If $$f(t) = \int_0^{\pi} \frac{2x \, dx}{1 - \cos^2 t \sin^2 x}$$, $$0 < t < \pi$$, then the value of $$\int_0^{\frac{\pi}{2}} \frac{\pi^2 dt}{f(t)}$$ equals _________

Using the property $$\int_0^a g(x) dx = \int_0^a g(a-x) dx$$:

$$f(t) = \int_0^\pi \frac{2(\pi-x)}{1-\cos^2 t \sin^2 x} dx$$.

 Adding the two forms: $$2f(t) = \int_0^\pi \frac{2\pi}{1-\cos^2 t \sin^2 x} dx \implies f(t) = \pi \int_0^\pi \frac{dx}{1-\cos^2 t \sin^2 x}$$.

Using symmetry: $$f(t) = 2\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{\sec^2 x - \cos^2 t \tan^2 x} = 2\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{1 + \tan^2 x \sin^2 t}$$.

let $$u = \tan x \sin t$$: $$f(t) = \frac{2\pi}{\sin t} [\tan^{-1}(\tan x \sin t)]_0^{\pi/2} = \frac{2\pi}{\sin t} \cdot \frac{\pi}{2} = \frac{\pi^2}{\sin t}$$.

 $$\int_0^{\pi/2} \frac{\pi^2}{\pi^2/\sin t} dt = \int_0^{\pi/2} \sin t \, dt = [-\cos t]_0^{\pi/2} = 0 - (-1) = \mathbf{1}$$