Curvature and turning number of $t \mapsto (\cos(t), \sin(3t))$ 
Is the closed curve with period $2 \pi$
$$
\delta(t) := (\cos(t), \sin(3t))
$$
  regularly homotopic to the positively traversed unit circle $(\cos(t), \sin(t))$?

From this question I now know that the answer is indeed yes and one answer even gives the regular homotopy.
Before finding this answer I tried to solve this task using the theorem of Whitney-Graustein, which says that two closed curves in $\mathbb R^2$ are regularly homotopic to each other if and only their turning number is equal.
Definition.
The turning number of a closed curve $(\gamma, \tau)$ is
$$ \tag{1}
n = \frac{1}{2 \pi} \int_0^{\tau} \kappa(s) ds,
$$
where $\kappa$ is the curvature of $\gamma$.
For curves (not necessarily parametrised with respect to arc lenght) we derived their curvature to be
$$ \tag{1}
\kappa(t)
= \frac{\det(\gamma'(t), \gamma''(t))}{| \gamma'(t) |^3}
$$
I got 
\begin{equation*}
        \delta'(t)
        = (- \sin(t), 3 \cos(3 t))
        \quad \text{and} \quad
        \delta''(t)
        = - (\cos(t), 9 \sin(3 t))
    \end{equation*}
and thus
\begin{equation*}
        \kappa(t)
        = \frac{9 \sin(3t) \sin(t) + 3 \cos(3t) \cos(t)}{(\sin^2(t) + 9 \cos^2(3t))^{3/2}}
        = \frac{6 \cos(2 t) - 3 \cos(4 t)}{(\sin^2(t) + 9 \cos^2(3t))^{3/2}}
    \end{equation*}
As $\kappa(t + \pi) = \kappa(t)$ holds we have $\int_0^{2 \pi} \kappa(s) ds = 2 \int_{0}^{\pi} \kappa(s) ds$ and thus the turning number is
    \begin{equation*}
    n
    = \frac{2}{2 \pi} \int_{0}^{\pi} \kappa(s) ds
    = \frac{1}{\pi}\int_{0}^{\pi} \frac{6 \cos(2 s) - 3 \cos(4 s)}{(\sin^2(s) + 9 \cos^2(3s))^{3/2}} ds
    = 1.92492, \end{equation*}
where the last equality is given by WolframAlpha.
Clearly this is incorrect as the turning number is a integer.
Is my understanding of curvature incorrect?
 A: In the definition you're using, the turning number is given by an integral with respect to $s$, arclength. Your parameter $t$ (although you switched randomly to both $x$ and $s$) is not an arclength parameter. So, if you correct by using $ds = \dfrac{ds}{dt}dt$, you get the correct answer that $n=1$.
A: The winding number of a curve $\gamma$ parametrized as $I\ni t \mapsto (x(t), y(t))= x(t) + i y(t) = z(t)$ can be calculated by the formula
$$\frac{1}{2 \pi i}\int_{\gamma} \frac{d z}{z}=\frac{1}{2 \pi i} \int_I \frac{z'(t)}{z(t)} dt$$
In your case the result is $-1$.
This can also be inferred by inspecting  the curve

You can also check that your curve is homotopic to the curve $t\mapsto (\cos t, - \sin t)$ by the obvious homotopy.
The analogous curve $(\cos t, \sin 5t)$ has index $1$. Worth looking at pictures...
$\bf{Added:}$ I mistook the turning number for the winding number. However, if $\gamma\colon I \to \mathbb{C}$ is a curve then the turning number of $\gamma$
equals the winding number of $\gamma'$, the derivative of $\gamma$. Indeed, one checks that this is correct for the arclength parametrization, and that it is independent on the parametrization:
Indeed, if $\eta(s) = \gamma(\phi(s))$, then $\eta'(s) = \gamma'(\phi(s)) \phi'(s)$, and $\eta''(s)= \gamma''(\phi(s))(\phi'(s))^2 + \gamma'(\phi(s))\phi''(s)$ so
$$\int_J\frac{\eta''(s)}{\eta'(s)} = \int_J\frac{\gamma''(\phi(s))}{\gamma'(\phi(s))}\cdot \phi'(s) + \int_J \frac{\phi''(s)}{\phi'(s)}= \int_I\frac{\gamma''(s)}{\gamma'(s)}+ 0$$
$\bf{Added}$ It is not hard to see that
$$\frac{1}{2\pi i} \int_I\frac{\gamma''(t)}{\gamma'(t)}dt = \frac{1}{2\pi}
\int_I\frac{\gamma''(t) \times \gamma'(t)}{\|\gamma'(t)\|^2} dt$$
in the second integral $\gamma$ is considered taking values in $\mathbb{R}^2$.
