Perimeter of Ellipse

Theorem
Let $K$ be an ellipse whose major axis is of length $2 a$ and whose major axis is of length $2 b$.

The perimeter $\mathcal P$ of $K$ is given by:
 * $\displaystyle \mathcal P = 4 a \int_0^{\pi / 2} \sqrt{1 - k^2 \sin^2 \theta} \ \mathrm d \theta$

where:
 * $k = \dfrac {\sqrt {a^2 - b^2}} a$

The definite integral:
 * $\displaystyle \mathcal P = \int_0^{\pi / 2} \sqrt{1 - k^2 \sin^2 \theta} \ \mathrm d \theta$

is the complete elliptic integral of the second kind.

Proof
Let $K$ be aligned in a cartesian coordinate plane such that:


 * The major axis of $K$ is aligned with the X-axis
 * The minor axis of $K$ is aligned with the Y-axis.

Then from Equation of Ellipse in parametric form:
 * $x = a \cos \theta, y = b \sin \theta$

Thus:

From Arc Length for Parametric Equations, the length of one quarter of the perimeter of $K$ is given by:

Since $\cos \theta = \sin \left({\dfrac \pi 2 - \theta}\right)$ we can write for any real function $f \left({x}\right)$:


 * $\displaystyle \int_0^{\pi / 2} f \left({\cos \theta}\right) \ \mathrm d \theta = \int_0^{\pi / 2} f \left({\sin \left({\frac \pi 2 - \theta}\right)}\right) \ \mathrm d \theta$

So substituting $t = \dfrac \pi 2 - \theta$ this can be converted to:

justifying the fact that $\cos$ can be replaced with $\sin$ in $(1)$ above, giving:
 * $\displaystyle \mathcal P = 4 a \int_0^{\pi / 2} \sqrt{1 - k^2 \sin^2 \theta} \ \mathrm d \theta$