Official solution
\[\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}x\,{{{\mathop{\rm cosec}\nolimits} }^{4/3}}x\,dx} \]
\[ = \int {{{{{\sec }^2}x} \over {{{\tan }^{4/3}}x}}dx} \]
Let tan x = t, sec2 x dx = dt
\[ = \int {{{dt} \over {{t^{4/3}}}}} \]
\[I = - 3\left( {{t^{ - 1/3}}} \right)\]
\[ \Rightarrow \left( { - 3{{\left( {\tan x} \right)}^{{{ - 1} \over 3}}}} \right)_{\pi /6}^{\pi /3}\]
\[ \Rightarrow 3\left( {{3^{{1 \over 3}}} - {1 \over {{3^{{1 \over 6}}}}}} \right)\]
= 37/6 - 35/6