Parametrizing a curve in arclength

Question

I have met this question in an ODE class

The function $$ S(x) = \int_{0}^{x} \sqrt{1+\sin^2 t} \space dt \quad 0 \leq x \leq 2\pi $$ is the arclength of the curve $y=\cos x$. Now we can reparametrize this curve by arclength, so we are asked to find the differential equation for $x(s)$ satisfying $ S(x(s)) = s $.

How does one find this differential equation? Do I plug $x(s)$ into the integral function $S$ and differentiate with respect to $s$ via the fundamental theorem of calculus? It kind of seems too simple so I was curious. We are supposed to get an ODE that we can solve numerically later on. Thanks to all helpers.

Answer 1

clearly $S'(x)={\sqrt{1+\sin^2 x}}.....(1)$

as $S(x(s))=s$

we have after differentiating . wrt '$s$

$S'(x(s)).x'(s)=1......(2)$

let $x(s)=y$

using ..(1),(2)

$\frac{dy}{ds}=\frac{1}{\sqrt{1+\sin^2y}}$