Integrate: $\int_0^{2\pi} \frac{d\theta}{1+\cos^2\theta}$

Question

I'm trying to integrate this here fella:

$\int_0^{2\pi} \frac{d\theta}{1+\cos^2\theta}$

from examples in Ablowitz I know that for $|A|^2>|B|^2$ and $A>0$, $\int_0^{2\pi} \frac{d\theta}{A+B\cos\theta}$ has solution $\frac{2*\pi}{\sqrt{A^2+B^2}}$

Since that is the only similar example I can find, the only approach would be to substitute A and B accordingly, but in my case I cannot substitute A=1 and B=$cos\theta$ since for $\theta=0$ or $\theta=2\pi*n$, A=B so the inequality $|A|>|B|$ is not strict.

Does this mean that the integral diverges? An answer here claims so: How to evaluate $\int_0^{2\pi} \frac{d\theta}{A+B\cos\theta}$? But I wanted to see if anyone has any other approaches to this problem.

Many thanks

Answer 1

$$\int_{0}^{2\pi}\frac{d\theta}{1+\cos^2\theta} = 4\int_{0}^{\pi/2}\frac{d\theta}{1+\cos^2\theta} = 4\int_{0}^{+\infty}\frac{dt}{2+t^2}=\color{red}{\pi\sqrt{2}}.$$ It is enough to exploit simmetry and use the substitution $\theta=\arctan t$.

The integral obviously cannot diverge: the integrand function is bounded between $\frac{1}{2}$ and $1$, hence the integral is between $\pi$ and $2\pi$.

Answer 2

Use $$\cos^2\theta = \frac{1}{2} + \frac{1}{2} \cos(2\theta)$$ to write as $$\int_0^{2\pi} \frac{d\theta}{\frac{3}{2} + \frac{1}{2} \cos(2\theta)} = \frac{1}{2} \int_0^{4\pi} \frac{dt}{\frac{3}{2} + \frac{1}{2} \cos t} = \int_0^{2\pi} \frac{dt}{\frac{3}{2} + \frac{1}{2} \cos t}$$ which is in the form that you can solve with that Ablowitz formula.

Answer 3

What did you try so far? You can use some trigonometric identities (I didn't try this approach) or you can write $cos(x) = \dfrac{e^{ix}+e^{-ix}}{2}$ and use the substitution $z=e^{ix}$, turning the integral to a complex integral of a closed curve.

Edit: You mentioned convergence\divergence (When you brought up those other integrals).It's worth mentioning that this integral clearly converge since $\dfrac{1}{1+\cos^2(x)}$ is continuous on the closed set $[0,2\pi]$.

Answer 4

\begin{align*} \int_0^{2\pi}\frac{{\rm d}x}{1+\cos^2 x}&=\int_0^{2\pi}\sum_{n=0}^{\infty}(-1)^n\cos^{2n}x{\rm d}x\\ &=\sum_{n=0}^{\infty}(-1)^n\int_0^{2\pi}\cos^{2n}{\rm d}x\\ &=2\pi\sum_{n=0}^{\infty}(-1)^n\binom{2n}{n}\left(\frac{1}{2}\right)^{2n}\\ &=\frac{2\pi}{\sqrt{1+4x^2}}\bigg|_{x=\frac{1}{2}}\\ &=\sqrt{2}\pi. \end{align*}