Prove that $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=2\pi$. Probably equation $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=\int_0^{2\pi} \frac{ab}{a^2\sin^2t+b^2\cos^2t}\mathrm dt$ is useful. Double integration is also a tool, but I don't know next step.
 A: A complex analysis approach: this is the real part of $\frac{1}{2\pi i} \int_C \frac{dz}{z}$ where $C$ is the ellipse $\gamma(t) = (a\cos t, b\sin t)$. Since this computes the winding number of $C$ around the origin, the answer is $1$, independent of $a$ and $b$.
A: Substitute $y = \tan t$
\begin{align}
\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}dt=&
\ 4ab \int_0^{\pi/2} \frac{ \sec^2 t}{a^2+b^2\tan^2t}dt\\
=& \ 4ab\int_0^\infty \frac {1}{a^2+b^2y^2}dy\\
=& \ 4\arctan \left(\frac{b\tan y}a\right) \bigg|_0^\infty = 2\pi
\end{align}
A: I thought it might be instructive to present an approach that uses the identities, $\cos^2(t)=\frac{1+\cos(2t)}{2}$ and $\sin^2(t)=\frac{1-\cos(2t)}{2}$ to streamline the analysis.  Proceeding, we can write
$$\begin{align}
\int_0^{2\pi}\frac{ab}{a^2\cos^2(t)+b^2\sin^2(t)}\,dt&=\int_0^{2\pi}\frac{2ab}{(a^2+b^2)+(a^2-b^2)\cos(2t)}\,dt\\\\
&\overbrace{=}^{t\mapsto t/2}\int_0^{4\pi} \frac{ab}{(a^2+b^2)+(a^2-b^2)\cos(t)}\,dt\\\\
&=4ab\int_0^{\pi}\frac{1}{(a^2+b^2)+(a^2-b^2)\cos(t)}\,dt\\\\
&\overbrace{=}^{\tan(t/2)\mapsto t}4\int_0^\infty \frac{ab}{a^2+b^2 t^2}\,dt\\\\
&=2\pi
\end{align}$$
and we are done!
