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Anybody know a general way to approach this sin/cos integral?

  1. Apr 2, 2009 #1
    Anyone know a nifty change of variables or trigonometric identity that will make this integral relatively easy to do:

    [tex]
    \int_0^T \sin(at+b) \cos(ct+d) dt
    [/tex]

    'Cause from where I'm standing, that's pretty awful...

    Thanks!
     
  2. jcsd
  3. Apr 2, 2009 #2

    Ben Niehoff

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    Is the integration over a whole period? That would simplify things significantly.
     
  4. Apr 2, 2009 #3
    well if you use the identities for sin(A+B) and cos(A+B) you can change it into four simpler integrals.

    Could you integrate cos(ax)sin(bx) with respect to x? if you can then using the above identities helps.
     
  5. Apr 2, 2009 #4
    I would choose the exponential form which makes it a matter of algebraic manipulation, but using trigonometric identities is equality valid.
    [tex]sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}[/tex]
    and [tex]cos(\theta) = = \frac{e^{i\theta} + e^{-i\theta}}{2}[/tex]
    substituting in the original equation, you'll eventually end up with an expression of the form
    [tex]\frac{e^{ix} - e^{-ix} + e^{iy} - e^{-iy}}{4i}[/tex]
    getting that back to the trigonometric form, you should end up with two easy integrals, namely the integral of 1/2sin(x) + 1/2sin(y)
     
    Last edited: Apr 2, 2009
  6. Apr 2, 2009 #5

    arildno

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    Remember that:
    [tex]\sin(u)\cos(v)=\frac{1}{2}(\sin(u+v)+\sin(u-v))[/tex]
     
  7. Apr 2, 2009 #6
    All of these are great suggestions. But I just wound up doing it by parts. It wasn't too bad.

    Thanks though. :smile:
     
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