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How to evaluate sin(t-tau)cos(tau) wrt tau(0t)?

  1. Aug 25, 2008 #1
    How to integrate sin(t-tau)cos(tau)d(tau) where tau = 0..t

    Maple gives answer (1/2)*t*sin(t)

    How can we compute it by hand?
  2. jcsd
  3. Aug 25, 2008 #2


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    I would start with a trig identity. sin(t- tau)= sin(t)cos(tau)- cos(t)sin(tau). Now the integral is
    [tex]\int_0^t (sin(t)cos^2(tau)- cos(t)cos(tau)sin(tau)) dtau[/tex]
    [tex]= sin(t)\int_0^t cos2(tau)dtau- cos(t)\int_0^t cos(tau)sin(tau)dtau[/tex]
    You can integrate the first by using cos2(tau)= (1/2(1+ cos(2tau)) and in the second, let u= sin(tau).
    Last edited by a moderator: Aug 26, 2008
  4. Aug 25, 2008 #3
    Any simple way to do it?

    If this happened in the exam, it will be a disaster!
  5. Aug 25, 2008 #4

    Ben Niehoff

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    You should learn the trig identities well enough so that they ARE simple to you. Then you can apply them in exams.

    If you forget the sum-of-angles formula, you can easily derive it from Euler's formula:

    [tex]e^{i(\theta + \phi)} = \cos (\theta + \phi) + i \sin (\theta + \phi)[/tex]

    [tex]e^{i\theta} e^{i\phi} = \cos (\theta + \phi) + i \sin (\theta + \phi)[/tex]

    [tex](\cos \theta + i \sin \theta) (\cos \phi + i \sin \phi) = \cos (\theta + \phi) + i \sin (\theta + \phi)[/tex]

    [tex](\cos \theta \cos \phi - \sin \theta \sin \phi) + i (\cos \theta \sin \phi + \sin \theta \cos \phi) = \cos (\theta + \phi) + i \sin (\theta + \phi)[/tex]

    from which you can read off the two identities:

    [tex]\cos (\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi[/tex]

    [tex]\sin (\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi[/tex]
  6. Aug 25, 2008 #5
    Unable to obtain (1/2)*t*sin(t) from this method.
    Would you please show a little steps?
  7. Aug 26, 2008 #6
    Which integral are you stuck on? Show your work.
  8. Aug 26, 2008 #7


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    Hi lkj-17 and HallsofIvy! :smile:
    Do it in one go with sinA cosB = (sin(A+B) + sin(A-B))/2:

    ∫sin(t-tau)cos(tau)dtau = ∫(sint + sin(t-2tau))dtau/2. :smile:
  9. Aug 26, 2008 #8


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    Yes, that was what was originally suggested. Apparently ikj-17 feels he is not capable of doing that.
  10. Aug 26, 2008 #9


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    No … the original suggestion was sin(A-B) = sinAcosB - cosAsinB, which is different, and only dealt with half the expression.

    sinA cosB = (sin(A+B) + sin(A-B))/2 deals with the whole thing in one go, and avoids all those nasty squares. :biggrin:

    ikj-17 may even be able to integrate ∫(sint + sin(t-2tau))dtau/2 from 0 to t just by looking at it! :wink:
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