Why are $\int_{0}^{\pi} d\theta\ \frac{\sin^2(\theta)}{1\pm 2 x \cos(\theta) + x^2}$ equal to the same thing?

Question

Consider the following two integrals for $x\geq 0$: $$ I_{\pm}(x) = \int_{0}^{\pi} d\theta\ \frac{\sin^2(\theta)}{1\pm 2 x \cos(\theta) + x^2} $$

These two integrals evaluate to the same thing: $$ I_{\pm}(x) = \begin{cases}\ \frac{\pi}{2} \ \ \ \ \ , \ x\in[0,1] \\ \ \frac{\pi}{2x^2}\ \ \ , \ x \in[1,\infty) \end{cases} $$

Do the integrals $I_{\pm}$ possess a symmetry where the fact that $I_{+} = I_{-}$ is immediately obvious? (taking $x \to -x$ or $\theta \to -\theta$ doesn't work and I am out of ideas)

Answer 1

Yes: replace $\theta$ by $\pi-\theta$.