Thread: computing integrals with sinx and sin(nx)

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.

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.

I shall try this. thank you very much.