Find natural (arc length) parametrization of a curve. If a curve is defined like this:
$$
\rho = 1 - \cos(x)
$$
find natural parametrization of it.  
So this is how the curve looks like when when I use polar coordinates $x = \rho \cos(t)$, $y = \rho\sin(t)$:
$$
\alpha(t) = \bigl((1 - \cos(t)) \cos(t), (1 - \cos(t)) \sin(t)\bigr).
$$
Now I found natural parameter to be $s = 4 \arcsin(\sqrt{s/8})$.
Can anyone tell me if this is good?
How does my curve look alike when I plug natural parameter in it?
Am I doing something wrong?
Sorry for the bad post. :/

The polar graph $\rho = f(\theta)$ parametrized by
$$
(x, y) = \bigl(f(t)\cos t, f(t) \sin t\bigr)
\tag{1}
$$
has speed
$$
\frac{ds}{dt} = \sqrt{f(t)^{2} + f'(t)^{2}}.
$$
For $f(t) = 1 - \cos t$, the speed is
$$
\sqrt{(1 - \cos t)^{2} + (\sin t)^{2}}
= \sqrt{2(1 - \cos t)}
= 2|\sin \tfrac{t}{2}|.
$$
If $0 \leq t \leq 2\pi$, then $\sin \frac{t}{2}$ is non-negative, so the arclength function is
$$
s = \int_{0}^{t} 2\sin \tfrac{u}{2}\, du
= 4(1 - \cos \tfrac{t}{2}),
$$
or $t = 2\arccos(1 - \frac{s}{4})$.
The double-angle formulas for the circular functions read
\begin{align*}
  \cos(2\theta) &= \cos^{2}\theta - \sin^{2}\theta, \\
  \sin(2\theta) &= 2\sin\theta \cos\theta,
\end{align*}
and $\cos(\arccos u) = u$ while $\sin(\arccos u) = \sqrt{1 - u^{2}}$.
Substituting $\theta = \arccos(1 - \frac{s}{4})$ gives
$$
\left.
\begin{aligned}
  \cos t &= (1 - \tfrac{s}{4})^{2} - 1 + (1 - \tfrac{s}{4})^{2}
  = 2(1 - \tfrac{s}{4})^{2} - 1, \\
  \sin t &= 2(1 - \tfrac{s}{4}) \sqrt{1 - (1 - \tfrac{s}{4})^{2}}
  = 2(1 - \tfrac{s}{4})\sqrt{\tfrac{s}{4}(2 - \tfrac{s}{4})}.
\end{aligned}
\right\}
\tag{2}
$$
The arc length parametrization is found by substituting (2) into (1). As a numerical check, the diagram shows the trig parametrization (red) overlaid by the putative arc length parametrization (blue).

