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Homework Help: How to integrate 0 to 1 (1-x^2)^n

  1. Nov 16, 2012 #1
    1. The problem statement, all variables and given/known data

    The reduction formula is:
    [tex] \int (1-x^2)^n dx = (1-x^2)^n x + 2n \int x^2(1-x^2)^{n-1} dx [/tex]

    and the question is:
    use this formula above how many times is necessary to prove:

    [tex] \int^{1}_{0} (1-x^2)^n dx = 2n \frac{2(n-1)}{3} \frac{2(n-2)}{5} ... \frac{4}{2n-3} \frac{2}{(2n-1)(2n+1)}[/tex]
    but I don't know how to get there.

    2. Relevant equations

    3. The attempt at a solution
    I tried to modify the reduction formula leaving it more recursive:

    [tex] \int (1-x^2)^n dx = (1-x^2)^n x + 2n \int x^2(1-x^2)^{n-1} dx [/tex]
    Integrating by parts:
    [tex] =x(1-x^2)^n+2n \int x^2(1-x^2)^{n-1}dx [/tex]
    let [tex] x^2 = -(1-x^2)+1 [/tex]
    [tex]=x(1-x^2)^n-2n \int(1-x^2)^n dx + 2n \int (1-x^2)^{n-1}dx [/tex]

    but I still don't know how to get there with this formula.

    Any guidance would be appreciated.
  2. jcsd
  3. Nov 16, 2012 #2
    use your last equality and induction.see attached.

    Attached Files:

    • 002.jpg
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  4. Nov 16, 2012 #3


    Staff: Mentor

    Since you have a definite integral perhaps you could tryparametric differentiation to get the answer.

    The only reference I have for this is Prof Nearings Mathematical Methods pdf at:


    section 1.2 pg 4.
  5. Nov 16, 2012 #4


    User Avatar
    Staff Emeritus
    Science Advisor
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    Gold Member

    Your last line says that

    [itex]\displaystyle \int (1-x^2)^n dx=x(1-x^2)^n-2n \int(1-x^2)^n dx + 2n \int (1-x^2)^{n-1}dx\ .[/itex]

    That has [itex]\displaystyle \ \int (1-x^2)^n dx\ \ [/itex] on both sides of the equation.

    Solve for [itex]\displaystyle \ \int (1-x^2)^n dx\ .[/itex]
  6. Nov 19, 2012 #5
    Hi hedipaldi, thanks for your help!

    How did you get this formula?
  7. Nov 19, 2012 #6
    substitute the limits in your last equality.
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