A) \[\frac{2n}{{{n}^{2}}-{{m}^{2}}}\]
B) 0
C) \[\frac{2n}{{{m}^{2}}-{{n}^{2}}}\]
D) \[\frac{2m}{{{n}^{2}}-{{m}^{2}}}\]
Correct Answer: A
Solution :
\[I=\frac{1}{2}\int_{0}^{\pi }{[\,\sin (m+n)x-\sin (m-n)x]dx}\] \[=-\frac{1}{2}\left[ \frac{\cos (m+n)x}{m+n}-\frac{\cos (m-n)x}{m-n} \right]_{0}^{\pi }\] \[=-\frac{1}{2}\left[ \left\{ \frac{{{(-1)}^{m+n}}}{m+n}-\frac{{{(-1)}^{m-n}}}{m-n}- \right\}-\left\{ \frac{1}{m+n}-\frac{1}{m-n} \right\} \right]\] Since n ? m is odd, therefore \[n+m\]must be odd, so \[{{(-1)}^{m+n}}={{(-1)}^{m-n}}=-1\]. Also, since \[|m|\ne |n|,m+n\ne 0,m-n\ne 0\] \ \[I=\frac{1}{m+n}-\frac{1}{m-n}=\frac{m+n-m-n}{{{m}^{2}}-{{n}^{2}}}=\frac{2n}{{{n}^{2}}-{{m}^{2}}}\].
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