Official solution
\[\int {{{\sec }^{{2 \over 3}}}} x\cos e{c^{{4 \over 3}}}xdx\]
= \[\int {{{{{\sec }^{{2 \over 3}}}x} \over {\cos e{c^{{2 \over 3}}}x}}\cos e{c^2}xdx} \]
= \[\int {{1 \over {{{\cot }^{{2 \over 3}}}x}}\cos e{c^2}xdx} \]
Let cot x = t3
\[ \Rightarrow \] - cosec2x dx = 3t2dt
= \[ - 3\int {{{{t^2}dt} \over {{t^2}}}} \]
= -3t + C
= -3\[{{{\cot }^{{1 \over 3}}}x}\] + C
= -3\[{{{\tan }^{ - {1 \over 3}}}x}\] + C