91Ó°ÊÓ

The given function is $u=\ln (\sec \mathrm{φ}+\tan \mathrm{φ})$.

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

Factorise $\sec \mathrm{φ}+\tan \mathrm{φ}$.

$$\begin{gathered} \sec \mathrm{φ}+\tan \mathrm{φ}=\frac{1}{\cos \mathrm{ϕ}}+\frac{\sin \mathrm{ϕ}}{\cos \mathrm{ϕ}} \\ =\frac{1+\sin \mathrm{ϕ}}{0+\cos \mathrm{ϕ}} \\ =\frac{\sin \frac{\mathrm{π}}{2}+\sin \mathrm{ϕ}}{\cos 0+\cos \mathrm{ϕ}} \\ =\tan (\frac{\mathrm{ϕ}}{2}+\frac{\mathrm{π}}{4}) \end{gathered}$$

Substitute the above value in the equation mentioned below.

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

The value $u=\ln (\sec \mathrm{φ}+\tan \mathrm{φ})$.

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

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

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

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

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

Simplify further.

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

The value$u=ln(sec\mathrm{φ}+tan\mathrm{φ})$

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

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

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

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

The statements have been proven.