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 31
 Problem Statement

Boas Chapter 11, Section 12, Problem 22. Show that the exact solution when ##\alpha## is not small is
$$sin \frac \theta 2=sin\frac \alpha 2 sn \sqrt \frac g l t$$
where k = ##sin\frac \alpha 2## is the modulus of the elliptic function.
 Relevant Equations
 $$sn(u) = \int_0^\phi \frac{d\theta}{\sqrt{1k^{2}sin^{2}\theta}}=sin\phi$$
I understand how to reach
$$\int_0^\phi \frac{d\theta}{\sqrt{1k^{2}sin^{2}\theta}}=\sqrt \frac g l t$$
from physics but from there I don't get how to turn that into this new (for me) sn(u) form.
$$\int_0^\phi \frac{d\theta}{\sqrt{1k^{2}sin^{2}\theta}}=\sqrt \frac g l t$$
from physics but from there I don't get how to turn that into this new (for me) sn(u) form.