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.
$$\int^1_0 \cos^2\frac{(m+n)\pi x}{2}\sin^2\frac{(n-m)\pi x}{2}dx$$
 A: $$\sin{x}\cos{y}=\dfrac{1}{2}[\sin{(x+y)}+\sin{(x-y)}]$$
then
$$I=\int_{0}^{1}\dfrac{1}{4}[\sin{(n\pi x)}-\sin{(m\pi x)}]^2dx$$
and note
$$\int_{0}^{1}\sin^2{(n\pi x)}dx=\dfrac{1}{n\pi}\int_{0}^{n\pi}\sin^2{u}du=\dfrac{1}{\pi}\int_{0}^{\pi}\sin^2{u}du=\dfrac{1}{2\pi}\int_{0}^{\pi}(\sin^2{u}+\cos^2{u})du=\dfrac{1}{2}$$
and
$$\int_{0}^{1}\sin{(n\pi x)}\sin{(m\pi x)}dx=0(n\ne m)$$
proof: note
$$\sin{x}\sin{y}=\dfrac{1}{2}(\cos{(x-y)}-\cos{(x+y)})$$
so
$$\int_{0}^{1}\sin{(n\pi x)}\sin{(m\pi x)}dx=\dfrac{1}{2}\int_{0}^{1}[\cos{(n-m)\pi x}-\cos{(n+m)\pi x}]dx=0$$
so
$$\int_{0}^{1}\cos^2{\dfrac{(m+n)\pi x}{2}}\sin^2{\dfrac{(m-n)\pi x}{2}}dx=\dfrac{1}{4}(\dfrac{1}{2}-0+\dfrac{1}{2})=\dfrac{1}{4},m\neq n$$
A: 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. 
