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Evaluate (t-tau)*sin(a*tau) with respect to tau, tau = 0t

  1. Aug 25, 2008 #1
    Both Int {(t-tau)*sin(a*tau)}d(tau) and Int {(tau)*sin(a*(t-tau)}d(tau) will give the same answer (a*t-sin(a*t))/(a^2), where tau = 0..t

    Anybody can give a hint how to do the integration.

    Personally, I think neither integration by parts nor substitution are suitable methods.
  2. jcsd
  3. Aug 26, 2008 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    The first integral can be easily done with integration by parts. For the second integral, use the sum-of-angles formula and then integrate by parts.
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