dado $I_{n}=\int^1_{0}x^n\sqrt{1-x^2}dx$ , entonces encontrar el valor de $ \frac{I_{n}}{I_{n-2}}$

Question

Dado $\displaystyle I_{n}=\int^{1}_{0}x^n\sqrt{1-x^2}dx$ , entonces encontrar el valor de $\displaystyle \frac{I_{n}}{I_{n-2}}$

Intento: poner $x=\sin \theta$ y $dx = \cos \theta$

$\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 $

podría alguien ayudarme a solucionarlo, gracias

Answer 1

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