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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


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    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?

    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
    c_n = \int_{0}^{1} (1 - x^2)^{\frac{n-1}{n}} dx


    c_{2n+1} = \frac{n-1}{2n} C_{2n-3}

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