A) 0
B) \[\pi \]
C) \[\frac{\pi }{2}\]
D) \[2\pi \]
Correct Answer: A
Solution :
Let \[I=2\int_{0}^{\pi }{\sin mx\sin nx\,dx}=\int_{0}^{\pi }{[\cos (m-n)x-\cos (m+n)x]dx}\] =\[\left[ \frac{\sin (m-n)x}{(m-n)}-\frac{\sin (m+n)x}{(m+n)} \right]_{0}^{\pi }\] \[=\left[ \frac{\sin (m-n)\pi }{(m-n)}-\frac{\sin (m+n)\pi }{(m+n)} \right]=0\]. Since, \[\sin (m-n)\pi =0=\sin (m+n)\pi \] for \[m\ne n\].
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