How to solve $ \int_0^{\pi} \sin{2x}\sin{x} dx $?

How to solve $$\int_0^{\pi} \sin{2x}\sin{x} dx$$

Edit: Sorry! I should have described more. This is not a homework. Recently, Out of the blue I got interest in physics and started reading and solving problems. This is part of a physics problem where I got stuck (because I forgot all high school formulae.). Thanks all of you guys for wonderful solutions.

Plotting $ \sin{2x}\sin{x} $ in $[0,\pi]$ shows that it's symmetrical with respect to $x=\pi/2$ and so the integral is zero.

HINT: Use this formula $$\cos C - \cos D = 2 \sin\frac{(C+D)}{2} \cdot \sin\frac{(D - C)}{2}$$

Two more ways: use Euler's formula $\sin ax=(e^{iax}-e^{-iax})/2i$, or integrate by parts twice to get an equality where your sought integral appears twice and can be solved for.

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