Let $\mathrm{I}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \mathrm{~d} x$
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{\cos ^{\frac{2}{3}} x \cdot \sin ^{\frac{4}{3}} x}$
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{\frac{\sin ^{\frac{4}{3}} x}{\cos ^{\frac{4}{3}} x} \cdot \cos ^2 x}$
$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec ^2 x}{\tan ^{\frac{4}{3}} x} \mathrm{~d} x$
$\begin{aligned} & \text { Put } \tan x=\mathrm{t} \\ & \Rightarrow \sec ^2 x \mathrm{~d} x=\mathrm{dt} \\ \therefore \quad \mathrm{I} & =\int_{\frac{1}{\sqrt{3}}} \mathrm{t}^{\frac{\sqrt{3}}{3}} \frac{\mathrm{dt}}{\mathrm{x}^{-1}} \\ & =\left[-3 \mathrm{t}^{-\frac{1}{3}}\right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \\ & =-3\left[(\sqrt{3})^{\frac{-1}{3}}-\left(\frac{1}{\sqrt{3}}\right)^{\frac{-1}{3}}\right] \\ & =-3\left(3^{-\frac{1}{6}}-3^{\frac{1}{6}}\right) \\ & =3^{\frac{7}{6}}-3^{\frac{5}{6}}\end{aligned}$