Revisit the properties of integrals

$\begin{aligned} I & = \int_0^\pi \frac x{\pi^2 - \cos^2 x} dx & \small \color{#3D99F6} \text{Using identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\pi \left(\frac x{\pi^2 - \cos^2 x}+\frac {\pi-x}{\pi^2 - \cos^2 x} \right) dx & \small \color{#3D99F6} \text{Note that }\cos (\pi - \theta) = - \cos \theta \\ & = \frac 12 \int_0^\pi \frac \pi{\pi^2 - \cos^2 x} dx & \small \color{#3D99F6} \text{The integral is symetrical at }\frac \pi 2 \\ & = \pi \int_0^\frac \pi 2 \frac 1{\pi^2 - \cos^2 x} dx & \small \color{#3D99F6} \text{Multiply up and down by }\sec^2 x \\ & = \pi \int_0^\frac \pi 2 \frac {\sec^2 x}{\pi^2\sec^2 x - 1} dx & \small \color{#3D99F6} \text{Let }t = \tan x \implies dt = \sec^ x \ dx \\ & = \pi \int_0^\infty \frac 1{\pi^2t^2 + \pi^2 - 1} dt \\ & = \frac \pi {\pi^2-1} \int_0^\infty \frac 1{\frac {\pi^2}{\pi^2 - 1}t^2 + 1} dt & \small \color{#3D99F6} \text{Let }u = \frac {\pi t}{\sqrt{\pi^2-1}} \implies du = \frac \pi{\sqrt{\pi^2-1}} dt \\ & = \frac 1{\sqrt{\pi^2-1}} \int_0^\infty \frac 1{u^2 + 1} du \\ & = \frac 1{\sqrt{\pi^2-1}} \tan^{-1} u \ \bigg|_0^\infty \\ & = \frac \pi{2\sqrt{\pi^2-1}} \approx 0.527 \end{aligned}$

Therefore, the answer is $\boxed{0<I<1 \text{ and } I = \dfrac \pi{2\sqrt{\pi^2-1}}}$