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Cyclical Integration by Parts, going round and round

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Integrate By Parts (i.e. not using formulas)
    ∫e3xcos(2x)dx


    3. The attempt at a solution
    I keep going around in circles, I know at some point I should be able to subtract the original integral across the = and then divide out the coefficient and thats the final answer. But I keep on getting different answers.

    If some one could show the proper u=x1 du=x2 and v=x2 dv=x4 and state how many times I will have to repeat this and what u=x1 du=x2 and v=x2 dv=x4 should be for each step in the serial progression that is integration by parts. Please and Thanks
     
  2. jcsd
  3. Feb 2, 2012 #2

    jbunniii

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    You should be able to integrate by parts twice. Then there is a nice trick. Can you show us what you have tried so far?
     
  4. Feb 2, 2012 #3
    y=∫e3xcos(2x)dx where u=e3x du=3e3x & v=(1/2)sin(2x) dv=cos(2x)

    After integrating by parts twice i get this

    y=e3xsin(2x)/2 - e3xsin(2x)/2 +sin2xe3x/2 -∫e3xcos(2x)dx


    the 2nd time I integrated by parts it was

    u=sin2x du=2cos(2x)

    v=(1/3)e3x dv=e3x
     
  5. Feb 2, 2012 #4

    jbunniii

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    OK good. Now for the trick.

    You have defined

    [tex]y = \int e^{3x} \cos(2x) dx[/tex]

    Now look at the right hand side of the result you got. The same integral appears again. So using the definition of y, you have

    y = other stuff - y

    Can you see what to do now?
     
  6. Feb 2, 2012 #5
    Yeah I can see it but my answer didn't match up with what Wolfram gave.

    they gave 1/13 e^(3 x) (3 cos(2 x)+2 sin(2 x))

    And what am I missing why do your equations look so much...prettier?
     
  7. Feb 2, 2012 #6
    And what did you get for your answer?
    It's possible your answer is still equivalent to what WA gives you.
     
  8. Feb 2, 2012 #7
    Well yes or coarse you silly goose. I guess i need to plug in some numbers and see if they work out. Ill report back shortly.
     
  9. Feb 2, 2012 #8
    Its not coming out the same, I don't get it?
     
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