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September 2, 2014

September 2, 2014

Posted by **Neutral** on Sunday, June 22, 2008 at 3:28pm.

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 4:15pmfirst 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^2-b^2) if (a-b) is odd

0 if (a-b) is even

so here:

2/(1-n^2) if n is even

0 if n is odd

- Fourier Sine Series Q -
**Anonymous**, Sunday, June 22, 2008 at 4:38pmSo 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 5:17pmyes :)

- Fourier Sine Series Q -

- Fourier Sine Series Q -
- thanks Damon -
**Neutral**, Sunday, June 22, 2008 at 4:55pmOkay, I got it, thanks Damon.

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