int_0^1 x^{3/2} sqrt{1-x} , dx is equal to | vedclass

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(D) Let $I = \int_0^1 x^{3/2} \sqrt{1-x} \, dx$. Substitute $x = \sin^2 	heta$,then $dx = 2 \sin 	heta \cos 	heta \, d	heta$. When $x = 0$,$	heta

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$\frac{\pi}{16}$

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(D) Let $I = \int_0^1 x^{3/2} \sqrt{1-x} \, dx$. Substitute $x = \sin^2 	heta$,then $dx = 2 \sin 	heta \cos 	heta \, d	heta$. When $x = 0$,$	heta = 0$,and when $x = 1$,$	heta = \frac{\pi}{2}$. Substituting these into the integral: $I = \int_0^{\pi/2} (\sin^2 	heta)^{3/2} \sqrt{1-\sin^2 	heta} \cdot (2 \sin 	heta \cos 	heta) \, d	heta$ $I = \int_0^{\pi/2} \sin^3 	heta \cdot \cos 	heta \cdot 2 \sin 	heta \cos 	heta \, d	heta$ $I = 2 \int_0^{\pi/2} \sin^4 	heta \cos^2 	heta \, d	heta$. Using the Wallis formula $\int_0^{\pi/2} \sin^m 	heta \cos^n 	heta \, d	heta = \frac{(m-1)(m-3)\dots(n-1)(n-3)\dots}{(m+n)(m+n-2)\dots} \cdot \frac{\pi}{2}$ (if both $m, n$ are even): $I = 2 \cdot \left[ \frac{(3 \cdot 1) \cdot (1)}{(6 \cdot 4 \cdot 2)} \cdot \frac{\pi}{2} ight]$ $I = 2 \cdot \left[ \frac{3}{48} \cdot \frac{\pi}{2} ight] = 2 \cdot \frac{3\pi}{96} = \frac{6\pi}{96} = \frac{\pi}{16}$.

$\int_0^1 x^{3/2} \sqrt{1-x} \, dx$ is equal to

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