Integrate $\int_0^{2 \pi } \frac{1}{(a+b \cos^2 (x))^2} \, \mathrm{d}x$

Question

I'm having a trouble with this integral expression: $$\int_0^{2 \pi } \frac{1}{(a+b \cos^2 (x))^2} \, \mathrm{d}x$$ I want to solve to using residue but it seems hard.

Answer 1

Check the following ideas:

$$a+b\cos^2x=a\left(1+\left(\sqrt\frac ba\;\cos x\right)^2\right)$$

Using the substitution on $\;[0,\pi]\;$:

$$t:=\sqrt\frac ba\;\cos x\;,\;\;dt=-\sqrt\frac ba\;\sin xdx\implies dx=-\sqrt\frac ab\frac{dt}{\sqrt{1-t^2}}\;:$$

$$\frac1a\int_{\sqrt\frac ba}^{-\sqrt\frac ba}-\sqrt\frac ab\frac{dt}{\sqrt{1-t^2}}\cdot\frac{dt}{1+t^2}=\frac2{\sqrt{ab}}\int_0^{\sqrt\frac ba}\frac{dt}{(1+t^2)^{3/2}}=\left.\frac2{\sqrt{ab}}\frac t{\sqrt{1+t^2}}\right|_0^{\sqrt\frac ba}=$$

$$=\frac2{\sqrt{ab}}\left(\sqrt\frac ba\frac1{\sqrt{1+\frac ba}}\right)=\frac2{\sqrt a\sqrt{a+b}}$$

Finally, just multiply by two.