Using Cauchy's integral formula to evaluate $\int_0^{2\pi} \frac{1}{a^2 \cdot {\cos}^2(t) + b^2 \cdot {\sin}^2(t)} dt$

Question

I have to solve this integral using Cauchy's integral formula. I tried to substitute it with several different attempts but without a solution. Can anyone help? $$\int_0^{2\pi} \frac{1}{a^2 \cdot {\cos}^2(t) + b^2 \cdot {\sin}^2(t)} dt$$

Answer 1

First, note that $\cos(t)=\frac{e^{it}+e^{-it}}{2}$ and $\sin(t)=\frac{e^{it}-e^{-it}}{2i}$. Substituting this into the integral yields: $$\int_o^{2\pi}\frac{1}{a^2\cdot(\frac{e^{it}+e^{-it}}{2})^2+b^2\cdot(\frac{e^{it}-e^{-it}}{2i})^2}dt$$ Simplifying yields: $$=\int_0^{2\pi}\frac{4e^{2it}}{(a^2-b^2)e^{4it}+2(a^2+b^2)e^{2it}+(a^2-b^2)}dt$$ Note that this is the same as integrating over the unit circle twice, i.e. we have: $$=\int_{\partial D(0,1)}\frac{2}{i}\cdot\frac{1}{(a^2-b^2)z^2+2(a^2+b^2)z+(a^2-b^2)}dz$$ The poles of the integrand occur at: $\frac{a^2+b^2\pm\sqrt{(a^2+b^2)^2-(a^2-b^2)^2}}{a^2-b^2}=\frac{(a\pm b)^2}{a^2-b^2}$. Here you should apply any assumptions you have on $a$ and $b$ to determine whether either pole occurs in the unit disc, and neither pole occurs on its boundary. For the sake of example, suppose only the pole at $\frac{(a- b)^2}{a^2-b^2}$ is inside of the unit disc. Then using the calculus of residues, we have: $$=2\pi i\cdot\frac{2}{i}\cdot\frac{-(a^2-b^2)}{4ab}\cdot 2=\frac{2\pi(b^2-a^2)}{ab}$$