<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] 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}
</s>