Fourier B-coefficients for a tricky f(x)

[blaksheep423]

Homework Statement

compute the Fourier coefficient B_n:

B_n = 2/L$$\int$$f(x)*sin(k_n x) dx
where k_n = n*pi/L and the integral runs over L (pardon the appearance, I am not very good with LaTeX).

Homework Equations

f(x) = 1/2(1-cos(2*pi*x/a) for 0 < x < a
f(x) = 0 for a < x < L

The Attempt at a Solution

I have a pretty solid grip on Fourier transforms and integrating to find the coefficients. Using quite a bit of trig identities and splitting the integral into about 5 parts I was able to come up with a very long solution, a solution which Maple 12 verified was true. I won't bother putting it here because I'm fairly sure that my professor is not looking for something that large and unwieldy. We are expected to use the coefficients for later parts of the assignment. Are there any sneaky tricks I don't know about for integrating sin*cos when the arguments inside are different?

 

[kuruman]

I don't know if this is "sneaky", but I do such integrals by converting sines and cosines to complex exponentials using

sinx= (e^ix-e^-ix)/2i
cosx= (e^ix+e^-ix)/2

When you multiply things out, you end up with a bunch of exponentials which are trivial to integrate. Once you evaluate the integrals, you can go back from complex exponentials to sines and cosines.