TITLE: A trigonometic integral with complex technicals
QUESTION [5 upvotes]: Let $a,b \in \mathbb{R}^+$. Show that $$\int_0^{2\pi}\frac{1}{a^2\cos^2(t)+b^2\sin^2(t)}dt=\frac{2\pi}{ab}$$ Help please!
Thanks.

REPLY [2 votes]: Hints:
$$\frac1{a^2\cos^2t+b^2\sin^2t}=\frac1{a^2\cos^2t}\frac1{1+\left(\frac ba\tan t\right)^2}=\frac1{ab}\frac{\left(\frac ba\tan t\right)'}{1+\left(\frac ba\tan t\right)^2}\implies$$
$$\implies\int\limits_0^{2\pi}\frac1{a^2\cos^2t+b^2\sin^2t}dt=\left.\frac1{ab}\arctan\left(\frac ba\tan t\right)\right|_0^{2\pi}$$
But ...observe closely the point $\;\pi/2\;,\;\;3\pi/2\;$...