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Integration by Parts (1) Help & Advice needed!

  1. Sep 13, 2007 #1
    Picture. I don't know where I've gone wrong :-[

    Any advice on how to tackle Integration by Parts? What should I subst. with what?

    Thanks!

    [​IMG]
     
  2. jcsd
  3. Sep 13, 2007 #2
    You have the right idea. This is one of those funny integrals where the answer "unfolds" to you as you move the "integral" itself to a certain side.

    I'll give you a hint, you have

    integral of something = crap + integral of something * constant

    that integral of something, on your last line of work, happens to be the same! Now from this point, it's all algebra. Move your integral of something to one side, leave that integral of something byitself, divide by whatever constant you got, that is your answer.
     
  4. Sep 14, 2007 #3

    Gib Z

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    ie [tex] (1+\frac{9}{4})I = \frac{1}{2}e^{2\theta} \sin 3\theta - \frac{3}{4} e^{2\theta} \cos 3\theta[/tex] :)
     
  5. Sep 14, 2007 #4

    HallsofIvy

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    Your basic problem is that you don't have any "=" in your last line. That's the right hand side of what you first wrote. What's the left hand side?
     
  6. Sep 14, 2007 #5
    lmao ... i just learned how to do it Calc. 3 style :D

    some drawbacks tho, but i'm still happyyy!
     
  7. Sep 15, 2007 #6
    lol what the heck? how do you do this with calc 3

    share with us
     
  8. Sep 15, 2007 #7

    dextercioby

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    Use Euler's formula [itex] e^{ix} =\cos x +i\sin x [/itex] and rewrite the "sine" function in terms of the complex exponential. Integrate the simple products of exponentials.
     
  9. Sep 15, 2007 #8

    Gib Z

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    It seems like that method would actually take more lines of working though...I mean, if you want to just make things more complex than they need to be, replace the integrand with its Taylor series, integrate term by term and then apply a transformation to the series to see what its closed form is.
     
  10. Sep 15, 2007 #9

    dynamicsolo

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    The complex-exponential approach requires roughly the same amount of writing as integrating by parts, but you *do* get the integrals for both [tex] e^{2\theta} \sin 3\theta[/tex] and [tex]e^{2\theta} \cos 3\theta[/tex] simultaneously. The Taylor series integration is fine, too, but the set-up for the integration and the unraveling of the final expression from the resultant series is a big pain, relatively speaking.

    Working with complex exponentials is tidy enough, but the preparation for the concept is a bit too involved to visit on first-time calculus students...
     
  11. Sep 15, 2007 #10
    i'll be glad to share :) i didn't realize that the problem i posted is one of the drawbacks, but with other types of problems, it's very useful :)

    take this for example

    [tex]\int x^{2}\cos{x}dx[/tex] let (and take the derivative till it becomes 0) [tex]\int^{'} = x^{2}[/tex] and (take the integral with respects to the derivative till it becomes 0) [tex]\int = cos{x}dx[/tex]

    1 [tex]+\int^{'}=x^{2}[/tex] \\ [tex]\int = cos{x}[/tex]
    2 [tex]-\int^{'}=2x[/tex] \\ [tex]\int = sin{x}[/tex]
    3 [tex]+\int^{'}=2[/tex] \\ [tex]\int = -cos{x}[/tex]
    4 [tex]-\int^{'}=0[/tex] \\ [tex]\int = -sin{x}[/tex]

    now just combine and multiply the signs the derivative of 1 with the integral of 2

    final answer shoud be

    [tex]x^{2}\sin{x}+2x\cos{x}-2\sin{x}+C[/tex]

    if none of your variables will go to 0, it kinda sucks, but there are other methods :D i can't wait to learn'em.
     
    Last edited: Sep 15, 2007
  12. Sep 15, 2007 #11

    HallsofIvy

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    Your last line, on your working, is
    [tex] \frac{1}{2}e^{2\theta}sin(3\theta)- \frac{3}{4}e^{2\theta}cos(3\theta)-\frac{9}{4}\int e^{2\theta}sin(3\theta)}[/tex]
    and you were stuck there.

    My point was that if you wrote out the whole thing:
    [tex]\int e^{2\theta}sin(3\theta)= \frac{1}{2}e^{2\theta}sin(3\theta)- \frac{3}{4}e^{2\theta}cos(3\theta)-\frac{9}{4}\int e^{2\theta}sin(3\theta)}[/tex]
    that might give you some insight! Can you solve that equation for the integral?

    And I don't believe for a moment that writing only part of an equation is "Calculus III style".
     
  13. Sep 15, 2007 #12
    But on the question above, you can't apply this method because none of the variables will go to 0. How did you solve that one?
     
  14. Sep 15, 2007 #13
    x^2 can

    if it's like the question i asked then idk how
     
  15. Sep 15, 2007 #14

    Gib Z

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    Is that Calc III? Because It's nothing special. Its just repeated integration by parts, or sometimes called Tabular integration.
     
  16. Sep 15, 2007 #15
    yes but i'm in Calc 2, i have to do it the hell way.
     
  17. Sep 16, 2007 #16

    HallsofIvy

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    Have you read my last post?
     
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