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Evaluate.

$$\int_0^1{} \frac{x^{2n+1}}{\sqrt{1-x^2}}dx$$

I've done substitution twice and have gotten to the following:

$$-\int(1-v^2)^ndv$$

I don't know how to advance from here though. Any help would be greatly appreciated!

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  • $\begingroup$ The last is just a polynomial - expanding it using the binomial theorem and integrating it would be simple $\endgroup$ – vrugtehagel Feb 15 '16 at 21:32
  • $\begingroup$ How would you be able to do that when n isn't know? $\endgroup$ – b-ballboy8 Feb 15 '16 at 21:36
  • $\begingroup$ I'll post an answer evaluting the last - are you sure that it's not a definite integral (since you're integrating over $[0,1]$ on the first one)? $\endgroup$ – vrugtehagel Feb 15 '16 at 21:37
  • $\begingroup$ The definite integral goes from 1 to 0 for the second one. u=x^2 and then v=sqrt(1-u). $\endgroup$ – b-ballboy8 Feb 15 '16 at 21:41
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We have \begin{align} \int(1-v^2)^ndv&=\int\sum_{k=0}^n\binom{n}{k}(-v^2)^{k}1^{n-k}dv\\ &=\sum_{k=0}^n\binom{n}{k}(-1)^k\int v^{2k}dv\\ &=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{2k+1}v^{2k+1} \end{align} We are integrating from $1$ to $0$, so \begin{align} \int_1^0(1-v^2)^ndv&=-\left[\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{2k+1}v^{2k+1}\right]_0^1\\ &=-\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{2k+1}1^{2k+1}\\ &=-\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{2k+1} \end{align}

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  • $\begingroup$ Didn't you forget the "-"? You wrote the expansion of $(1+v^2)^n$ $\endgroup$ – Francesco Gemma Feb 15 '16 at 21:47
  • $\begingroup$ @FrancescoGemma, yes, fixed. Thanks! $\endgroup$ – vrugtehagel Feb 15 '16 at 21:50
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Let $u=1-x^2$, then the integral becomes $$ \frac{1}{2}\int_0^1(1-u)^nu^{-1/2}\;du $$ This is $\frac{1}{2}B(1/2,n+1)$, where $B(x,y)$ is the Beta function. It turns out that $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$. So if $n$ is a non-negative integer, then $$\frac{1}{2}B(1/2,n+1)=\frac{1}{2}\frac{\Gamma(\frac{1}{2})\Gamma(n+1)}{\Gamma(n+\frac{3}{2})}=\frac{1}{2}\cdot\frac{n!\sqrt{\pi}}{2^{-n-1}(2n+1)(2n-1)\cdots 3\cdot 1\sqrt{\pi}}$$ $$=\frac{2^{n}n!}{(2n+1)(2n-1)\cdots 3\cdot 1}$$

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