I'm trying to derive the asymptotic solution for the following Jacobian elliptic function.
$x = \int^{\mathrm{sn}(x,k)}_{0} \frac{dt}{\sqrt{(1 - t^{2}) (1 - k^{2}t^{2})}} $
where $-1 \leq x \leq 1$ and $0 \leq k \leq 1$. If $k \ll 1$, we are assuming that $\mathrm{sn}(x,k) \sim s_{0} + k^{2}s_{1}$. For $k=0$, we get $\mathrm{sn}(x,0) = \mathrm{sin}(x)$, from which we can derive $s_{0} = \mathrm{sin}(x)$.
I know that the solution for $s_{1}$ is $s_{1} = -\frac{1}{4}(x - \mathrm{sin}(x) \mathrm{cos}(x))\mathrm{cos}(x) $. To reach this solution, I took the Taylor expansion of the function inside the integral with respect to $k$ and set $k=0$ within each term in the expansion, giving me the following integral and solution
$x = \int^{s_{0} + k^{2}s_{1}}_{0} \frac{dt}{\sqrt{(1 - t^{2}) }} + k \int^{s_{0} + k^{2}s_{1}}_{0} \frac{dt}{(1 - t^{2})^{ \frac{3}{2} } } = \mathrm{arcsin}(t') |^{s_{0} + k^{2}s_{1}}_{0} + \frac{t'}{\sqrt{(1 - t'^{2}) }} |^{s_{0} + k^{2}s_{1}}_{0} $
which reduces to
$x = \mathrm{arcsin}(s_{0} + k^{2}s_{1}) + \frac{k(s_{0} + k^{2}s_{1}) }{\sqrt{(1 - (s_{0} + k^{2}s_{1})^{2}) }} $
Plugging in $s_{0} = \mathrm{sin}(x)$, we arrive at
$x = \mathrm{arcsin}(\mathrm{sin}(x) + k^{2}s_{1}) + \frac{k(\mathrm{sin}(x) + k^{2}s_{1}) }{\sqrt{(1 - (\mathrm{sin}(x) + k^{2}s_{1})^{2}) }} $
At this point, I'm unsure how to solve the equation to find $s_{1}$, or if taking the initial Taylor expansion was even the correct move.