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## Homework Statement

The problem is f(x) = sin2πx - (1/πsquare)*sinπx

and its given Bn sin (nπx) = f(x)

Question is find Bn.

## Homework Equations

Bn = 2/L ∫ (sin2πx - (1/πsquare)*sinπx) * sin(nπx/L) where L is 1

## The Attempt at a Solution

I did

[/B]

∫ 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?