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.

Clarifying @Blue's answer, taking $[0,2\pi]=[0,\pi/2]\cup(\pi/2,3\pi/2]\cup(3\pi/2,2\pi]$ and regarding that: $$\lim_{x\rightarrow (2k+1)\pi/2}\tan(x)=\pm\infty\qquad (k=0,1)$$
then:
$$\int_{0}^{2\pi}\frac{\sec^2(t)}{a^2\tan^2(t)+b^2}dt=\int_{0}^{\infty}\frac{du}{u^2+1}+\int_{-\infty}^{\infty}\frac{du}{u^2+1}+\int_{-\infty}^{0}\frac{du}{u^2+1}=\arctan(u)\biggl.\biggr|_{0}^{\infty}+\arctan(u)\biggl.\biggr|_{-\infty}^{\infty}+\arctan(u)\biggl.\biggr|_{-\infty}^{0}=\pi/2+\pi+\pi/2=2\pi$$.
