I am trying to solve this problem (1.56) from Tapp's Differential Geometry of Curves and Surfaces:
Let $m$ , $n$ be positive numbers. Find the rotation index of the Lissajous curve $\gamma\colon[0, 2\pi] \to \mathbb{R} $ defined as:
$$ \gamma(t) = (\cos(mt), \sin(nt)) $$
The definition of rotation index is given by this equation:
$$ ri = \frac{1}{2\pi}\int_{a}^{b} \kappa_{s}(t)\ |\gamma'(t)|dt = \frac{1}{2\pi}(\theta(b)-\theta(a)) $$
I tried to calculate the integral, but you get a really hard expression that I think it can not be solved. I also computed the curvature at $2\pi$ and $0$ ang I got this:
$$ \kappa(0)=\kappa(2\pi) = \frac{m^2}{n^2} $$
I also computed the tangent unit vector at this points, and I got:
$$ T(2\pi) = (0,1) = (\cos(\theta(2\pi)),\sin(\theta(2\pi))) $$
For some angle function $\theta$ that always exists for regular closed curves. $T(0)= (0,1)$, which makes sense because $\gamma(t)$ is a closed curve. I tried to think of an angle function that could work but I just couldn´t get anthing. Can someby please help me... am I doing anything wrong? Am I missing something. Thank you so much.