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Help to prove a reduction formula?

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Let In = [tex]\int[/tex][tex]^{1}_{-1}[/tex] (1-x[tex]^{2}[/tex])[tex]^{n}[/tex] dx.

    Use integration by parts to show that In = ([tex]\frac{2n}{2n+1}[/tex]) In-1 for n [tex]\geq[/tex]1.

    (The integral above is supposed to be between the limits -1 and 1... sorry I couldn't figure out how to make the limits appear properly.)

    2. Relevant equations

    Integration by parts.

    3. The attempt at a solution

    I have tried several different ways.

    First of all I tried letting u=(1-x[tex]^{2}[/tex])[tex]^{n}[/tex] and dv/dx = 1. Hence, du/dx = (1-x[tex]^{2}[/tex])[tex]^{n-1}[/tex](-2x) and v=x. Using this I got In = n [tex]\int[/tex][tex]^{1}_{-1}[/tex] 2x2(1-x[tex]^{2}[/tex])[tex]^{n-1}[/tex] dx. This didn't seem to be very helpful.

    I then tried writing In as [tex]\int[/tex][tex]^{1}_{-1}[/tex] (1+x)[tex]^{n}[/tex](1-x)[tex]^{n}[/tex] dx, and letting u be (1+x)[tex]^{n}[/tex] and dv/dx be (1-x)[tex]^{n}[/tex]. I thought this was a pretty clever idea, but it didn't give me what I wanted. I got In = [tex]\frac{n}{n+1}[/tex] [tex]\int[/tex][tex]^{1}_{-1}[/tex] (1-x2)n-1(1-x)2 dx, which can be written equivalently as [tex]\frac{n}{n+1}[/tex] [tex]\int[/tex][tex]^{1}_{-1}[/tex] (1-x)n+1(1+x)n-1 dx. I then tried doing another integration by parts, letting u=(1+x)n-1 and dv/dx=(1-x)n+1, and this gave me In = ([tex]\frac{n}{n+1}[/tex])([tex]\frac{n-1}{n+2}[/tex]) [tex]\int[/tex][tex]^{1}_{-1}[/tex] (1-x2)n-2(1-x)4 dx, or equivalently ([tex]\frac{n}{n+1}[/tex])([tex]\frac{n-1}{n+2}[/tex]) [tex]\int[/tex][tex]^{1}_{-1}[/tex] (1+x)n-2(1-x)n+2 dx , but I still didn't seem to be any closer.

    The other thing I tried was writing In = [tex]\int[/tex][tex]^{1}_{-1}[/tex] ([tex]\sqrt{1-x^2[/tex])2n dx, and letting u = ([tex]\sqrt{1-x^2}[/tex])2n and dv/dx = 1. This gave me a slightly more interesting result of In = 2n [tex]\int[/tex][tex]^{1}_{-1}[/tex] x2 ([tex]\sqrt{1-x^2}[/tex])2n-2 dx , i.e. In = 2n [tex]\int[/tex][tex]^{1}_{-1}[/tex] x2 ([tex]1-x^2}[/tex])n-1 dx, but I wasn't sure what to do next.

    Please help... what should I be doing?
     
    Last edited: Dec 6, 2008
  2. jcsd
  3. Dec 6, 2008 #2

    HallsofIvy

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    I think it would make a lot more sense to let u= 1-x2 and dv= (1- x2)n-1dx.
     
  4. Dec 6, 2008 #3

    But then I would have to integrate (1- x2)n-1dx, and I don't know how to do that when n is unknown. Am I missing something? I tried to integrate it using MAPLE, and it gave me something useless to do with the hypergeometric function.
     
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