Using the arc length function to find a parameterization of C in terms of s The problem: Find the arc length function $s(t)$ for the curve defined by $\vec r(t)$. Then use this result to find a parametrization of $C$ in terms of $s$.
$$\vec r(t) = a\cos^3t\,\hat i + a\sin^3t\,\hat j+ \hat k,\quad 0\le t\le 2\pi$$
Attempt at Solution:
For the first part, I have produced 
$$\vec r'(t) = -3a \cos^2t \sin t\,\hat i + 3a\sin^2t \cos t \,\hat j$$
and
$$|\vec r'(t)| = 3a\sin t \cos t$$
then
$$\int^t_0|\vec r'(u)| du = \left. -\frac 32a\cos^2u \right|^t_0 = \frac 32a\sin^2t$$
At this point I need to find a parametrization in terms of $s$. There may be some identities that could help make this cleaner, but, given my current knowledge, this is what I have come up with:
$$t = \arcsin\left(\sqrt \frac 23 \sqrt s\right)$$
If I was successful in solving for t, then the answer would be:
$$\vec r(t(s)) = a \cos^3 \left[\arcsin\left(\sqrt \frac 23 \sqrt s\right)\right]\,\hat i + a\sin^3\left[\arcsin\left(\sqrt \frac 23 \sqrt s\right)\right]\,\hat j + \hat k$$
I hope I have given enough information and I greatly appreciate any advice or direction.
 A: I agree with you nearly up to this point:
$$ds \ = \ 3a \sin t \cos t  \ , $$
but because we are taking the square-root $ \ \sqrt{\cos^2 t \ \sin^2 t} \ $ to obtain the infinitesimal arclength element , it would be better that we write the result as  $ \ ds \ = \ 3a \cdot | \sin t \cos t \ |  \ =  \frac{3}{2} a \ | \sin 2t \ |  \ . $
The curve itself is our "old friend", the astroid (here, lying in the plane $ \ z = 1 \ $ ) , so the sinusoidal nature of the arclength differential $ \ ds \ $ is perhaps not too surprising.  As André Nicolas describes, however, this presents a small complication when we prepare to integrate this in order to obtain the arclength function $ \ s(t) \ . $  Integration in the first quadrant correctly gives us $ \ s(t) = \frac{3}{4} a \ ( 1-  \cos 2t ) \ $ , but from there on, accumulation of arclength is not automatically brought about properly.  We must adjust the integration in each quadrant by adding in the arclength from "previous" quadrants and adjust the integrand, with its absolute value in the expression, appropriately in alternate quadrants.
We thus obtain a piecewise-defined arclength function, given here for $ \ a = 1 \ , $
$$s(t) \ = \ \begin{array}{cc}\frac{3}{4}  -  \frac{3}{4} \cos 2t \ = \ \frac{3}{2}  \sin^2 t \ , \ 0 \le t \le \frac{\pi}{2} \\ \frac{9}{4}  +  \frac{3}{4} \cos 2t \ = \ \frac{3}{2} + \frac{3}{2}  \cos^2 t \ , \ \frac{\pi}{2} \le t \le \pi \\ \frac{15}{4}  -  \frac{3}{4} \cos 2t \ = \ 3 + \frac{3}{2}  \sin^2 t \ , \ \pi \le t \le \frac{3 \pi}{2} \\ \frac{21}{4}  +  \frac{3}{4} \cos 2t \ = \ \frac{9}{2} + \frac{3}{2}  \cos^2 t \ , \ \frac{3 \pi}{2} \le t \le 2 \pi \end{array} \ \ . $$
Graphs of the "unit" astroid and its arclength function [the total arclength of which is 6] (sketched in dark blue) are presented below.  I hope this is of some help and provides an appreciation of the special nature of the problem.

EDIT: I had to make a small correction above.  The factor $  \ \sin 2t \ $ alternates sign in alternate quadrants.
