I find the derivation of the shape of the catenary curve by Jeremy Tatum in the libretexts version of his book quite elegant.
I reuse an image from an earler catenary question:
Tatum defines:
$\mu$ mass per unit of length of chain
$g$ gravitational force per unit of mass
$s$ length from midpoint to current coordinate
$\psi$ angle at current coordinate
$T_0$ Force component in the horizontal direction
then
$$ \tan \psi = \frac{\mu g s}{T_0} \tag 1 $$
for brevity a quantity $a$ is defined:
$$ a = \frac{T_0}{\mu g} \tag 2 $$
So:
$$ \tan \psi = \frac{dy}{dx} = \frac{s}{a} \tag 3 $$
from which, by derivation and rearranging:
$$ \frac{ds}{dx} = a \frac{d^2y}{dx^2} \tag 4 $$
Then Tatum uses $ds = (1 + \tfrac{dy}{dx}^2)^\tfrac{1}{2}dx$ to arrive at:
$$ \left(1 + \tfrac{dy}{dx}^2\right)^\tfrac{1}{2} = a \frac{d^2y}{dx^2} \tag 5 $$
Then Tatum capitalizes on the relation $\cosh^2 = \sinh^2 + 1$
With the condition that $\tfrac{dy}{dx} = 0$ where $x=0$
$$ \frac{dy}{dx} = \sinh (\frac{x}{a}) \tag 6 $$
Hence:
$$ y = a \cosh(\frac{x}{a}) + C \tag 7 $$
Tatum calls the move from (5) to (6) 'integration', but it seems to me that that step isn't actually integrating, but a substitution.
But: if I'm substituting anyway, then I can just as well go straight from (5) to (7)
I very much like the idea of capitalizing on the fact that $\sinh$ is the derivative of $\cosh$, and vice versa. The differential equation enforces the $\cosh$ as solution anyway, so as soon as you recognize that the differential equation is solved.
So I would love to learn about a way of going from (5) to (7) straigh away, explicitly motivated.
(The earlier question about the derivation of the catenary curves here on math.stackexchange gives in effect the same derivation as Tatum gives.)
If you allow $z(x)=y'(x)$, then (5) is $$ az'(x) = \sqrt{1+z^2}. $$
By separating variables: $$ \frac{dz}{\sqrt{1+z^2}} = \frac{dx}{a}, $$
which can be integrated to $$ \mathop{\mathrm{arcsinh}} z = \frac{x-x_0}a. $$