 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 yieldsAlso $b=a\sqrt{1-e^2}=a(1-e^2/2+O(e^4))$, so the familiar symmetric approximation $\boxed{L=\pi (a+b)+O(a{e}^4)}$ 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 $e^4$.