# What is the arc length of f(x)=cosx on x in [0,pi]?

Jun 1, 2018

$2 E \left(- 1\right)$, where $E$ is the complete elliptic integral of the second kind. There is no simpler way of expressing this analytically. Numerically, this is approximately equal to 3.82.

#### Explanation:

Arc length of a curve $f \left(x\right)$ is given by the formula $s = {\int}_{{x}_{0}}^{{x}_{1}} \sqrt{1 + f ' {\left(x\right)}^{2}} \mathrm{dx}$. Note that it is often a difficult problem to calculate this integral, and many classes of relatively simple curves do not have simple solutions for it.

This formula is derived by adding up infinitesimal arc lengths along the curve. Here, for example, is an online explanation: http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx

In this case
$s = {\int}_{0}^{\pi} \sqrt{1 + {\sin}^{2} x} \mathrm{dx}$

Now here we have run straight into the problem I warned about above! This is not an integral that is solvable with elementary functions. Indeed, it is an integral that is part of a famous problem from mathematical history - the calculation of the arc length of the ellipse. The solution of this equation is a transcendental function, one that is simply defined to solve this particular equation, the (complete) Elliptic Integral of the Second Kind: $E \left({k}^{2}\right)$.

This is defined as
$E \left({k}^{2}\right) = {\int}_{0}^{\frac{\pi}{2}} \sqrt{1 - {k}^{2} {\sin}^{2} \theta} d \theta$
Note that $\sin$'s symmetry around $\frac{\pi}{2}$ means that then
$2 E \left({k}^{2}\right) = {\int}_{0}^{\pi} \sqrt{1 - {k}^{2} {\sin}^{2} \theta} d \theta$
which now matches our limits of integration.

Taking ${k}^{2} = - 1$ (i.e. $k = i$), we see that
$2 E \left(- 1\right) = {\int}_{0}^{\pi} \sqrt{1 + {\sin}^{2} \theta} d \theta$, the arc length of $\cos \theta$ over the interval we seek.

Numerically, this is approximately 3.82.