Write and solve the Euler equations to make the following integrals stationary. Change the independent variable, if needed, to make the Euler equation simpler.

${\mathbf{\int }}_{{\mathbf{\varphi }}_{\mathbf{1}}}^{{\mathbf{\varphi }}_{\mathbf{2}}}\sqrt{\mathbf{\theta }{\mathbf{\text{'}}}^{\mathbf{2}}\mathbf{+}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\theta }}\mathit{d}\mathit{\varphi }\mathbf{,}\mathbf{}\mathit{\theta }\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{d}\mathbf{\theta }}{\mathbf{d}\mathbf{\varphi }}$

The given integral is${\int }_{{\mathrm{\varphi }}_1}^{{\mathrm{\varphi }}_2}\sqrt{\mathrm{\theta }{\text{'}}^2+{\sin }^2\mathrm{\theta }}d\varphi ,\theta \text{'}=\frac{\mathrm{d\theta }}{\mathrm{d\varphi }}$.The given integral is to be made stationary by using Euler equations.

In the calculus of variations andclassical mechanics, the Euler equations is a system of second-orderordinary differential equations whose solutions arestationary points of the givenaction functional.

The given integral is${\int }_{{\mathrm{\varphi }}_1}^{{\mathrm{\varphi }}_2}\sqrt{\mathrm{\theta }{\text{'}}^2+{\sin }^2\mathrm{\theta }}d\varphi ,\theta \text{'}=\frac{\mathrm{d\theta }}{\mathrm{d\varphi }}$

Euler equation for polar coordinates$\left(\theta ,\varphi \right)$is$\frac{d}{d\theta }\left(\frac{\partial F}{\partial \varphi \text{'}}\right)-\frac{\partial F}{\partial \varphi }=0$ where $\theta \text{'}=\frac{d\theta }{d\varphi }$

$$\begin{gathered} \theta \text{'}=\frac{d\theta }{d\varphi } \\ \Rightarrow \theta \text{'}=\frac{1}{\varphi \text{'}} \end{gathered}$$

Use the two equations above to change the given integral

$$\begin{gathered} \int \sqrt{\theta {\text{'}}^2+{\sin }^2\theta }d\varphi =\int \sqrt{\theta {\text{'}}^2+{\sin }^2\theta }\varphi \text{'}d\theta \\ =\int \sqrt{\varphi {\text{'}}^{-2}+{\sin }^2\theta }\varphi \text{'}d\theta \\ =\int \sqrt{1+\varphi {\text{'}}^2+{\sin }^2\theta }d\theta \end{gathered}$$

Now, let$F=\sqrt{1+\varphi {\text{'}}^2+{\sin }^2\theta }$.

Use the Euler equation , $\frac{d}{d\theta }\left(\frac{\partial F}{\partial \varphi \text{'}}\right)-\frac{\partial F}{\partial \varphi }=0$

Calculate the required derivatives

$$\begin{gathered} \frac{\partial F}{\partial \varphi \text{'}}=\frac{\varphi \text{'}{\sin }^2\theta }{\sqrt{1+\varphi {\text{'}}^2{\sin }^2\theta }} \\ \frac{\partial F}{\partial \varphi }=0 \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{d}{d\theta }\left[\frac{\varphi \text{'}{\sin }^2\theta }{\sqrt{1+\varphi {\text{'}}^2{\sin }^2\theta }}\right]=0 \\ \Rightarrow \frac{\varphi \text{'}{\sin }^2\theta }{\sqrt{1+\varphi {\text{'}}^2{\sin }^2\theta }}=C \\ \Rightarrow \varphi \text{'}=\frac{C}{\sin \theta \sqrt{{\sin }^2\theta -C^2}} \end{gathered}$$

Where C is constant.

Integratelocalid="1665053818462" $\varphi \text{'}=\frac{C}{\sin \theta \sqrt{{\sin }^2\theta -C^2}}$to get the desired result

$$\begin{gathered} \varphi \text{'}=\int \frac{C}{sin\theta \sqrt{si{n}^2\theta -C^2}}d\theta \\ =-\frac{CArcTanh\left({\displaystyle \frac{\sqrt{2}\sqrt{-C^2}cos\theta }{\sqrt{1-2C^2-cos2\theta }}}\right)}{\sqrt{-C^2}}+B \end{gathered}$$

Wherelocalid="1665053113180" $B$is integration constant.

The solution is real if

$1-2C^2-\cos 2\theta \le 0$

This means that spectrum of $\theta$is bound and it can be moved by value of $C$

Therefore, $\varphi =\int \frac{C}{\sin \theta \sqrt{{\sin }^2\theta -C^2}}d\theta =-\frac{CArcTanh\left({\displaystyle \frac{\sqrt{2}\sqrt{-C^2}\cos \theta }{\sqrt{1-2C^2-\cos 2\theta }}}\right)}{\sqrt{-C^2}}+B$, where$B$is the integration constant.