= Solution
For small $e$, the <binomial series> gives the uniform expansion $\sqrt{1-e^2\sin^2\theta}=1-(e^2/2)\sin^2\theta+O(e^4)$. Since $\int_0^{2\pi}\sin^2\theta\,d\theta=\pi$, integration yields
$$
\boxed{L=2\pi a\left(1-\frac{e^2}{4}\right)+O(ae^4)
=\frac{\pi(3a^2+b^2)}{2a}+O\left(\frac{(a^2-b^2)^2}{a^3}\right)}.
$$
Also $b=a\sqrt{1-e^2}=a(1-e^2/2+O(e^4))$, so the familiar symmetric approximation $\boxed{L=\pi(a+b)+O(ae^4)}$ is equivalent to this order. This is the first correction in the <ellipse perimeter expansion near a circle>; the two approximate formulas need not agree at order $e^4$.
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