Hi everyone,

Is there a trig identity for combining

$\displaystyle

sin(x)*sin(ax)

$

and if not, what's the best way to integrate this?

Thanks,

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- Mar 27th 2010, 10:57 AM #1

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- Nov 2008
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- Mar 27th 2010, 11:07 AM #2
No there is no identity that I know of, the solution is to integrate by parts twice and realise that the second integral is equal to the question being asked. Also you should be very careful with your

**+**and**-**signs

$\displaystyle I = \int \sin(x) \sin(ax)$

$\displaystyle u = \sin(ax) \: \rightarrow \: u' = a \cos(ax)$

$\displaystyle v' = \sin(x) \: \rightarrow \: v = -\cos(x)$

$\displaystyle I = -\sin(ax) \cos(x) + a\int \cos(x) \cos(ax)$

Integrate by parts on the integral we have on the RHS

$\displaystyle u = \cos(ax) \: \rightarrow \: u' = -a \sin(ax)$

$\displaystyle v' = \cos(x) \: \rightarrow \: v = \sin(x)$

$\displaystyle \int \cos(x) \cos(ax) = (\sin(x) \cos(ax) + a\int \sin(x) \sin(ax))$

Putting this into the original integration by parts.

$\displaystyle I = -\sin(ax) \cos(x) + a(\sin(x) \cos(ax) + a\int \sin(x) \sin(ax))$

Note that we have the question, I, here.

$\displaystyle I = -\sin(ax) \cos(x) + a(\sin(x) \cos(ax) + aI) = -\sin(ax) \cos(x) + a\sin(x) \cos(x) + a^2I$

$\displaystyle I(1-a^2) = -\sin(ax) \cos(x) + a\sin(x) \cos(x)$

Isolate I and this expression will be equal to the question

- Mar 27th 2010, 11:12 AM #3

- Mar 27th 2010, 11:34 AM #4

- Joined
- Nov 2008
- Posts
- 8

- Mar 27th 2010, 11:42 AM #5

- Mar 27th 2010, 12:30 PM #6

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