Question

$\underset{0}{\overset{1}{\int }}{x}^4(1-x^2{)}^{\frac{3}{2}}dx$ is equal to

• A. $\frac{3\pi }{256}$
• B. $\frac{\pi }{256}$
• C. $\frac{3\pi }{128}$
• D. $\frac{\pi }{64}$

Answer 1

The correct option is A $\frac{3\pi }{256}$

Let, $I=\underset{0}{\overset{1}{\int }}{x}^4(1-x^2{)}^{\frac{3}{2}}dx$
Put $x=\sin \theta$
$\Rightarrow dx=\cos \theta d\theta$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}{\sin }^4\theta \cdot {\cos }^4\theta d\theta$
Applying wallis formula
$I_{m,n}=\frac{(m-1)(m-3)(m-5)\cdots (n-1)(n-3)(n-5)}{(m+n)(m+n-2)(m+n-4)\cdots }\cdot \frac{\pi }{2}\text{, when both}m,n\text{is even}$
$\Rightarrow I_{4,4}=\frac{(3\cdot 1)(3\cdot 1)}{8\cdot 6\cdot 4\cdot 2}\cdot \frac{\pi }{2}=\frac{3\pi }{256}$