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Question: If \(| m | \neq | n |\) then \(\int _ { 0 } ^ { \pi } \cos m x \sin n x d x\) is...

If ∣m∣≠∣n∣| m | \neq | n | then ∫0πcos⁡mxsin⁡nxdx\int _ { 0 } ^ { \pi } \cos m x \sin n x d x is

A

2nn2−m2\frac { 2 n } { n ^ { 2 } - m ^ { 2 } }

B

0

C

2nm2−n2\frac { 2 n } { m ^ { 2 } - n ^ { 2 } }

D

2mn2−m2\frac { 2 m } { n ^ { 2 } - m ^ { 2 } }

Answer

2nn2−m2\frac { 2 n } { n ^ { 2 } - m ^ { 2 } }

Explanation

Solution

I=12∫0π[sin⁡(m+n)x−sin⁡(m−n)x]dxI = \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } [ \sin ( m + n ) x - \sin ( m - n ) x ] d x

=−12[cos⁡(m+n)xm+n−cos⁡(m−n)xm−n]0π= - \frac { 1 } { 2 } \left[ \frac { \cos ( m + n ) x } { m + n } - \frac { \cos ( m - n ) x } { m - n } \right] _ { 0 } ^ { \pi }

Since n – m is odd, therefore n+mn + m must be odd, so (−1)m+n=(−1)m−n=−1( - 1 ) ^ { m + n } = ( - 1 ) ^ { m - n } = - 1 .

Also, since ∣m∣∄n∣,m+n≠0,m−n≠0| m | \nexists n \mid , m + n \neq 0 , m - n \neq 0

∴ I=1m+n−1m−n=m+n−m−nm2−n2=2nn2−m2I = \frac { 1 } { m + n } - \frac { 1 } { m - n } = \frac { m + n - m - n } { m ^ { 2 } - n ^ { 2 } } = \frac { 2 n } { n ^ { 2 } - m ^ { 2 } } .