The value of the definite integral ${\int }_{\pi /24}^{5\pi /24}\frac{dx}{1+\sqrt[3]{\tan 2x}}$ is

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

$\Rightarrow I={\int }_{\pi /24}^{5\pi /24}\frac{(\cos 2x{)}^{1/3}}{(\cos 2x{)}^{1/3}+(\sin 2x{)}^{1/3}}dx ...\left(\mathrm{i}\right)$

$\Rightarrow I={\int }_{\pi /24}^{5\pi /24}\frac{{\left(\cos \left\{2\left(\frac{\pi }{4}-x\right)\right\}\right)}^{\frac{1}{3}}}{{\left(\cos \left\{2\left(\frac{\pi }{4}-x\right)\right\}\right)}^{\frac{1}{3}}+{\left(\sin \left\{2\left(\frac{\pi }{4}-x\right)\right\}\right)}^{\frac{1}{3}}}dx$  $\left\{{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right\}$

So, $I={\int }_{\pi /24}^{5\pi /24}\frac{(\sin 2x{)}^{1/3}}{(\sin 2x{)}^{1/3}+(\cos 2x{)}^{1/3}}dx ...\left(\mathrm{ii}\right)$

Adding the equation (i)+(ii), we get

Hence, $2I={\int }_{\pi /24}^{5\pi /24}dx$ 

$\Rightarrow 2I={\left[x\right]}_{\frac{\mathrm{\pi }}{24}}^{\frac{5\mathrm{\pi }}{24}}$

$\Rightarrow 2I=\left[\frac{5\mathrm{\pi }}{24}-\frac{\mathrm{\pi }}{24}\right]$

$\Rightarrow 2I=\frac{4\pi }{24}\Rightarrow I=\frac{\pi }{12}$