For a Fourier method problem, I need to solve the following integral for the closed forms of coefficients $\displaystyle c_n$

$\displaystyle \int_0^\pi \! sin^2(x)sin(nx) \, \mathrm{d} x.$

For n even, the integral is 0 since sin(nx) would be odd about $\displaystyle \pi/2$

For n odd, I let n = 2m + 1 and hence my integral with the sin(a+b) identity gives

$\displaystyle \int_0^\pi \! sin^2(x)sin(2mx)cos(x) + sin^2(x)cos(2mx)sin(x) \, \mathrm{d} x.$

but here I'm not sure what to do next. perhaps there is an easier way to approach this problem. any suggestions would be greatly appreciated.