You could try integrating by parts with:
u = Sin x
v' = Sin x * Sin nx = 0.5[Cos (n-1)x - Cos (n+1)x]
I haven't tried it, but solving for this particular v is a common task in fourier problems.
For a Fourier method problem, I need to solve the following integral for the closed forms of coefficients
For n even, the integral is 0 since sin(nx) would be odd about
For n odd, I let n = 2m + 1 and hence my integral with the sin(a+b) identity gives
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.