Definite Integration Question 100
Question: $ \int _{-\pi }^{\pi }{{{(\cos px-\sin qx)}^{2}}dx} $ is equal to (where $ p $ and $ q $ are integers)
[IIT 1992]
Options:
A) $ -\pi $
B) 0
C) $ \pi $
D) $ 2\pi $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int _{-\pi }^{\pi }{({{\cos }^{2}}px+{{\sin }^{2}}qx-2\sin qx\cos px)dx} $
$ =\int _{-\pi }^{\pi }{({{\cos }^{2}}px+{{\sin }^{2}}qx)dx-2\int _{-\pi }^{\pi }{\sin qx\cos pxdx}} $
$ =2\int_0^{\pi }{({{\cos }^{2}}px+{{\sin }^{2}}qx)dx-0} $
$ y=4x-x^{2} $ .
BETA
