Show that total curvature of ellipse is $2\pi$ I'm trying to show that the total curvature $$K=\int_C\kappa\,ds$$ is equal to $2\pi$ over the ellipse $C$ with axes $a,b$ (and $\kappa$ is curvature).
I computed: 
$$x(t)=(a\cos t,b\sin t,0) \\ v(t)=(-a\sin t,b\cos t,0) \\ a(t)=-(a\cos t,b\sin t,0) \\ v(t)\times a(t)=\begin{vmatrix}i&j&k \\ -a\sin t&b\cos t&0 \\ -a\cos t&-b\sin t&0\end{vmatrix}=(0,0,ab) \\ ||v(t)\times a(t)||=ab \\ ||v(t)||^2=a^2\sin^2t+b^2\cos^2t \\ K=\int_0^{2\pi}\frac{||v\times a||}{||v||^3}||v||\,dt=\int_0^{2\pi}\frac{ab}{a^2\sin^2t+b^2\cos^2t}dt$$ and now I'm stuck. 
Please stick to multivariable-calc level methods, I am not familiar with differential geometry.

Trig substitution:
$$\frac{ab}{a^2\sin^2t+b^2\cos^2t}dt= \frac{ab}{a^2\tan^2t+b^2}d(\tan t)$$
