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Integration tough cookie

  1. Jan 16, 2013 #1
    1. The problem statement, all variables and given/known data
    see attachment.

    2. Relevant equations

    3. The attempt at a solution
    ok so the equation is in the attachment. My question is how did ∫2e....etc. in the first line get solved and get the answer in the second line. I tried integration by parts and set 2e^-2/t as the u and sin(3t)dt as dv and then did the uv-∫vdu, but then i get a ecos(3t) form for the ∫vdu and if i kept doing integration by parts i get never ending esin/cos. Anybody can help me or tell me how to solve this much appreciated.

    Attached Files:

  2. jcsd
  3. Jan 16, 2013 #2

    Ray Vickson

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    You cannot set u = 2e^(-2/t); instead, set u = 2 e^(-t/2).

    In problems like this, the integral I can be found using integration by parts twice: you will get an equation of the form I = f1(t) + f2(t)*I, so you can solve for I. I am surprised you have not seen this before; it is one of the standard 'tricks'.
  4. Jan 17, 2013 #3
    well i apologize if my lack of intelligence offended you, but thanks anyway.
  5. Jan 17, 2013 #4


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    Ray Vickson said nothing about your intelligence. He only said that he was surprised that you had not seen, before, a particular, but commonly used, method for this kind of problem. That has to do with experience, not intelligence.
    Last edited by a moderator: Jan 17, 2013
  6. Jan 17, 2013 #5


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    You may have a typo in your working of the problem.

    The stated problem in the attachment has [itex]\displaystyle \ \ e^{-t/2}\ .\ [/itex] I your Original Post, you are letting u=2e-2/t rather than letting u=2e-t/2 .

    In either case, du/dt ≠ e .

    [itex]\displaystyle \frac{d}{dt} e^{-t/2}=\frac{-1}{2}e^{-t/2}\ .\ [/itex]

    Considering the sign and the coefficient of the cosine term, I'm convinced that you should go the other way with integration by parts,
    Let u=sin(3t) and dv=2e-2/tdt .​
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