How to prove this integral identity?
$$ \int _0 ^1 \left ( \sqrt{1-x^2}\right )^n dx = \prod_{k=1} ^n \frac {2k}{2k+1} $$
↑ This identity is false. It should be corrected to $ \int _0 ^1 \left ( {1-x^2}\right )^n dx = \prod_{k=1} ^n \frac {2k}{2k+1} $
I'm sorry for causing confusion
Hint: $$I_{n+1}:=\int _0 ^1 \left ( \sqrt{1-x^2}\right )^n dx =\int _{-\frac{\pi}{2}} ^0 \left ( \sqrt{1-\sin^2 t}\right )^n \cos t dt=\int _{-\frac{\pi}{2}} ^0 \cos^{n+1} t dt=$$ integrating by parts $$\left[\sin t \cos^{n}t\right]_{-\frac{\pi}{2}} ^0+n\int _{-\frac{\pi}{2}} ^0 \sin t\cos^{n-1} t \sin t dt=n\int _{-\frac{\pi}{2}} ^0 \cos^{n-1} t (1-\cos^2 t) dt$$ $$=n\int _{-\frac{\pi}{2}} ^0 \cos^{n-1} t dt-n\int _{-\frac{\pi}{2}} ^0 \cos^{n+1} t dt=nI_{n-1}-nI_{n+1}$$ Hence: $$I_{n+1}=\frac{n}{n+1}I_{n-1}$$
Hint
Start with a change of variable $x=\sin(y)$. So $$I_n=\int _0 ^1 \left ( \sqrt{1-x^2}\right )^n~ dx = \int _0 ^{\frac{\pi}{2}} \cos^{n+1}(y) ~dy$$ and now use integration by reduction method as described in
http://en.wikipedia.org/wiki/Integration_by_reduction_formulae
I am sure that you can take from here.
Use the change of variables $x = \sin t$. You get the Wallis integral.
If you don't know haw to handle it, start with an integration by parts.
