given $I_{n}=\int^1_{0}x^n\sqrt{1-x^2}dx$, then finding value of $ \frac{I_{n}}{I_{n-2}}$

Question

given $\displaystyle I_{n}=\int^{1}_{0}x^n\sqrt{1-x^2}dx$, then finding value of $\displaystyle \frac{I_{n}}{I_{n-2}}$

Attempt: put $x=\sin \theta$ and $dx = \cos \theta$

$\displaystyle I_{n} = \int^{\frac{\pi}{2}}_{0}\sin^{n}\theta \cdot \cos^2 \theta d \theta = \int^{\frac{\pi}{2}}_{0}\sin^{n}\theta (1-\sin^2 \theta)d \theta $

could some help me how to solve it, thanks

Have a look at http://math.stackexchange.com/questions/112687/integrating-int-sinnx-dx — Claude Leibovici, Mar 01 '17 at 08:21
Sorry, see beta function and it's properties https://en.wikipedia.org/wiki/Beta_function — Nosrati, Mar 01 '17 at 09:28

score 3 · Accepted Answer · answered Mar 01 '17 at 08:18

$I_n=\dfrac12\beta(\dfrac{n+1}{2},\dfrac32)$ so $$I_n=\dfrac12\beta(\dfrac{n+1}{2},\dfrac32)=\dfrac12\beta(\dfrac{n-1}{2},\dfrac32)\dfrac{n-1}{n+2}=\dfrac12\beta(\dfrac{n-3}{2},\dfrac32)\dfrac{n-3}{n}\dfrac{n-1}{n+2}=I_{n-2}\dfrac{n-3}{n}\dfrac{n-1}{n+2}$$

given $I_{n}=\int^1_{0}x^n\sqrt{1-x^2}dx$, then finding value of $ \frac{I_{n}}{I_{n-2}}$

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