Question

The value of ${\int }_0^1{x}^2(1-x^2{)}^{3/2}dx$ is

• A. $\frac{1}{32}$
• B. $\frac{\pi }{8}$
• C. $\frac{\pi }{16}$
• D. $\frac{\pi }{32}$

Answer 1

The correct option is D $\frac{\pi }{32}$

Let

$I={\int }_0^1{x}^2(1-x^2{)}^{3/2}dx$

(Let $x=\sin \theta \Rightarrow dx=\cos \theta d\theta )$
$I={\int }_0^{\pi /2}{\sin }^2\theta ({\cos }^2\theta {)}^{3/2}\cdot \cos \theta d\theta )$

$I={\int }_0^{\pi /2}{\sin }^2\theta \cdot {\cos }^4\theta d\theta$

By Gamma function,
$I=\frac{\frac{2+1}{2}\frac{4+1}{2}}{2\frac{2+4+2}{2}}=\frac{\frac{3}{2}\frac{5}{2}}{2\Gamma 4}$

$=\frac{\frac{1}{2}\sqrt{\frac{1}{2}}\cdot \frac{3}{2}\cdot \frac{1}{2}\cdot \sqrt{\frac{1}{2}}}{2\cdot 3\cdot 2}$

$=\frac{3\cdot \sqrt{\pi }\cdot \sqrt{\pi }}{2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 2}=\frac{\pi }{32}$