Answer
\[L=4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-e^2\sin^2\theta}\;\;d\theta\]
Work Step by Step
\[x=a\sin\theta\;,\;y=b\cos\theta\;\;\ ,\;0\leq \theta< 2\pi\]
Differentiate $x$ with respect to $\theta$:
\[\frac{dx}{d\theta}=a\cos\theta\]
Differentiate $y$ with respect to $t$:
\[\frac{dy}{d\theta}=-b\sin\theta\]
Therefore the arc length is given by:
\[L=\int_{0}^{2\pi}\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta\]
\[L=\int_{0}^{2\pi}\sqrt{\left(a\cos\theta\right)^2+\left(-b\sin\theta\right)^2}\;\;d\theta\]
\[L=\int_{0}^{2\pi}\sqrt{\left(a^2\cos^2\theta\right)+\left(b^2\sin^2\theta\right)}\;\;d\theta\]
\[L=\int_{0}^{2\pi}\sqrt{\left(a^2(1-\sin^2\theta\right)+\left(b^2\sin^2\theta\right)}\;\;d\theta\]
\[L=\int_{0}^{2\pi}\sqrt{a^2-(a^2-b^2)\sin^2\theta}\;\;d\theta\]
\[L=a\int_{0}^{2\pi}\sqrt{1-\frac{(a^2-b^2)}{a^2}\sin^2\theta}\;\;d\theta\]
Since ellipse is symmetric in four quadrants
\[\Rightarrow L=4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-\frac{(a^2-b^2)}{a^2}\sin^2\theta}\;\;d\theta\]
Since $c^2=a^2-b^2$
\[\Rightarrow L=4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-\frac{c^2}{a^2}\sin^2\theta}\;\;d\theta\]
Since $e=\displaystyle\frac{c}{a}$
\[\Rightarrow L=4a\int_{0}^{\frac{\pi}{2}}\sqrt{1-e^2\sin^2\theta}\;\;d\theta\]
Hence Proven.
