Integrating the Euler-Lagrange equation Let us have a Lagrangian $L(y,y') = f(y)\sqrt{1+y'^2}$, where $y=y(x)$. The corresponding Euler-Lagrange equation is
$$\frac{f'}{\sqrt{1+y'^2}} - f\frac{y''}{(1+y'^2)^{3/2}}=0$$
This expression should be possible to recast in the following form
$$\frac{d}{dy}\left(\frac{f}{\sqrt{1+y'^2}}\right) = 0$$
which can be integrated to get the stationary solution for $y'(x)$. I am not able to figure out how to get this equation from the first one above. Any idea?
 A: I just found the solution to this problem. We are given the Lagrangian  $L(y,y')=f(y)\sqrt{1+y'^2}$. Since this is not an explicit function of $x$, the Euler-Lagrange equation simplifies into this form:
$$\frac{d}{dx}\left(L-y'\frac{\partial L}{\partial y'}\right)=0$$
(see the Wikipedia page "Beltrami identity") for the derivation. In our case, this will give us
$$\frac{d}{dx} \left( f\frac{1}{\sqrt{1+y'^2}} \right) = 0\,.$$
Again, the term in the bracket is not an explicit function of $x$ and thus this total differential can be written as $d/dx = y'(\partial/\partial y)+y''(\partial/\partial y')$. Instead of doing the derivative $d/dx$ above,  we use the right-hand side of this identity. This yields:
$$y'\left[ f' \frac{1}{\sqrt{1+y'^2}} - f \frac{y''}{(1+y'^2)^{3/2}}\right] = 0\,.$$
The last step is to multiply the last two equations by $dx/dy\equiv1/y'$. This will change the variable with respect to which we differentiate in the first and will remove the prefactor $y'$ in the second. The results are
$$\frac{d}{dy} \left( f\frac{1}{\sqrt{1+y'^2}} \right) = 0$$
$$f' \frac{1}{\sqrt{1+y'^2}} - f \frac{y''}{(1+y'^2)^{3/2}} = 0$$
The second equation was obtained using the indentity (total differential) and is thus necessarily the same as the first. However, from the second, it is not obvious how one would obtain the stationary solution $y'(x)$, while this is straightforward from the first.
