91Ó°ÊÓ

The statements given are mentioned below:

$$\begin{gathered} u=F(\mathrm{ϕ},k) \\ u=\underset{0}{\overset{\mathrm{ϕ}}{∫}}\frac{1}{\sqrt{1-k^2{\sin }^2\mathrm{θ}}}d\mathrm{θ} \\ u=s{n}^{-1}\left(\sin \mathrm{ϕ}\right) \end{gathered}$$

The elliptic form of the integral is defined as $\mathit{K}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\frac{\mathbf{π}}{\mathbf{2}}}{\mathbf{∫}}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}{\mathbf{sin}}^{\mathbf{2}}\mathbf{θ}}}\mathit{d}\mathit{θ}$.

For $k=0$.

$$\begin{gathered} u=F(\mathrm{ϕ},0) \\ u=\underset{0}{\overset{\mathrm{ϕ}}{∫}}d\mathrm{θ} \\ u=\mathrm{ϕ} \end{gathered}$$

It is given that$u=s{n}^{-1}\left(\sin \mathrm{Ï•}\right)$

Substitute $u=\mathrm{Ï•}$in the above equation.

$$\begin{gathered} u=s{n}^{-1}\left(\sin \mathrm{Ï•}\right) \\ snu=\sin \mathrm{Ï•} \\ snu=\sin u \end{gathered}$$

Formula states that $cnu=\sqrt{1-{\left(\sin \mathrm{Ï•}\right)}^2}$.

Simplify the formula mentioned above.

$$\begin{gathered} cnu=\sqrt{1-{\left(\sin \mathrm{Ï•}\right)}^2} \\ cnu=\sqrt{1-{\left(\sin u\right)}^2} \\ cnu=\cos u \end{gathered}$$

Formula states that $dnu=\sqrt{1-{\left(k\sin \mathrm{Ï•}\right)}^2}$

Simplify the formula mentioned above.

$$\begin{gathered} dnu=\sqrt{1-{\left(k\sin \mathrm{Ï•}\right)}^2} \\ dnu=1 \end{gathered}$$

For$k=1$.

$$\begin{gathered} u=F(\mathrm{ϕ},1) \\ u=\underset{0}{\overset{\mathrm{ϕ}}{∫}}\frac{1}{\cos \mathrm{θ}}d\mathrm{θ} \\ u=\ln (\sec \mathrm{ϕ}+\tan \mathrm{ϕ}) \end{gathered}$$

It is proved earlier that$snu=\sin \mathrm{Ï•}$.

Formula states that $\tanh u=\sin \mathrm{Ï•}$.

Hence, $snu=\tanh u$.

Formula states that $cnu=\sqrt{1-s{n}^{2}u}$.

$$\begin{gathered} cnu=\sqrt{1-s{n}^{2}u} \\ cnu=\sqrt{1-{\tanh }^{2}u} \\ cnu=\frac{1}{\cosh u} \\ cnu=sechu \end{gathered}$$

Formula states that $dnu=\sqrt{1-s{n}^{2}u}$.

$dnu=sechu$

Hence, the statements have been proven.