Alternatives to the classical pendulum ODE

It is possible to find another way for the solution of $$y''+k\sin(y)=0 \qquad \text{with} \qquad y=y(x)\qquad \text{and} \qquad k>0$$ Switch variables and then write $$\frac {x''}{[x']^3}=k\sin(y)$$ The usual reduction of order gives $$x'=\pm\frac{1}{\sqrt{2 k \cos (y)+c_1}}$$ So $$x+c_2=\pm\frac{2}{\sqrt{c_1+2 k}}F\left(\frac{y}{2}|\frac{4 k}{c_1+2 k}\right)$$ where appears the elliptic integral of the first kind.

This can be inversed easily leading to the amplitude for the Jacobi elliptic functions.

To solve the ODE in the form of $$2y''+ p^2\sin y = 0,\quad p=\sqrt{\dfrac l{2g}},$$ can be used the integrating factor $\;y'\not=0,\;$ and that leads to the equation $$y'^2 - p^2\cos y = y'^2\big|_{y=0} -p^2 = (k^2-1)p^2,$$ $$\dfrac{\text dy}{\text dx} = \pm p\sqrt{k^2 - 1-\cos y}\,,$$ with the solution $$\pm p(x\bigg|_{y=0}+x) = \dfrac2{\sqrt{k^2-2}}\operatorname{F}\left(\dfrac y2\,\bigg|-\dfrac2{k^2-2}\right),$$ where $\;F\;$is the elliptic integral of the first kind.