Using the King's property, evaluate $\int_0^{\pi/2} \dfrac{dx}{1 + \sqrt[3]{\tan x}}$:

Let $I = \int_0^{\pi/2} dx/(1+\sqrt[3]{\tan x}) = \int_0^{\pi/2} \sqrt[3]{\cos x}/(\sqrt[3]{\cos x}+\sqrt[3]{\sin x})\,dx$. Apply King: $I = \int_0^{\pi/2} \sqrt[3]{\sin x}/(\sqrt[3]{\sin x}+\sqrt[3]{\cos x})\,dx$. Adding gives $2I = \int_0^{\pi/2} 1\,dx = \pi/2$, so $I = \pi/4$. Beautiful symmetry trick that works for ANY $f$, not just cube root.