Question

$\underset{\mathbf{0}}{\overset{\mathbf{1}}{\mathbf{\int }}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{t}}^{\mathbf{2}}}\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{t}}^{\mathbf{2}}}{\mathbf{4}}}}}\mathit{d}\mathit{t}$

Short answer

The value of integral in elliptic form is $F\left({\sin }^{-1}\left(\frac{1}{2}\right),\frac{1}{2}\right)$.

Step-by-step solution

Given Information

The value of integration is $\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-{\mathrm{t}}^2}\sqrt{1-{\displaystyle \frac{{\mathrm{t}}^2}{4}}}}dt$.

Definition of elliptic form.

The elliptic form of the integral is defined as$\mathit{F}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{1}}{\mathbf{\int }}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{t}}^{\mathbf{2}}}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}{\mathbf{t}}^{\mathbf{2}}}}\mathit{d}\mathit{t}$.

Find the value of Integral.

The value of integration is $\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-{\mathrm{t}}^2}\sqrt{1-{\displaystyle \frac{{\mathrm{t}}^2}{4}}}}dt$.

The formula for the beta function is $F=\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-{\mathrm{t}}^2}\sqrt{1-{\mathrm{k}}^2{\mathrm{t}}^2}}dt$.

Equate the above equation with the value of I, the value of I becomes follows.

$$\begin{gathered} l=\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-{\mathrm{t}}^2}\sqrt{1-{\displaystyle \frac{t^2}{4}}}}dt \\ l=F\left({\sin }^{-1}\left(\frac{1}{2}\right),\frac{1}{2}\right) \end{gathered}$$

Hence, the value of integral in elliptic form is $l=F\left({\sin }^{-1}\left(\frac{1}{2}\right),\frac{1}{2}\right)$.