Hi guys; I'm just dealing with fourier series and they evaluate integrals such as ∫sin(nπx/L)dx from 0 to L as (L/nπ)[1-(-1)^n]. Can someone please tell me how to get to this conclusion or point me in the direction of a resource that will show me? Additionally I need to solve similar integrals like ∫sin^2(nπx/L)dx and ∫xsin(nπx/L) so I really need to know the technique. Thanks in advance.
∫sin(nπx/L)dx from 0 to L = (L/nπ)[1-(-1)^n]
The Attempt at a Solution
I would traditionally evaluate ∫sin(nπx/L)dx from 0 to L as [L-Lcos(nπ)]/nπ