The value of int_{0}^{1} x^{2}(1-x^{2})^{3/2} | vedclass

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(D) Let $I = \int_{0}^{1} x^{2}(1-x^{2})^{3/2} dx$. Substitute $x = \sin 	heta$,then $dx = \cos 	heta d	heta$. When $x = 0$,$	heta = 0$,and when $

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$\frac{\pi}{32}$

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(D) Let $I = \int_{0}^{1} x^{2}(1-x^{2})^{3/2} dx$. Substitute $x = \sin 	heta$,then $dx = \cos 	heta d	heta$. When $x = 0$,$	heta = 0$,and when $x = 1$,$	heta = \frac{\pi}{2}$. $I = \int_{0}^{\pi/2} \sin^{2} 	heta (\cos^{2} 	heta)^{3/2} \cos 	heta d	heta = \int_{0}^{\pi/2} \sin^{2} 	heta \cos^{4} 	heta d	heta$. Using the Wallis formula or Beta function $\int_{0}^{\pi/2} \sin^{m} 	heta \cos^{n} 	heta d	heta = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$. Here $m=2, n=4$,so $I = \frac{\Gamma(3/2) \Gamma(5/2)}{2 \Gamma(4)} = \frac{(\frac{1}{2} \sqrt{\pi}) (\frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi})}{2 \cdot 6} = \frac{\frac{3}{8} \pi}{12} = \frac{3\pi}{96} = \frac{\pi}{32}$.

The value of $\int_{0}^{1} x^{2}(1-x^{2})^{3/2} dx$ is

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