If the solve the problem


$\int \sin m x \cos n x d x, m \neq n$


We know $2 \sin A \cos B=\sin (A+B)+\sin (A-B)$

$\therefore \sin m x \cos n x=\frac{\sin (m+n) x+\sin (m-n) x}{2}$

$\therefore$ The above equation becomes

$\Rightarrow \int \frac{1}{2}(\sin (\mathrm{m}+\mathrm{n}) \mathrm{x}+\sin (\mathrm{m}-\mathrm{n}) \mathrm{x}) \mathrm{dx}$

We know $\int \sin \mathrm{ax} \mathrm{dx}=\frac{-1}{\mathrm{a}} \cos \mathrm{ax}+c$

$\Rightarrow \frac{1}{2}\left(\frac{-1}{m+n} \cos (m+n) x-\frac{1}{(m-n)} \cos (m-n) x\right)$

$\Rightarrow \frac{1}{2}\left(\frac{-(m-n) \cos (m+n) x-(m+n) \cos (m-n) x}{m^{2}-n^{2}}\right)$

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