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Homework Help: Integrate polymonial of any degree

  1. Jun 6, 2010 #1
    how to integrate [tex](1-x^2)^n[/tex] for [tex]n \in N[/tex] ? Limits of integral are from 0 to 1, but i dont think that matter.

    (i tried to use latex for int, but it wasn't working).
     
  2. jcsd
  3. Jun 6, 2010 #2

    Cyosis

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    Homework Helper

    Use the binomial expansion.
     
  4. Jun 6, 2010 #3
    [tex]\int_{0}^{1} (1-x^2)^n dx[/tex]

    using binomial expansion:

    [tex]\int \sum_{k=0}^{n}( {n \choose k} (-x^2)^k) dx[/tex] = [tex]\sum_{k=0}^{n}( {n \choose k} \frac{x (-x^2)^k)}{2 k+1})[/tex]

    and going to definite integral:

    [tex]\int_{0}^{1} (1-x^2)^n dx = \sum_{k=0}^{n}( {n \choose k} \frac{(-1)^k)}{2 k+1})[/tex]

    is this correct? Can anything else be done there?

    And either "preview post" or latex is not working correctly on this forum.

    How to solve this using integration by parts?

    @edit
    ok, nvm, done already.
     
    Last edited: Jun 6, 2010
  5. Jun 6, 2010 #4

    Gib Z

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    That's correct. It can be written as [tex]\frac{\sqrt{\pi} \Gamma(1+n)}{2\Gamma(\frac{3+2n}{2})}[/tex] but that might be too advanced.
     
  6. Jun 6, 2010 #5
    [tex]
    c_n = \int_{0}^{1} (1 - x^2)^{\frac{n-1}{n}} dx
    [/tex]

    Therefore,

    [tex]
    c_{2n+1} = \frac{n-1}{2n} C_{2n-3}
    [/tex]

    I'm sure if you were to solve that explicitly, you would get what Gib Z got.
     
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