<s>[INST] How to prove $ \int _0 ^1 \left ( \sqrt{1-x^2}\right )^n dx = \prod_{k=1} ^n \frac {2k}{2k+1} $ 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
 [/INST] 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: $$I_{n+1}=\frac{n}{n+1}I_{n-1}$$
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