Struggling with the integrability of $\int_{\frac{\pi}{2}}^{\pi}(\tan(x))^{\frac{1}{3}}\text{d}x$ I know quite a lot tools to determine the integrability of functions, but in this case I really don't know where to start:
$$\int_{\frac{\pi}{2}}^{\pi}(\tan(x))^{\frac{1}{3}}\text{d}x$$

On the integration range $\tan x$ is negative, hence I assume that $x^{1/3}$ is defined as $-(-x)^{1/3}$ for negative $x$. Given that, by setting $x=\pi-z$, then $z=\arctan t$, we have:
$$ I = -\int_{0}^{\pi/2}\tan(z)^{1/3}\,dz = -\int_{0}^{+\infty}\frac{t^{1/3}}{1+t^2}\,dt$$
and the last integral is convergent, since:
$$0\leq \int_{0}^{1}\frac{t^{1/3}}{1+t^2}\,dt \leq \int_{0}^{1}t^{1/3}\,dt = \frac{3}{4}$$
and:
$$0\leq \int_{1}^{+\infty}\frac{t^{1/3}}{1+t^2}\,dt \leq \int_{1}^{+\infty}t^{-5/3}\,dt = \frac{3}{2}.$$
We may also compute it exactly:
$$ \int_{0}^{+\infty}\frac{t^{1/3}}{1+t^2}\,dt=\int_{0}^{+\infty}\frac{3u^3}{1+u^6}\,du = \frac{\pi}{\sqrt{3}}.$$
