Definite Integration Question 499
Question: If $ (n-m) $ is odd and $ |m|\ne |n|, $ then $ \int_0^{\pi }{\cos mx\sin nx}dx $ is
Options:
A) $ \frac{2n}{n^{2}-m^{2}} $
B) 0
C) $ \frac{2n}{m^{2}-n^{2}} $
D) $ \frac{2m}{n^{2}-m^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\frac{1}{2}\int_0^{\pi }{[\sin (m+n)x-\sin (m-n)x]dx} $
$ =-\frac{1}{2}[ \frac{\cos (m+n)x}{m+n}-\frac{\cos (m-n)x}{m-n} ]_0^{\pi } $
$ =-\frac{1}{2}[ { \frac{{{(-1)}^{m+n}}}{m+n}-\frac{{{(-1)}^{m-n}}}{m-n}- }-{ \frac{1}{m+n}-\frac{1}{m-n} } ] $
Since n - m is odd, therefore $ n+m $ must be odd, so $ {{(-1)}^{m+n}}={{(-1)}^{m-n}}=-1 $ . Also, since $ |m|\ne |n|,m+n\ne 0,m-n\ne 0 $
$ I=\frac{1}{m+n}-\frac{1}{m-n}=\frac{m+n-m-n}{m^{2}-n^{2}}=\frac{2n}{n^{2}-m^{2}} $ .
BETA
