King's rule. The sum of the limits is $\frac{6\pi }{24}=\frac{\pi }{4}$. Substituting $x\to \frac{\pi }{4}-x$, since $\tan \left(\frac{\pi }{2}-2x\right)=\cot 2x$:

$I={\int }_{\pi /24}^{5\pi /24}\frac{dx}{1+\sqrt[3]{\cot 2x}}={\int }_{\pi /24}^{5\pi /24}\frac{\sqrt[3]{\tan 2x}}{1+\sqrt[3]{\tan 2x}}dx$

Adding to the original integral, the integrand becomes $1$:

$2I={\int }_{\pi /24}^{5\pi /24}dx=\frac{\pi }{6}\Rightarrow I=\frac{\pi }{12}$.