Let $m$ and $n$ be positive integers. First consider the case where $mn.$ By the product identity $sin(x)sin(y)=\frac{1}{2}[cos(x-y)-cos(xy)],$ the integral can be rewritten as follows.

Now, consider the case where $m=n.$ Note that the integral can be rewritten as follows.

${\displaystyle \underset{-}{\overset{}{}}}sin(mx)sin(nx)dx={\displaystyle \underset{-}{\overset{}{}}}sin(mx)sin(mx)dx$

$={\displaystyle \underset{-}{\overset{}{}}}si{n}^2(mx)dx$

By the half angle formula $si{n}^2(x)=\frac{1-cos(2x)}{2},$

${\displaystyle \underset{-}{\overset{}{}}}si{n}^2(mx)dx={\displaystyle \underset{-}{\overset{}{}}}\frac{1}{2}(1-(\frac{cos(2x)}{2}))dx$

$=[\frac{x}{2}{]}_{-}^{}-[\frac{sin(2x)}{(4x)}{]}_{-}^{}=$

$28$