$-[\sin \pi-\sin 0]$
 
But $ \sin \pi=\sin 0=0$
 
Therefore I=0

 

answered Feb 11, 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
...