If$\mathit{u}\mathbf{=}\mathit{l}\mathit{n}\mathbf{(}\mathit{s}\mathit{e}\mathit{c}\mathit{\phi }\mathbf{+}\mathit{t}\mathit{a}\mathit{n}\mathit{\phi }\mathbf{)}$, then φ is a function of u called the Gudermannian of u, $\mathit{\phi }\mathbf{=}\mathit{g}\mathit{d}\mathit{u}$. Prove that: .

$\mathit{u}\mathbf{=}\mathit{l}\mathit{n}\mathit{t}\mathit{a}\mathit{n}\mathbf{(}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\frac{\mathbf{\varphi }}{\mathbf{2}}\mathbf{)}\mathbf{,}\mathit{t}\mathit{a}\mathit{n}\mathit{g}\mathit{d}\mathit{u}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{u}\mathbf{,}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{d}\mathit{u}\mathbf{=}\mathit{t}\mathit{a}\mathit{n}\mathit{h}\mathit{u}\mathbf{,}\frac{\mathbf{d}}{\mathbf{du}}\mathit{g}\mathit{d}\mathit{u}\mathbf{=}\mathbf{sech}\mathit{u}$

The given function is $u=\ln (\sec \phi +\tan \phi )$.

The elliptic form of the integral is defined as $\mathbf{K}\mathbf{\left(}\mathbf{k}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\frac{\mathbf{\pi }}{\mathbf{2}}}{\mathbf{\int }}}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{k}}^{\mathbf{2}}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}}\mathbf{d}\mathbf{\theta }$.

Factorise $\sec \phi +\tan \phi$.

$$\begin{gathered} \sec \phi +\tan \phi =\frac{1}{\cos \varphi }+\frac{\sin \varphi }{\cos \varphi } \\ =\frac{1+\sin \varphi }{0+\cos \varphi } \\ =\frac{\sin \frac{\pi }{2}+\sin \varphi }{\cos 0+\cos \varphi } \\ =\tan (\frac{\varphi }{2}+\frac{\pi }{4}) \end{gathered}$$

Substitute the above value in the equation mentioned below.

$$\begin{gathered} u=\ln (\sec \phi +\tan \phi ) \\ u=\mathrm{lntan}(\frac{\varphi }{2}+\frac{\pi }{4}) \end{gathered}$$

The value $u=\ln (\sec \phi +\tan \phi )$.

$$\begin{gathered} e^u=(\sec \phi +\tan \phi ) \\ {e}^u=\frac{1+\sin \varphi }{\cos \varphi } \\ {e}^{-u}=\frac{\cos \varphi }{1+\sin \varphi } \end{gathered}$$

Formula states that $\sinh \varphi =\frac{e^u-e^{-u}}{2}$.

$$\begin{gathered} \sinh u=\frac{e^u-e^{-u}}{2} \\ \sinh u=\frac{1}{2}\left(\frac{1+\sin \varphi }{\cos \varphi }-\frac{\cos \varphi }{1+\sin \varphi }\right) \\ \sinh u=\frac{1}{2}\left(\frac{{\left(1+\sin \varphi \right)}^2-\cos \varphi }{\cos \varphi \left(1+\sin \varphi \right)}\right) \\ \sinh u=\tan gdu \end{gathered}$$

Formula states that$\tanh \varphi =\frac{e^u-e^{-u}}{e^u+e^{-u}}$.

$$\begin{gathered} \tanh u=\frac{e^u-e^{-u}}{e^u+e^{-u}} \\ \tanh u=\frac{\left(\frac{1+\sin \varphi }{\cos \varphi }-\frac{\cos \varphi }{1+\sin \varphi }\right)}{\left(\frac{1+\sin \varphi }{\cos \varphi }+\frac{\cos \varphi }{1+\sin \varphi }\right)} \\ \tanh u=\frac{\left(\frac{{\left(1+\sin \varphi \right)}^2-\cos \varphi }{\cos \varphi \left(1+\sin \varphi \right)}\right)}{\left(\frac{{\left(1+\sin \varphi \right)}^2+\cos \varphi }{\cos \varphi \left(1+\sin \varphi \right)}\right)} \\ \tanh u=\frac{{\left(1+\sin \varphi \right)}^2-\cos \varphi }{{\left(1+\sin \varphi \right)}^2+\cos \varphi } \end{gathered}$$

Simplify further.

localid="1664310207234" $\tanh u=\sin gdu$

The value$u=ln(sec\phi +tan\phi )$

$$\begin{gathered} \frac{d\phi }{du}=\frac{1}{\frac{du}{d\phi }} \\ \frac{d\phi }{du}=\frac{\sec \phi +\tan \phi }{\sec \phi \tan \phi +{\sec }^2\phi } \\ \frac{d\phi }{du}=\frac{1}{\sec \phi } \\ \frac{d\phi }{du}=\cos \phi \end{gathered}$$

Formula states that$\cosh \varphi =\frac{e^u+e^{-u}}{2}$.

$$\begin{gathered} \cosh u=\frac{e^u+e^{-u}}{2} \\ \cosh u=\frac{1}{2}\left(\frac{1+\sin \varphi }{\cos \varphi }+\frac{\cos \varphi }{1+\sin \varphi }\right) \\ \cosh u=\frac{1}{2}\left(\frac{{\left(1+\sin \varphi \right)}^2+\cos \varphi }{\cos \varphi \left(1+\sin \varphi \right)}\right) \\ \cosh u=\frac{1}{\cos \phi } \end{gathered}$$

Hence$\frac{d}{du}gdu=sechu$

The statements have been proven.