How to integrate $$\int_{0}^{\infty} \frac{1}{x^2 (\tan^2 x + \cot^2 x)} \,dx$$
Let $z = e^{ix}$. We write the Fourier series $$ \frac1{\tan^2x+\cot^2x} = \frac1{\left(\frac{z-z^{-1}}{i(z+z^{-1})}\right)^2 + \left(\frac{i(z+z^{-1})}{z-z^{-1}}\right)^2} $$
$$ = -\frac1{\left(\frac{1-z^2}{1+z^2}\right)^2 + \left(\frac{1+z^2}{1-z^2}\right)^2} $$
$$ = -\frac{(1+z^2)^2(1-z^2)^2}{(1-z^2)^4+(1+z^2)^4} $$
$$ = -\frac{(1-z^4)^2}{2(1+6z^4+z^8)} $$
$$ = -\frac12 + \frac{4z^4}{1+6z^4+z^8} $$
$$ = -\frac12 + \frac1{\sqrt2[1+(3-2\sqrt2)z^4]} - \frac1{\sqrt2[1+(3+2\sqrt2)z^4]} $$
$$ = -\frac12 + \frac1{\sqrt2[1+(3-2\sqrt2)z^4]} - \frac{(3-2\sqrt2)z^{-4}}{\sqrt2[1+(3-2\sqrt2)z^{-4}]} $$
Apply the Lobachevsky integral formula
\begin{align} &\int_{0}^{\infty} \frac{1}{x^2 (\tan^2 x + \cot^2 x)} \,dx\\ =& \int_0^\infty\frac{\sin^2x}{x^2}\cdot\frac{\sec^2x}{\tan^4x+1}dx=\int_0^{\pi/2}\frac{\sec^2x}{\tan^4x+1}dx=\frac{\pi}{2\sqrt2} \end{align}