Definite Integration Question 59
Question: The value of the integral $ \int _{-\pi }^{\pi }{{{(\cos ax-\sin bx)}^{2}}dx} $ , (a and b are integer) is
[UPSEAT 2001]
Options:
A) $ -\pi $
B) 0
C) $ \pi $
D) $ 2\pi $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int _{-\pi }^{\pi }{{{(\cos ax-\sin bx)}^{2}}dx} $
$ I=\int _{-\pi }^{\pi }{({{\cos }^{2}}ax+{{\sin }^{2}}bx-2\cos ax\sin bx)dx} $
$ I=\int _{-\pi }^{\pi }{({{\cos }^{2}}ax+{{\sin }^{2}}bx)dx}-\int _{-\pi }^{\pi }{2\cos ax\sin bxdx} $
$ I=2\int_0^{\pi }{({{\cos }^{2}}ax+{{\sin }^{2}}bx)dx}-0 $
$ I=2\int_0^{\pi }{( \frac{1+\cos 2ax}{2}+\frac{1-\cos 2bx}{2} )dx} $
$ I=\int_0^{\pi }{( 2+\cos 2ax-\cos 2bx )dx}=2\pi . $
BETA
