Posted by Neutral on Sunday, June 22, 2008 at 3:28pm.
I have the function f(x) = cos(x) on the interval from 0 to pi and I need to comput the Fourier sine series.
I have the integral of cos(x) multiplied by sin(nx), I can't figure out a way to integrate them! The "n" gets in the way, what do I do?

Fourier Sine Series Q  Damon, Sunday, June 22, 2008 at 4:15pm
first sketch a graph of cos x and sin x, sin 2x, sin 3x, sin 4x
in the interval from 0 to pi
You will notice that for n = 1 for example, the plus contribution between 0 and pi/2 cancels the negative contribution between pi/2 and pi
in fact for cos 1x * sin n x dx, only even values of n will contribute for the integral from 0 to pi.
The general rule for this definite integral is:
integral from 0 to pi of
sin ax cos bx dx
is:
2a/(a^2b^2) if (ab) is odd
0 if (ab) is even
so here:
2/(1n^2) if n is even
0 if n is odd

Fourier Sine Series Q  Anonymous, Sunday, June 22, 2008 at 4:38pm
So basically, that is the coefficient "bn" that I place in the summation from n=1 to infinity of sin(nx)?
If that is correct, thanks, I was going into a complicated page of integration . . .

Fourier Sine Series Q  Damon, Sunday, June 22, 2008 at 5:17pm
yes :)

thanks Damon  Neutral, Sunday, June 22, 2008 at 4:55pm
Okay, I got it, thanks Damon.
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