<s>[INST] 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.
 [/INST] 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!
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