${\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/3} {\sec }^{2/3}{\mathrm{xcosec}}^{4/3}\mathrm{xdx}=$

$$\begin{gathered} \mathrm{I}={\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/3} {\sec }^{2/3}{\mathrm{xcosec}}^{4/3}\mathrm{xdx}={\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/3} \frac{{\sec }^2\mathrm{x}}{{\sec }^{4/3}\mathrm{x}}{\mathrm{cosec}}^{4/3}\mathrm{xdx} \\ ={\int }_{\mathrm{\pi }/6}^{\mathrm{\pi }/3} \frac{{\sec }^2\mathrm{x}}{{\tan }^{4/3}\mathrm{x}}\mathrm{dx} \\ \mathrm{Put} \tan \mathrm{x}=\mathrm{t}. \text{ Then }{\sec }^2\mathrm{xdx}=\mathrm{dt}.\mathrm{x}=\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{3}\Rightarrow \mathrm{t}=\frac{1}{\sqrt{3}},\sqrt{3} \\ \therefore \mathrm{I}={\int }_{1/\sqrt{3}}^{\sqrt{3}} \frac{\mathrm{dt}}{{\mathrm{t}}^{/3}}={\left[\frac{{\mathrm{t}}^{-1/3}}{-1/3}\right]}_{1/\sqrt{3}}^{\sqrt{3}}={\left[-\frac{3}{{\mathrm{t}}^{1/3}}\right]}_{1/\sqrt{3}}^{\sqrt{3}} \\ =-\frac{3}{3^{1/6}}+\frac{3}{3^{-1/6}}=3^{7/6}-3^{5/6}. \end{gathered}$$