A trigonometic integral with complex technicals 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.
 A: Use this to conclude what you want:
$$\int_0^{2 \pi} \dfrac{dt}{1+r \sin^2(t)} = \sum_{k=0}^{\infty} (-r)^k \int_0^{2\pi} \sin^{2k}(t)dt = 2 \pi \sum_{k=0}^{\infty}(-r/4)^k \dbinom{2k}k = \dfrac{2\pi}{\sqrt{1+r}}$$
where we used the fact that
$$\int_0^{2\pi} \sin^{2k}(t) dt = \dfrac{2\pi}{4^k}\dbinom{2k}k$$
and
$$\dfrac1{\sqrt{1+r}} = \sum_{k=0}^{\infty}(-r/4)^k \dbinom{2k}k$$
We have
$$a^2 \cos^2(t) + b^2 \sin^2(t) = a^2 + (b^2-a^2)\sin^2(t) = a^2\left(1+\left(\left(\dfrac{b}a\right)^2 - 1\right)\sin^2(t)\right)$$
Hence, in our case, we have $r = \left(\dfrac{b}a \right)^2 - 1$ and hence $\sqrt{1+r} = \dfrac{b}a$.
$$I = \int_0^{2\pi} \dfrac{dt}{a^2 \cos^2(t) + b^2 \sin^2(t)} = \dfrac1{a^2} \times \dfrac{2\pi}{b/a} = \dfrac{2\pi}{ab}$$
A: 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 $\;\pi/2\;,\;\;3\pi/2\;$...
