Answer
The parametrization of the ellipse $c\left( t \right) = \left( {a\sin t,b\cos t} \right)$, where $0 \le t \le 2\pi $ has length:
$L = 4aG\left( {\frac{\pi }{2},k} \right)$
Work Step by Step
The ellipse ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$ can be parametrized by $c\left( t \right) = \left( {a\sin t,b\cos t} \right)$, where $0 \le t \le 2\pi $.
We have
$x\left( t \right) = a\sin t$, ${\ \ \ }$ $x'\left( t \right) = a\cos t$,
$y\left( t \right) = b\cos t$, ${\ \ \ }$ $y'\left( t \right) = - b\sin t$.
Using Eq. (3) of Theorem 1, the arc length for $0 \le t \le \theta $ is
$s\left( \theta \right) = \mathop \smallint \limits_0^\theta \sqrt {{{\left( {a\cos t} \right)}^2} + {{\left( { - b\sin t} \right)}^2}} {\rm{d}}t = \mathop \smallint \limits_0^\theta \sqrt {{a^2}{{\cos }^2}t + {b^2}{{\sin }^2}t} {\rm{d}}t$
It may be written as
$s\left( \theta \right) = \mathop \smallint \limits_0^\theta \sqrt {{a^2}\left( {{{\cos }^2}t + \frac{{{b^2}}}{{{a^2}}}{{\sin }^2}t} \right)} {\rm{d}}t = a\mathop \smallint \limits_0^\theta \sqrt {1 - {{\sin }^2}t + \frac{{{b^2}}}{{{a^2}}}{{\sin }^2}t} {\rm{d}}t$
$s\left( \theta \right) = a\mathop \smallint \limits_0^\theta \sqrt {1 - \left( {1 - \frac{{{b^2}}}{{{a^2}}}} \right){{\sin }^2}t} {\rm{d}}t = a\mathop \smallint \limits_0^\theta \sqrt {1 - {k^2}{{\sin }^2}t} {\rm{d}}t$,
where $k = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} $.
Write $G\left( {\theta ,k} \right) = \mathop \smallint \limits_0^\theta \sqrt {1 - {k^2}{{\sin }^2}t} {\rm{d}}t$.
So, the length of the ellipse for $0 \le t \le \theta $ is
$s\left( \theta \right) = aG\left( {\theta ,k} \right)$.
By symmetry, the length of the ellipse for $0 \le t \le 2\pi $ is four times the length for $0 \le t \le \frac{\pi }{2}$. Hence,
$L = 4aG\left( {\frac{\pi }{2},k} \right)$.
