sin(wt)cos(nwt) dt

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- Nov 3rd 2009, 09:37 AMrichard7893how do you do this integral
sin(wt)cos(nwt) dt

- Nov 3rd 2009, 09:44 AMJameson
- Nov 3rd 2009, 09:52 AMrichard7893
But if you do this your du=-nwsin(

**nwt**). By doing this you cant substitute for sin(wt) because du=-nwsin(**nwt**) - Nov 3rd 2009, 10:41 AMJameson
Ooops! Sorry I missed that. Eyes are wearing out from looking at the computer.

Hmmm, my first thought is to try the product to sum conversation for sin(a)cos(b), which looks like:

$\displaystyle \sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$

That should split up into things you can integrate separately easily.