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Prove that

$$\int_0^1 x^{k}(1-x)^kdx=\frac{k!k!}{(2k+1)!}.$$ (Edit: Actually the proof can be found here http://en.wikipedia.org/wiki/Beta_function )

How would you show this

$\text{Beta}(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta, \qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0$ ?


One attempt is to expand using Binomial theorem, integrate terms by terms (there are finitely many of them), and then sum up each individual term (which is the difficult part).

Edit: Sorry actually this is a beta function. But can we do this without resorting to Beta function? Thanks.

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    $\begingroup$ Resembles? That integral is $B(k+1,k+1)$. $\endgroup$ – Daniel Fischer Nov 26 '13 at 22:30
  • $\begingroup$ Yeah you're right. Somehow I was looking at the wrong definition. $\endgroup$ – Christmas Bunny Nov 26 '13 at 22:32
  • $\begingroup$ That said, integration by parts also reaches the target without much problems. $\endgroup$ – Daniel Fischer Nov 26 '13 at 22:32
  • $\begingroup$ For the edited question, make the substitution $x=\sin^2(\theta)$. $\endgroup$ – Meow Nov 26 '13 at 22:42
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Hint. Integration by parts is always your best friend. Or, just observe that

$$ \frac{1}{n+1} \times \frac{t^{n+1} - 1}{t - 1} = \int_{0}^{1} \{ (1-x) + x t \}^{n} \, dx = \sum_{k=0}^{n} \binom{n}{k} t^{k} \int_{0}^{1} x^{k}(1-x)^{n-k} \, dx. $$

This equates two polynomials of $t$. So you can compare both sides to reach the conclusion.

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