91Ó°ÊÓ

The Integral is$\underset{0}{\overset{1}{∫}}{x}^2{(1-x^2)}^{\frac{3}{2}}$.

Beta function is defined as $\mathit{β}\mathbf{(}\mathit{p}\mathbf{,}\mathit{q}\mathbf{)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{1}}{\mathbf{∫}}}{\mathit{t}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\left(1-t\right)}^{\mathbf{q}\mathbf{-}\mathbf{1}}\mathit{d}\mathit{t}$.

The Integral is$\underset{0}{\overset{1}{∫}}{x}^2{(1-x^2)}^{\frac{3}{2}}$.

The formula for the beta function is$\mathrm{β}(p,q)=\underset{0}{\overset{1}{∫}}{t}^{p-1}{(1-t)}^{q-1}dt$.

Let $x^2=y$and derivate with respect to x.

$\begin{array}{l}{x}^2=y\\ 2xdx=dy\\ dx=\frac{1}{2\sqrt{y}}dy\end{array}$

Substitute the value of $x^2$ in the integral.

The equation becomes as follows.

$\begin{array}{l}I=\underset{0}{\overset{1}{∫}}y{(1-y)}^{\frac{3}{2}}\frac{1}{2\sqrt{y}}dy\\ I=\underset{0}{\overset{1}{∫}}\sqrt{y}{(1-y)}^{\frac{3}{2}}\frac{1}{2}dy\\ I=\frac{1}{2}\mathrm{β}(\frac{3}{2},\frac{5}{2})\\ \end{array}$

The relation between $\mathrm{β}$and $\mathrm{Γ}$is the equation mentioned below.

$\mathrm{β}(m,n)=\frac{\mathrm{Γ}\left(m\right)\mathrm{Γ}\left(n\right)}{\mathrm{Γ}(m+n)}$

Substitute the values given below in the equation mentioned above.

$\begin{array}{c}m=\frac{3}{2}\\ n=\frac{5}{2}\end{array}$

The equation becomes as follows.

$\begin{array}{c}\mathrm{β}(\frac{3}{2},\frac{5}{2})=\frac{\mathrm{Γ}\left(\frac{3}{2}\right)\mathrm{Γ}\left(\frac{5}{2}\right)}{\mathrm{Γ}(\frac{3}{2}+\frac{5}{2})}\\ \mathrm{β}(\frac{3}{2},\frac{5}{2})=\frac{\mathrm{Γ}\left(\frac{3}{2}\right)\mathrm{Γ}\left(\frac{5}{2}\right)}{\mathrm{Γ}\left(4\right)}\\ I=\frac{1}{2}×\frac{\mathrm{Γ}\left(\frac{3}{2}\right)\mathrm{Γ}\left(\frac{5}{2}\right)}{\mathrm{Γ}\left(4\right)}\end{array}$

Evaluate the value of gamma function $\mathrm{Γ}\left(\frac{3}{2}\right),\mathrm{Γ}\left(\frac{5}{2}\right)$ and $\mathrm{Γ}\left(4\right)$.

$$\begin{gathered} \mathrm{Γ}\left(\frac{5}{2}\right)=\left(\frac{3}{2}\right)\mathrm{Γ}\left(\frac{3}{2}\right) \\ \mathrm{Γ}\left(\frac{5}{2}\right)=\left(\frac{3}{2}\right)×\left(\frac{1}{2}\right)\mathrm{Γ}\left(\frac{1}{2}\right) \\ \mathrm{Γ}\left(\frac{5}{2}\right)=\left(\frac{3}{2}\right)×\left(\frac{1}{2}\right)\sqrt{\mathrm{π}} \\ \mathrm{Γ}\left(\frac{5}{2}\right)=\left(\frac{3}{4}\right)\sqrt{\mathrm{π}} \end{gathered}$$

Simplify further

$$\begin{gathered} \mathrm{Γ}\left(\frac{3}{2}\right)=\left(\frac{1}{2}\right)\mathrm{Γ}\left(\frac{1}{2}\right) \\ \mathrm{Γ}\left(\frac{3}{2}\right)=\left(\frac{1}{2}\right)\sqrt{\mathrm{π}} \\ \mathrm{Γ}\left(4\right)=3! \\ \mathrm{Γ}\left(4\right)=6 \end{gathered}$$

Substitute above value in the equation mentioned below.

$$\begin{gathered} I=\frac{1}{2}\left[\frac{\mathrm{Γ}\left({\displaystyle \frac{3}{2}}\right)\mathrm{Γ}\left(\frac{5}{2}\right)}{\mathrm{Γ}\left(4\right)}\right] \\ I=\frac{1}{2}\left[\frac{\left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\mathrm{π}}{6}\right] \\ I=\frac{\mathrm{π}}{32} \end{gathered}$$

Hence, the solution is mentioned below.

The value of integral is $\frac{\mathrm{Ï€}}{32}$.