Question

Evaluate $\int \sin mx\sin nxdx$ on $R,m\ne n,m$ and $n$ are positive integers.

Answer 1

We have,

$\int \sin \left(mx\right)\sin \left(nx\right)dx$

On multiplying and divide by $2$ and we get,

$\frac{1}{2}\int 2\sin \left(mx\right)\sin \left(nx\right)dx=\frac{1}{2}\int [\cos (mx-nx)-\cos (mx+nx)]dx$

$=\frac{1}{2}\int \cos (m-n)xdx-\frac{1}{2}\int \cos (m+n)xdx$

$=\frac{1}{2}[\frac{\sin (m-n)x}{m-n}-\frac{\sin (m+n)}{m+n}]+c$

Where $m\ne n$

Hence, this is the answer.