1. ## sin(x)*sin(ax)

Hi everyone,

Is there a trig identity for combining
$
sin(x)*sin(ax)
$

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

Thanks,

2. Originally Posted by brian311
Hi everyone,

Is there a trig identity for combining
$
sin(x)*sin(ax)
$

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

Thanks,
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

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

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

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

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

Integrate by parts on the integral we have on the RHS

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

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

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

Putting this into the original integration by parts.

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

Note that we have the question, I, here.

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

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

Isolate I and this expression will be equal to the question

3. $\sin x \sin ax = \frac{1}{2} \Big(\cos (1-a)x - \cos (1+a)x \Big)$

4. Originally Posted by e^(i*pi)
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

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

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

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

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

Integrate by parts on the integral we have on the RHS

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

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

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

Putting this into the original integration by parts.

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

Note that we have the question, I, here.

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

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

Isolate I and this expression will be equal to the question

I think there may have been a small typo, I'm getting:
$I(1-a^2) = -\sin(ax) \cos(x) + a\sin(x) \cos(ax)$

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

Thanks again!

5. Originally Posted by brian311

I think there may have been a small typo, I'm getting:
$I(1-a^2) = -\sin(ax) \cos(x) + a\sin(x) \cos(ax)$

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

Thanks again!
You're right, it's hard to keep track of without seeing it infront of me on a sheet of paper

6. You don't have to integrate by parts.

$\int \sin x \sin ax = \frac{1}{2} \int \Big(\cos (1-a)x - \cos (1+a)x \Big) \ dx$

$= \frac{\sin(1-a)x}{2(1-a) } - \frac{\sin (1+a)x}{2(1+a)} +C$

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