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Homework Help: Definite integration by parts!

  1. Mar 31, 2007 #1
    Hello there. I feel like this isn't the right answer, but I'd like some verification as to where exactly I went wrong! 1. The problem statement, all variables and given/known data is [tex]\int_{0}^{pi}x^2cos x dx[/tex]

    3. The attempt at a solution went something like this:
    [tex]u=x^2 dv=cos x dx
    du=2x dx v=\int_{0}^{pi}cos x dx= sin x[/tex]

    The integral was then:
    [tex]\int_{0)^{pi}x^2cos x dx= x^2 sin x\right]_{0}^{pi}-\int_{0}^{pi}sin x 2x dx[/tex]

    to solve:

    [tex]=x^2 sin x + cos x x^2[/tex]
    [tex]=pi^2 sin pi + cos 0 o^2[/tex]
    [tex]9.87 times 0=1+0[/tex]
    [tex]=1[/tex]

    Thanks for all your help in advance! :smile:





     
    Last edited: Mar 31, 2007
  2. jcsd
  3. Mar 31, 2007 #2

    cristo

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    Ive just cleaned up your tex so I can check your work.
     
  4. Mar 31, 2007 #3

    cristo

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    From here: [tex]\int_{0}^{\pi}x^2cos x dx= \left[x^2 sin x\right]_{0}^{\pi}-\int_{0}^{\pi}sin x 2x dx[/tex]

    You need to apply integration by parts again to the second term on the RHS. I don't quite know what you've done in your original post.
     
  5. Mar 31, 2007 #4

    D H

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    You started just fine:

    [tex]\int x^2 \cos x\,dx = x^2\sin x - \int \sin x\,2x\,dx[/tex]

    You're problem is the second integration:

    [tex]\int \sin x\,2x\,dx \ne -\cos x\,x^2[/tex]
     
  6. Apr 3, 2007 #5
    Do I need to integrate by parts again?
    [tex]\int \sin x\,2x,dx=2/int,x,sinx,dx[/tex]
    so
    [tex]= \left[x^2 sin x\right]_{0}^{\pi}-2\int_{0}^{\pi}xsin x dx[/tex]
    is this right?
     
  7. Apr 3, 2007 #6

    Gib Z

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    No :(

    [tex]\int \sin x 2x dx[/tex].
    u=sin x, dv = 2x dx
    du=cos x dx, v=x^2

    Subbing these straight into integration by parts it should be
    [tex]\sin (x) \cdot x^2 - \int x^2 \cos x dx[/tex].

    Hmm Isnt that interesting, your original integral appeared again, how could this help i wonder :P
     
  8. Apr 3, 2007 #7

    Gib Z

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    Actually I did it wrong, you should be able to see why the way I did it doesn't help.

    instead, in integration by parts, let u=2x and dv= sin x dx
    that way, du= 2 dx, and v= -cos x, which gives us
    [tex]-2x\cos x + 2\int \cos x dx[/tex].
    The integral in that is just 2 sin x, so the solution to your original integral is easy from here.
     
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