I have the following curve equation:
$$r(t) = \frac{c^2}{a}\cos^3(t)i + \frac{c^2}{b}\sin^3(t)j, \quad c^2 = a^2 - b^2, \quad 0 \leq t \leq 2\pi$$
I need to find the path traced by a particle moving along this path during the interval of $t$ specified. It is quite clear that the path traced is the integral of speed $\lVert r'(t) \rVert$ over the interval $[0, 2\pi]$. Thus, I first find the speed:
$$\lVert r'(t)\rVert = \frac{3c^2\cos t \sin t}{2ab}\sqrt{b^2 + c^2\sin^2t}$$
which seems to be continuous everywhere on $[0, 2\pi]$, and the target integral is then:
$$\Lambda(0, 2\pi) = \frac{3}{2ab}\int_{0}^{2\pi}2c^2\cos t \sin t \sqrt{b^2 + c^2\sin^2t} dt $$ which can then be evaluated with the substitution. However, the integral I get is:
$$\frac{3}{2ab}\sqrt{b^2 + c^2\sin^2t}^3 + C$$
If I use this with its limits $2\pi$ and $0$, the arc length is 0!!! I plotted the curve, and this path DOES exist, and seems continuous. What am I doing wrong here?
PS. The answer in Apostol 14.13#4 seems to be $\frac{4(a^3 - b^3)}{ab}$.
If you plot your $\|r'(t)\|$ for $t \in (0,2\pi)$, you will see it takes on negative values and is $\pi$-periodic, so $\int_0^{2\pi} \|r'(t)\|dt = 0$ as you have found.
I suggest using symmetry to claim that $$ L = 4\int_0^{\pi/2} \left\|r'(t) \right\| dt $$