Evaluate: `intsec^(4//3)x" cosec"^(8//3)xdx`

Explanation

We have,
`I=intsec^(4//3)x" cosec "^(8//3)xdx=int(1/(cos^(4//3)xsin^(8//3)x)) dx = int(cos^(-4//3)xsin^(-8/3))xdx`
Divide `N^(')` and `D^(')` by `cos^(4)x`
`=int(sec^(4)x)/(tan^(8//3)x)dx= int(1+tan^(2)x)/(tan^(8//3)x)dx = int(1+tan^(2)x)/(tan^(8//3)x)sec^(2)xdx=int(1+tan^(2)x)/(tan^(8//3)x)d(tanx) = int(1+t^(2))/(t^(8//3))dt` where t=tanx
`=int(t^(-8//3)+t^(-2//3))dt = -3/5 t^(-5//3) + 3t^(1//3) + C = -3/5 tan^(-5//3)x+3tan^(1//3)+C`