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$\int_0^1 x^{3 / 2} \sqrt{1-x} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{\pi}{16}$
Let $I=\int_0^1 x^{3 / 2} \sqrt{1-x} d x$
Put $x=\sin ^2 \theta \Rightarrow d x=2 \sin \theta \cos \theta d \theta$
$$
\begin{aligned}
\therefore I & =\int_0^{\pi / 2} \sin ^3 \theta \cdot \sqrt{1-\sin ^2 \theta} 2 \sin \theta \cos \theta d \theta \\
& =2 \int_0^{\pi / 2} \sin ^4 \theta \cos ^2 \theta d \theta \\
& =2\left[\frac{3 \cdot 1 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}\right] \text { (using Wallis formula) } \\
& =\frac{\pi}{16}
\end{aligned}
$$
Put $x=\sin ^2 \theta \Rightarrow d x=2 \sin \theta \cos \theta d \theta$
$$
\begin{aligned}
\therefore I & =\int_0^{\pi / 2} \sin ^3 \theta \cdot \sqrt{1-\sin ^2 \theta} 2 \sin \theta \cos \theta d \theta \\
& =2 \int_0^{\pi / 2} \sin ^4 \theta \cos ^2 \theta d \theta \\
& =2\left[\frac{3 \cdot 1 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}\right] \text { (using Wallis formula) } \\
& =\frac{\pi}{16}
\end{aligned}
$$
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