The Euler-Lagrange equation and the Beltrami identity

[Cleonis]

TL;DR Summary

The operation that converts the Euler-Lagrange equation to the Beltrami identity is integration with respect to the y-coordinate. I'm looking for a transparent way to perform that conversion.

This question is specifically about deriving the Beltrami identity.

Just to give this question context I provide an example of a problem that is solved with Calculus of Variations: find the shape of a soap film that stretches between two coaxial rings.

For the surface area the expression to be integrated from start point to end point:
$$ F = 2 \pi \int_{x_0}^{x_1} y \ \sqrt{1 + (y')^2} \ dx \tag{1} $$
For the purpose of finding the function that minimizes that surface area the Euler-Lagrange equation is applied.

As we know, since the value of $F$ does not depend directly on the x-coordinate the Beltrami identity is applicable.

Comparison of the EL-equation and the Beltrami identity:Euler-Lagrange:
$$ \frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) = 0 \tag{2} $$
Beltrami:
$$ F - y' \frac{\partial F}{\partial y'} = C \tag{3} $$
We see that the process of conversion from EL-eq. to Beltrami consists of integration with respect to the y-coordinate

For the first term:
$$ \int \frac{\partial F}{\partial y} dy = F + C \tag{4} $$
with $C$ an arbitrary integration constant.

Question:
Is there a transparent way to evaluate the same integral for the second term?
$$ \int \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)dy + C \quad = ? = \quad \frac{dy}{dx} \frac{\partial F}{\partial y'} \tag{5} $$

The thing is: showing that (5) is indeed correct is worthwhile only if it can be done in a way that is more accessible than the usual way of obtaining the Beltrami identity.

 

[vanhees71]

The Beltrami identity comes from the independence of the Lagrangian of the independent variable, $x$. You can get it by just taking the total derivative of $L(y,y',x)$:
$$\mathrm{d}_x L=y' \partial_y L + y'' \partial_{y'} L + \partial_x L.$$
For the solutions of the Euler-Lagrange equations you can write this as
$$\mathrm{d}_x L=y' \mathrm{d}_x (\partial_{y'} L) + y'' \partial_{y'} L+ \partial_x L= \mathrm{d}_x (y' \partial_{y'} L)+\partial_x L$$
or
$$\mathrm{d}_x (y' \partial_{y'} L-L)=-\partial_x L.$$
If $L$ doesn't depend explicitly on $x$, you have $\partial_x L$ and
$$y' \partial_{y'}L-L=C=\text{const}$$
for all $y$ that are solutions of the Euler-Lagrange equation.