computing integrals with sinx and sin(nx)

**MacstersUndead** · Mar 14th 2012, 02:27 PM

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

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

For n even, the integral is 0 since sin(nx) would be odd about $\pi/2$
For n odd, I let n = 2m + 1 and hence my integral with the sin(a+b) identity gives

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

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.

**Kiwi_Dave** · Mar 14th 2012, 02:56 PM

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.

**MacstersUndead** · Mar 14th 2012, 02:58 PM

I shall try this. thank you very much.

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