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Integration by parts twice

  1. Dec 7, 2012 #1
    ∫ e^at. sinωt dt

    This is the second part of an electrical circuit DE problem from our notes (first part not required to solve the above integral) however in-between this integral and the answer our professor only told us that we would get to the answer by using integration by parts twice. I am curious to understand this missed step and have tried integration by parts twice but can't seem to get the same answer as our professor gave us which should work out as:

    e^at (asinωt - ωcosωt) / (ω^2 + a^2)

    I would really appreciate it if someone could help me here.. I basically tried integrating by parts twice and taking the original negative integral from the second integration by parts to the LHS and then using that to half the RHS and ended up with e^at(sintωt - cosωt)/2, which is different to our lecturers result.

    Thanks to anyone for their time!
     
    Last edited: Dec 7, 2012
  2. jcsd
  3. Dec 7, 2012 #2

    micromass

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    Could you show us step-by-step what you did??
     
  4. Dec 7, 2012 #3
    ∫ e^at. sinωt dt

    u = e^at, u' = ae^at

    v' = sinωt, v = -cosωt

    ∫ uv' = uv - ∫ u'v

    -e^at cosωt - ∫ - ae^at cosωt

    -e^at cosωt + ∫ ae^at cosωt


    now the second integration by parts,

    a = ae^at; a' = a^2 e^at;
    b' = cosωt; b = sinωt;

    ab - ∫ a'b

    ae^at sinωt - ∫ a^2 e^at sinωt

    divide everything by a^2

    e^at sinωt/a - ∫ e^at sinωt

    let original integral = y now - ∫ e^at sinωt = -y

    take that to RHS, 2y = e^at sinωt/a

    y = e^atsinωt/2a


    There is my solution, I can't find where I went wrong so please feel free to point out where I screwed up or if anyone can find the solution and post it I should be able to work it out from there
     
    Last edited: Dec 7, 2012
  5. Dec 7, 2012 #4
    Forgot to quote
     
  6. Dec 7, 2012 #5

    micromass

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    It seems to me that you're forgetting a factor [itex]\omega[/itex] here.
    If [itex]v=-\cos\omega t[/itex], then [itex]v^\prime= \omega \sin\omega t[/itex], no??

    You made the same mistake in your second integration by parts.
     
  7. Dec 7, 2012 #6
    Ahh I get it now. Thanks for pointing that out I should be okay from here (hopefully)
     
  8. Dec 7, 2012 #7
    Hi, how about this:

    ∫eat sinωt dt

    u=sinωt du=ωcosωt dt
    v=∫eat dt = 1/a eat

    ∫eat sinωt dt = 1/a eat sinωt - ∫1/a eat ωcosωt dt

    ∫eat sinωt dt = 1/a eat sinωt - ω/a ∫eat cosωt dt

    u=cosωt du= -ωsinωt dt
    v=∫eat dt = 1/a eat

    ∫eat sinωt dt = 1/a eat sinωt - ω/a[ 1/a eat cosωt - ∫-ω/a eat sinωt dt ]

    ∫eat sinωt dt = 1/a eat sinωt - ω/a[ 1/a eat cosωt + ω/a∫eat sinωt dt ]

    ∫eat sinωt dt = 1/a eat sinωt - ω/a2 eat cosωt - ω2/a2 ∫eat sinωt dt

    ∫eat sinωt dt + ω2/a2 ∫eat sinωt dt = 1/a eat sinωt - ω/a2 eat cosωt

    a2∫eat sinωt dt + ω2∫eat sinωt dt = aeat sinωt - ωeat cosωt

    2 + a2)∫eat sinωt dt = eat(asinωt - ωcosωt)

    => ∫eat sinωt dt = eat(asinωt - ωcosωt) / (ω2 + a2) + C

    Ross :)
     
  9. Dec 7, 2012 #8

    Dick

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    Yes, that's just great!
     
  10. Dec 9, 2012 #9
    Thanks for your solution, I couldn't manage to see where the omega squared came from until I read yours
     
  11. Dec 9, 2012 #10
    Great! Glad to help. That was fun. :)
     
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