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

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Aicharya · Accepted Answer

Let $I=\int_0^1 x^{3 / 2} \sqrt{1-x} d x$ Put $x=\sin ^2 	heta \Rightarrow d x=2 \sin 	heta \cos 	heta d 	heta$ $ \begin{aligned} 	herefore I & =\int_0^{\pi / 2} \sin ^3 	heta \cdot \sqrt{1-\sin ^2 	heta} 2 \sin 	heta \cos 	heta d 	heta \ & =2 \int_0^{\pi / 2} \sin ^4 	heta \cos ^2 	heta d 	heta \ & =2\left[\frac{3 \cdot 1 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}ight] 	ext { (using Wallis formula) } \ & =\frac{\pi}{16} \end{aligned} $

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

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