91Ó°ÊÓ

The given Integral is $\underset{0}{\overset{1}{∫}}\frac{x^4}{\sqrt{1-x^2}}$.

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

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 $\underset{0}{\overset{1}{∫}}\frac{x^4}{\sqrt{1-x^2}}$.

The equation becomes as follows.

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

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.

$\begin{array}{c}m=\frac{5}{2}\\ n=\frac{1}{2}\end{array}$

The equation becomes as follows.

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

Evaluate the value of gamma function $\mathrm{Γ}\left(\frac{5}{2}\right),\mathrm{Γ}\left(\frac{1}{2}\right)$ and $\mathrm{Γ}\left(3\right)$.

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

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

Substitute above values in the equation mentioned below.

$\begin{array}{c}I=\frac{1}{2}×\frac{\mathrm{Γ}\left(\frac{5}{2}\right)\mathrm{Γ}\left(\frac{1}{2}\right)}{\mathrm{Γ}\left(3\right)}\\ I=\frac{1}{2}×\frac{\frac{3}{4}\mathrm{π}}{2}\\ I=\frac{3\mathrm{π}}{16}\end{array}$

Hence, the value of integral is $\frac{3\mathrm{Ï€}}{16}$.