<s>[INST] Arc length of parametric equations Find the arc length of the parametrized path $x(t) = t^2/2$ and $y(t) = t^3/3$ for $1 \le t \le 3$.
I try to do the standard integral of $\sqrt{\mathrm dx^2 + \mathrm dy^2}$, but get stuck when I have to integrate $\sqrt{t^2 + t^4}$. 
Any ideas? 
 [/INST] $$
\int\mathrm{d}t \sqrt{t^2+t^4} = \int\mathrm{d}t\ \ t\sqrt{1+t^2}
$$
for positive t. Now let $\sqrt{1+t^2}=y-t$: 
$$
1+t^2=y^2+t^2-2yt
$$
$$
t=\frac{y^2-1}{2y}
$$
$$
\mathrm{d}t=\mathrm{d}y \frac{y^2+1}{2y^2}
$$
and
$$
\sqrt{1+t^2}=\frac{y^2+1}{2y}.
$$
Substituting yields:
$$
\int \mathrm{d}y \frac{y^2+1}{2y^2} \frac{y^2-1}{2y} \frac{y^2+1}{2y} = \frac{1}{8}\int \mathrm{d}y \frac{(y^4-1)(y^2+1)}{y^4}= \frac{1}{8}\int \mathrm{d}y \frac{y^6+y^4-y^2-1}{y^4}=\frac{1}{8}\int \mathrm{d}y\left( y^2+1-y^{-2}-y^{-4}\right)
$$
which is no big deal!
</s>