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Math Help - Reduction Formula

  1. #1
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    Reduction Formula

    Given that I_{n} = \int_{0}^{1}{x^n(1-x)^{\frac{1}{2}}dx} where n \geq 0 , prove that for n \geq 1

    I_{n} = \frac{2n}{2n+3} I_{n-1}
    The first thing that comes to mind is x = \sin^2u so dx = 2 \sin u \cos u du

    giving I_{n} = \int_{0}^{\frac{\pi}{2}}{sin^{2n} \cos u \ 2 \sin u \cos u \ du }

    \Rightarrow 2 \int_{0}^{\frac{\pi}{2}}{sin^{2n+1} \cos^2 u  \ du }

    my first thought is to use integration by part, with du = sin^{2n+1} \cos u and v = \cos u

    this doesn't get me very far sadly....
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  2. #2
    Senior Member
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    Just do integration by parts without the substitution:
    u=x^n, dv=\sqrt{1-x}dx
    Thus du=nx^{n-1}dx, v=-\frac{2}{3}(1-x)^{\tfrac{3}{2}}
    and:
    I_{n}=\int_{0}^{1}x^n\sqrt{1-x}\,dx
    I_{n}=-\frac{2}{3}x^n(1-x)^{\tfrac{3}{2}}|_{x=0}^1-\int_{0}^{1}(nx^{n-1})\cdot\left(-\frac{2}{3}(1-x)^{\tfrac{3}{2}}\right)\,dx
    I_{n}=\frac{2n}{3}\int_{0}^{1}x^{n-1}(1-x)^{\tfrac{3}{2}}\,dx
    I_{n}=\frac{2n}{3}\int_{0}^{1}x^{n-1}(1-x)\sqrt{1-x}\,dx
    I_{n}=\frac{2n}{3}\int_{0}^{1}(x^{n-1}-x^n)\sqrt{1-x}\,dx
    I_{n}=\frac{2n}{3}\left(\int_{0}^{1}x^{n-1}\sqrt{1-x}\,dx-\int_{0}^{1}x^n\sqrt{1-x}\,dx\right)
    I_{n}=\frac{2n}{3}(I_{n-1}-I_{n})
    and algebra can do the rest.

    --Kevin C.
    Last edited by TwistedOne151; March 22nd 2008 at 06:02 PM. Reason: LaTeX error
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