The problem is f(x) = sin2πx - (1/πsquare)*sinπx
and its given Bn sin (nπx) = f(x)
Question is find Bn.
Bn = 2/L ∫ (sin2πx - (1/πsquare)*sinπx) * sin(nπx/L) where L is 1
The Attempt at a Solution
∫ sin2πx * sin (nπx) - (1/πsquare)*sin square πx
then I tried changing it to cos but it doesn't make sense, for the first term ∫sin2πx * sin (nπx)
∫sin2πx * sin (nπx) = -1/2 ∫cos(2π+nπ)-cos(2π-nπ)
But how do i actually integrate the above with the n in it i am so confused...I know cosnπ = (-1)^n but I can only use that after final integration and even if I finally somehow manage to integrate it, it will be changed to sinnπ. Please can anyone help!
. Is there any simple way to do this?