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Math Help - Integration!

  1. #1
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    Integration!

    Hi all,

    Can someone please help me with the last part of this question? I am lost at how to apply the other parts of the question to this last part..

    Thank you so much!
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  2. #2
    Junior Member FelixFelicis28's Avatar
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    Re: Integration!

    Quote Originally Posted by Tutu View Post
    Hi all,

    Can someone please help me with the last part of this question? I am lost at how to apply the other parts of the question to this last part..

    Thank you so much!
    From part ii) you have derived what is called a reduction formula, OP.

    Clearly:

    \displaystyle \begin{aligned} \int_0^1 \frac{x^n}{\sqrt{1-x^2}} \ dx & = \frac{n-1}{n} \int_0^1 \frac{x^{n-2}}{\sqrt{1-x^2}} \ dx \\ & = \left(\frac{n-1}{n}\right)\cdot\left(\frac{n-3}{n-2}\right)\int_0^1\frac{x^{n-4}}{\sqrt{1-x^2}} \ dx \\ & \vdots \\ & = \prod_{r=0}^{\frac{n-2}{2}} \left(\frac{n-1 -2r}{n - 2r}\right) \int_0^1 \frac{dx}{\sqrt{1-x^2}}  \end{aligned}

    for even n.

    Use n=6 until you've reduced \int_0^1 \frac{x^6}{\sqrt{1-x^2}} \ dx in terms of \int_0^1 \frac{dx}{\sqrt{1-x^2}} .

    Edit: in fact, out of interest, we can find a closed form.

    Spoiler:
    For even n:

    \int_0^1 \frac{x^n}{\sqrt{1-x^2}} \ dx = \frac{(n-1)!}{\left(\frac{n-2}{2}\right)!\left(\frac{n}{2}\right)!2^{n-1}} \int_0^1 \frac{dx}{\sqrt{1-x^2}} = \frac{\pi (n-1)!}{\left(\frac{n-2}{2}\right)!\left(\frac{n}{2}\right)!2^n}
    Last edited by FelixFelicis28; September 22nd 2013 at 11:33 AM.
    Thanks from topsquark
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