<s>[INST] Evaluation of $\int^1_0 \cos^2\frac{(m+n)\pi x}{2}\sin^2\frac{(n-m)\pi x}{2}dx$. Just wondering whether the following integration is something special. To be more specific, is it equal to some constant real number, please? I found a integration table involving sine and cosine but it requires that sine and cosine have the same argument, which is clearly not the case here.
$$\int^1_0 \cos^2\frac{(m+n)\pi x}{2}\sin^2\frac{(n-m)\pi x}{2}dx$$
 [/INST] First notice that $$\cos\frac{(m+n)\pi x}{2}\sin \frac{(n-m)\pi x}{2} = \frac{1}{2} \left( \sin n\pi x -\sin m\pi x\right)$$
From this, we deduce that the given integral formula is equal to
$$\frac{1}{4}\int_0^1 {\left( \sin n\pi x -\sin m\pi x\right)}^2dx \\= \frac{1}{4}\int_0^1 { \sin^2 n\pi x +\sin^2 m\pi x - \sin n\pi x \sin m \pi x}dx \\ = \frac{1}{4}\int_0^1 { \frac{1-\cos 2n\pi x}{2} +\frac{1-\cos 2m\pi x}{2} + \frac{1}{2}\left( \cos (n+m)\pi x - \cos (n-m)\pi x\right)}dx$$
The calculations left are easy. 
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