<s>[INST] Evaluate The Arc Length 
Let $$\alpha(t)=(\alpha \cos^3t,\alpha \sin^3 t)$$ Evaluate the arc length when $\alpha>0$ and $t\in[0,2\pi]$

$$\alpha'(t)=(-3\alpha \cos^2{t} \sin{t} ,3\alpha \sin^2t\ \cos t)$$
$$||\alpha'(t)||=\sqrt{(-3\alpha \cos^2{t} \sin{t})^2+(3\alpha \sin^2t\ \cos t)^2}=\sqrt{9\alpha^2\cos^4{t} \sin^2{t}+9\alpha^2 \sin^4t\ cos^2t}=\sqrt{9\alpha^2\cos^2t\sin^2t(\cos^2t+\sin^2t)}=\sqrt{9\alpha^2\cos^4{t} \sin^2{t}+9\alpha^2 \sin^4t\cos^2t}=\sqrt{9\alpha^2\cos^2t\sin^2t}=3\alpha \cos t \sin t$$
So we have:
$$\int_{0}^{2\pi}||\alpha'(t)||dt=\int_{0}^{2\pi}(3\alpha \cos t  \sin t)\,dt=3\alpha \int_{0}^{2\pi} \cos t \sin t\, dt$$
$u=\sin t$ and $du=\cos t\, dt$
$$3\alpha \int_{0}^{2\pi} udu=3\alpha\frac{u^2}{2}|_{0}^{2\pi}=\frac{3}{2}\alpha{\sin t^2}|_{0}^{2\pi}=\frac{3}{2}(0-0)=0$$
But the answer is $6\alpha$ where did I get it wrong?
 [/INST] Here another way to do this, this time in the complex plane. Let the astroid be given by
$$z=a(\cos^2 t+i~b\sin^3 t),\quad t\in[0,2\pi]$$
The arc length is given by
$$
\begin{align}
s
&=\int_0^{2\pi}|\dot z|~dt\\
&=4\int_0^{\pi/2}|\dot z|~dt,\quad \text{by symmetry}\\
&=4a\int_0^{\pi/2}|-3\cos^2 t~\sin t+3\sin^2 t~\cos t|~dt\\
&=4a\int_0^{\pi/2}\sqrt{9\cos^4 t~\sin^2 t+9\sin^4 t~\cos^2 t}~dt\\
&=4a\int_0^{\pi/2}\sqrt{\cos^2 t~\sin^2 t}~dt\\
&=12a\int_0^{\pi/2} \sin t~d(\sin t)\\
&=6a \sin^2 t\biggr|_0^{\pi/2}\\
&=6a
\end{align}
$$
</s>