I have to show that $\int_{0}^{2\pi}\frac{d\theta}{(a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta)^{2}} =\frac{ \pi(a^{2}+b^{2})}{a^{3}b^{3}}$ where $a,b>0$. I have tried using double angle formulas and Euler's trig identities to simplify this in order to use either residues or Cauchy's integral formula, but everything I have tried has made it messier and messier. Either I'm trying to simplify some things too early or something, but any help or tips would be much appreciated.
Okay, I made the substitutions, and right now just focusing on the denominator, I got to $(a^{2}(z+z^{-1})^2-b^{2}(z-z^{-1})^2)^2$ (i took out the common factor of 4 that I got). Then here is where I'm stuck. It looks like I have a difference of 2 perfect squares, so I then wrote $((a(z+z^{-1})-b(z+z^{-1}))(a(z+z^{-1})+b(z+z^{-1})))^2$ which I then wrote as $(a(z+z^{-1})-b(z+z^{-1}))^2(a(z+z^{-1})+b(z+z^{-1}))^2$ If I did it right, which I hope I did, Now I'm not sure where to go from here into getting a rational function.