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How does one show that $I_n=\int\limits_0^1 (1-x^2)^ndx $ satisfies the recursion relation $I_n={2n\over 2n+1}I_{n-1}$? I don't think I have to explicitly evaluate the integral right?

Thanks.

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    @DidierPiau: Ah, well-spotted typo! Thanks!2011-11-03

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You just verify this integrating by parts: for $n\geq 1$ $ I_n = \int_0^1 (1-x^2)^n\,dx = x(1-x^2)^n\mid^1_0 + 2n\int\limits_{0}^1(1-x^2)x^2\,dx = $ $ = 0 +2n\int\limits_{0}^1(1-x^2)^{n-1}(x^2-1)\,dx + 2n I_{n-1} = 2nI_{n-1} - 2nI_n $ by expressing $I_n$ we have $I_n = \frac{2n}{2n+1}I_{n-1}$. There should be a typo in a $-$ sign as Didier told you.

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    @ChristianBlatter: nevermind)2011-11-03
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Start with $ \frac{\mathrm{d}}{\mathrm{d} x} \left( x \left(1-x^2\right)^n \right) = \left(2n+1\right) \left(1-x^2\right)^n - 2 n \left(1-x^2\right)^{n-1} $ Assuming $n > 0$, and integrating both sides from 0 to 1 we get: $ 0 = (2n+1) I_n - 2n I_{n-1} $ Clearly, $I_0 = \int_0^1 \mathrm{d} x = 1$.