The integral

Let, $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1 \cdot d x}{\cos ^{\frac{2}{3}} x \cdot \sin ^{\frac{4}{3}} x}$

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1 d x}{\cos ^{2} x \cdot \tan ^{\frac{4}{3}} x}=\int_{\frac{\pi}{6} \tan ^{\frac{\pi}{3}} x}^{\frac{\pi}{3}} \frac{\sec ^{2} x d x}{}$

Let $\tan x=u$

$I=\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} u^{-\frac{4}{3}} d u=\frac{3\left[u^{-\frac{1}{3}}\right]}{-1} \frac{1}{\sqrt{3}}$

$=-3\left[3^{-\frac{1}{6}}-\frac{1}{3^{\frac{-1}{6}}}\right]=-3\left(3^{\frac{-1}{6}}-3^{\frac{1}{6}}\right)$

$=3\left(3^{\frac{1}{6}}-3^{\frac{1}{6}}\right)=\left(3^{\frac{7}{6}}-3^{\frac{5}{6}}\right)$