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Integration by parts I(n-2)

  1. Aug 21, 2007 #1
    The problem statement, all variables and given/known data

    Hi, I'm trying to solve
    [​IMG]
    in terms of I(n-2) but I'm not exactly sure where to start/what to do :uhh:
     
  2. jcsd
  3. Aug 21, 2007 #2

    HallsofIvy

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    You titled this "integration" by parts. Doesn't that give you a hint where to start?

    Since it is much easier to differentiate a power of a function than to integrate, the choice of u= sinn(x) and dv= sin(x) seems simplest.

    Since the problem says "in terms of I(n-2)" I suspect you will need to do it twice.
     
  4. Aug 21, 2007 #3

    Gib Z

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    Dido what Halls said, with tiny corrections:

    [tex]u= \sin^{n-1} x, dv = \sin x dx[/tex] is what he might have meant.
     
  5. Aug 21, 2007 #4
    Bad mentor! Giving wrong hints, tut tut...

    Hints: only perform by parts once; getting [tex]I(n)[/tex] again on the right hand side is no bad thing; remember that for trig problems you'll usually need to use trig identities at some point.
     
  6. Aug 21, 2007 #5

    HallsofIvy

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    I was thinking right- my fingers hit the wrong keys!

    (and it is "ditto", not "dido"!)
     
    Last edited: Aug 21, 2007
  7. Aug 21, 2007 #6
    How would I diff u?
     
  8. Aug 21, 2007 #7

    Gib Z

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    Use the chain rule?
     
  9. Aug 21, 2007 #8
    Let u = (n-1) hence y=[sin(x)]^(u)
    then du/dn = 1

    but again, unsure how to get dy/du now..
     
  10. Aug 21, 2007 #9

    Gib Z

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    Instead, try letting u = sin x when differentiating :)
     
  11. Aug 21, 2007 #10
    [tex]\frac{d}{dx}\left(\sin^{n-1}(x)\right) = \left((n-2)\sin^{n-2}(x)\right) \left(\cos(x)\right)[/tex]

    Btw, you can't really do integration without being able to do differentiation like the back of your hand -- more practise at that would be helpful.
     
    Last edited: Aug 21, 2007
  12. Aug 21, 2007 #11
    I made a mistake before when doing chain rule. To prevent confusion, I will initially let f(x)=[sin(x)]^(n-1) and g'(x)=sinx

    Then I use chain rule,
    now I let u = sinx, hence f(x) = u^(n-1)

    dy/du=(n-1)(u^(n-2))
    and du/dx = cos(x)

    hence dy/dx= (n-1)(sinx^(n-2))(cosx)?
     
  13. Aug 21, 2007 #12
    Can you please explain how you got that/what steps did I do wrong above?
     
  14. Aug 21, 2007 #13
    You got it right -- I got it wrong -- looks like I need more practice... :redface:
     
  15. Aug 21, 2007 #14

    Gib Z

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    *in imitation of Mr. Burns on the Simpsons* Excellent..
     
  16. Aug 21, 2007 #15
    coolio!
    I sub all the relevant values into the "magical integration by parts formula", but the second part seems awfully tedious..

    [​IMG]

    Can (n-1) be taken out as a constant, leaving just the sin and cos terms for me to integrate by parts again?
     
  17. Aug 21, 2007 #16

    Gib Z

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    Yes it can, but also it wants you to express cos^2 x as 1-sin^2 x, get the original integral "I" on both sides :)
     
  18. Aug 21, 2007 #17
    ah ok!
    After converting (cosx)^2 using the trig identity I get..
    [​IMG]
    Where to from there?
     
  19. Aug 21, 2007 #18

    learningphysics

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    careful. you can't separate it like that:

    [tex]\int{f(x)*g(x)dx}\ne\int{f(x)dx}*\int{g(x)dx}[/tex]

    keep the [tex]sin^{n-2}x[/tex] inside the integral, and multiply it out by [tex]-1+sin^2x[/tex]
     
  20. Aug 21, 2007 #19
    ah ok, you mean expand to give..
    [​IMG]

    but then what?
     
  21. Aug 21, 2007 #20

    learningphysics

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    Remember... your right side right now just has [tex]\int{vdu}[/tex]... but it's supposed to be [tex]uv - \int{vdu}[/tex]

    To continue... remember that sums within integrals can be separated out into separate integrals (unlike products)... once that's done, you can substitute your I(n-2)... and use algebra to solve for [tex]\int{sin^n(x)dx}[/tex]
     
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