sin(x)*sin(ax)

**brian311** · March 27th 2010, 10:57 AM

Hi everyone,

Is there a trig identity for combining
$<br /> sin(x)*sin(ax)<br />$

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

Thanks,

**e^(i*pi)** · March 27th 2010, 11:07 AM

Originally Posted by brian311

Hi everyone,

Is there a trig identity for combining
$<br /> sin(x)*sin(ax)<br />$

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

**Random Variable** · March 27th 2010, 11:12 AM

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

**brian311** · March 27th 2010, 11:34 AM

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

Thanks for your help!

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

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

Thanks again!

**e^(i*pi)** · March 27th 2010, 11:42 AM

Originally Posted by brian311

Thanks for your help!

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

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

Thanks again!

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

**Random Variable** · March 27th 2010, 12:30 PM

You don't have to integrate by parts.

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

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

Thread: sin(x)*sin(ax)

LinkBack

Thread Tools

Display

sin(x)*sin(ax)

Search Tags