Answer
We prove:
$L \approx \dfrac{\pi }{2}\left( {3b + \dfrac{{{a^2}}}{b}} \right)$
Work Step by Step
We have from Exercise 90:
$G\left( k \right) = \dfrac{\pi }{2} - \dfrac{\pi }{2}\mathop \sum \limits_{n = 1}^\infty {\left( {\dfrac{{1\cdot3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot\cdot\cdot4\cdot\left( {2n} \right)}}} \right)^2}\dfrac{{{k^{2n}}}}{{2n - 1}}$, ${\ \ \ }$ converges for $\left| k \right| \lt 1$.
We use the first two terms of the series, so
$G\left( k \right) \approx \dfrac{\pi }{2} - \dfrac{\pi }{2}{\left( {\dfrac{1}{2}} \right)^2}{k^2}$
From Exercise 91, we also know that the arc length of the ellipse is $L = 4bG\left( k \right)$, where $k = \sqrt {1 - \dfrac{{{a^2}}}{{{b^2}}}} $.
So, ${k^2} = 1 - \dfrac{{{a^2}}}{{{b^2}}}$.
Substituting ${k^2}$ in $L = 4bG\left( k \right)$ gives
$L \approx 4b\left[ {\dfrac{\pi }{2} - \dfrac{\pi }{2}{{\left( {\dfrac{1}{2}} \right)}^2}\left( {1 - \dfrac{{{a^2}}}{{{b^2}}}} \right)} \right]$
$L \approx 4b\left( {\dfrac{\pi }{2} - \dfrac{\pi }{8} + \dfrac{\pi }{8}\dfrac{{{a^2}}}{{{b^2}}}} \right) = 4b\left( {\dfrac{{3\pi }}{8} + \dfrac{\pi }{8}\dfrac{{{a^2}}}{{{b^2}}}} \right)$
Hence, $L \approx \dfrac{\pi }{2}\left( {3b + \dfrac{{{a^2}}}{b}} \right)$.
