Definite Integration Question 470
Question: $ \int\limits_0^{\pi }{\frac{\sin ( n+\frac{1}{2} )x}{\sin x}}dx $ , $ (n\in N) $ equals
[Kurukshetra CEE 1998]
Options:
A) $ n\pi $
B) $ (2n+1)\frac{\pi }{2} $
C) $ \pi $
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
$ 2\sin \frac{x}{2}.( \frac{1}{2}+\cos x+\cos 2x+…..+\cos nx ) $
$ =\sin \frac{x}{2}+2\sin \frac{x}{2}\cos x+2\sin \frac{x}{2}\cos 2x+….+2\sin \frac{x}{2}\cos nx $
$ =\sin \frac{x}{2}+\sin \frac{3x}{2}-\sin \frac{x}{2}+\sin \frac{5x}{2}-\sin \frac{3x}{2}+….. $
$ +\sin ( n+\frac{1}{2} )x-\sin ( n-\frac{1}{2} )x $
$ =\sin ( n+\frac{1}{2} )x $
$ \frac{1}{2}+\cos x+\cos 2x+…..+\cos nx=\frac{\sin ( n+\frac{1}{2} )x}{2\sin ( \frac{x}{2} )} $
therefore $ \int_0^{\pi }{\frac{\sin ( n+\frac{1}{2} )x}{\sin ( \frac{x}{2} )}dx}=2( \int_0^{\pi }{\frac{1}{2}dx+\int_0^{\pi }{\cos xdx+…..+\int_0^{\pi }{\cos nxdx}}} ) $
$ =2( \frac{\pi }{2}+\sin x+…..+\frac{\sin nx}{n} )_0^{\pi }=\pi $ .
BETA
