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**Nexus** Here is a problem that I was wondering if anyone could help me out on.

An ellipse has parametric equations $\displaystyle x = a\cos{t}$, $\displaystyle y =b\sin{t}$ for $\displaystyle 0 \leq t \leq 2\pi$ where $\displaystyle a>b>0$.

Show that its perimeter is given by the integral $\displaystyle \int_{0}^{2\pi} {\sqrt{\frac{a^2 + b^2}{2} - \frac{a^2 - b^2}{2}cos2t}}dt$.

Hence show that, if the ellipse is close to circular, the perimeter is approximately equal to $\displaystyle 2\pi\sqrt{\frac{a^2 + b^2}{2}}\left(1 -\frac{1}{16}\left(\frac{a^2 - b^2}{a^2 + b^2}\right)^2 - \frac{15}{1024}\left(\frac{a^2 - b^2}{a^2 + b^2}\right)^4\right)$

I managed to show that the perimeter is given by the given integral but I cannot make headway with the approximation. I rearrange the integral to give this;

$\displaystyle {\sqrt{\frac{a^2 + b^2}{2}}\int_{0}^{2\pi} {\sqrt{1 - \frac{a^2 - b^2}{a^2 + b^2}cos2t}}dt$

I thought maybe I could integrate the binomial for the terms under the root symbol (using the maclaurin expansion of $\displaystyle \cos2t$ but then that would take me to an answer with higher powers of $\displaystyle \pi$ so I don't really know where to go with it.

Any thoughts would be appreciated!