Answer
An arc length parametrization of the cycloid:
${{\bf{r}}_1}\left( s \right) = (2{\cos ^{ - 1}}\left( {1 - \frac{s}{4}} \right) - \sin \left( {2{{\cos }^{ - 1}}\left( {1 - \frac{s}{4}} \right)} \right),$
$1 - \cos \left( {2{{\cos }^{ - 1}}\left( {1 - \frac{s}{4}} \right)} \right))$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t - \sin t,1 - \cos t} \right)$. The derivative is
${\bf{r}}'\left( t \right) = \left( {1 - \cos t,\sin t} \right)$
So,
$||{\bf{r}}'\left( t \right)|{|^2} = \left( {1 - \cos t,\sin t} \right)\cdot\left( {1 - \cos t,\sin t} \right)$
$||{\bf{r}}'\left( t \right)|{|^2} = {\left( {1 - \cos t} \right)^2} + {\sin ^2}t$
$||{\bf{r}}'\left( t \right)|{|^2} = 1 - 2\cos t + {\cos ^2}t + {\sin ^2}t$
$||{\bf{r}}'\left( t \right)|{|^2} = 2 - 2\cos t = 2\left( {1 - \cos t} \right)$
$||{\bf{r}}'\left( t \right)|| = \sqrt {2\left( {1 - \cos t} \right)} $
Recall the identity ${\sin ^2}\left( {\frac{t}{2}} \right) = \frac{1}{2}\left( {1 - \cos t} \right)$. So,
$||{\bf{r}}'\left( t \right)|| = \sqrt {4{{\sin }^2}\left( {\frac{t}{2}} \right)} = 2\sin \left( {\frac{t}{2}} \right)$
Evaluate the arc length function:
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = 2\mathop \smallint \limits_0^t \sin \left( {\frac{u}{2}} \right){\rm{d}}u$
$s\left( t \right) = - 4\cos \left( {\frac{u}{2}} \right)|_0^t = - 4\left( {\cos \left( {\frac{t}{2}} \right) - 1} \right)$
It follows that
$ - \frac{s}{4} + 1 = \cos \left( {\frac{t}{2}} \right)$
$t = 2{\cos ^{ - 1}}\left( {1 - \frac{s}{4}} \right)$
Thus, the inverse of $s\left( t \right)$ is $t = 2{\cos ^{ - 1}}\left( {1 - \frac{s}{4}} \right)$.
Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {t - \sin t,1 - \cos t} \right)$ gives the arc length parametrization:
${{\bf{r}}_1}\left( s \right) = (2{\cos ^{ - 1}}\left( {1 - \frac{s}{4}} \right) - \sin \left( {2{{\cos }^{ - 1}}\left( {1 - \frac{s}{4}} \right)} \right),$
$1 - \cos \left( {2{{\cos }^{ - 1}}\left( {1 - \frac{s}{4}} \right)} \right))$
