Computing $\int_0^1 \sin(3\pi x)\sin(2\pi x) \mathrm{d}x$ 
How do I show
  $$\int_0^1 \sin(3\pi x)\sin(2\pi x) \mathrm{d}x=0.$$

I tried using
$$\sin (x)  \sin (y)  = \cos ( x \pm y ) \mp \cos (x)  \cos (y)  $$
Then the integral becomes
$$\int_0^1\cos(3\pi x)\cos(2\pi x) \mathrm{d}x$$
Which doesn't help me. I also tried using Taylor series expansion of sinus but I don't see how it is helpful since we have the product of two infinite sums.
 A: It should be
$$\int\limits_0^1 \sin3\pi x\sin2\pi x \mathrm{d}x=\frac{1}{2}\int\limits_0^1( \cos\pi x-\cos5\pi x) \mathrm{d}x=$$
$$=\frac{1}{2}\left(\frac{1}{\pi}\sin\pi x-\frac{1}{5\pi}\sin5\pi x\right)_0^1=\frac{1}{2}(0-0)=0$$
A: $$ \int_0^1 \sin (2\pi x)\sin(3\pi x)\mathrm{d}x = \int_{\frac{-1}{2}}^{\frac{1}{2}} \sin(2\pi (x+\frac{1}{2}))\sin(3\pi(x+\frac{1}{2}))\mathrm{d}x$$
$$ = \int_{\frac{-1}{2}}^{\frac{1}{2}} \sin(2\pi x)\cos(3\pi x)\mathrm{d}x$$
As $x \mapsto \sin(2\pi x)\cos(3\pi x)$ is an odd function and $[-\frac{1}{2}, \frac{1}{2}]$ is centred around $0$, the integral is equal to 0.
A: Using integration by parts twice,
$\begin{align}J&=\int_0^1 \sin\left(3\pi x\right)\sin\left(2\pi x\right)\,dx\\
&=\left[-\frac{1}{2\pi}\sin\left(3\pi x\right)\cos\left(2\pi x\right)\right]_0^1+\frac{1}{2\pi}\int_0^1 \sin\left(3\pi x\right)\cos\left(2\pi x\right)\, dx\\
&=\frac{1}{2\pi}\int_0^1 \sin\left(3\pi x\right)\cos\left(2\pi x\right)\, dx\\
&=\left[\frac{1}{(2\pi)^2}\sin\left(3\pi x\right)\sin\left(2\pi x\right)\right]_0^1-\frac{1}{(2\pi)^2}J\\
&=-\frac{1}{(2\pi)^2}J
\end{align}$
Since $\displaystyle 1+\frac{1}{(2\pi)^2}\neq 0$ then $J=0$.
