Since this is a first time I hear about trochoids, I just computed the arc length using $$\begin{eqnarray*} x &=& a\phi - b \sin (\phi) \implies x'=a -b\cos(\phi)\\\ y &=& a - b \cos (\phi)\implies y'=b \sin (\phi) \end{eqnarray*}$$ $$L=\int \sqrt{(x')^2+(y')^2} \,d\phi=\int \sqrt{a^2+b^2-2 a b \cos (\phi )}=2 |a-b| \,E\left(\frac{\phi }{2}|-\frac{4 a b}{(a-b)^2}\right)$$ if $a \
eq b$.
Looking at the first argument of the elliptic integral, it seems that your are totally correct.