integral (sec^4/3 x . cosec^8/3 x) – please find the answer

Dear student
I = cos^-4/3 x + sin^-4/3x
Divide both numerator and denominator by cos4 x
I = sec4 x/ tan^8/3x
I = (1 + tan² x) sec² x/ tan^8/3x
Put tan x = t then
Sec² x dx = dt
I = ( 1+ t²) * dt/t^8/3
Now you can solve this.
In case of any difficulty please feel free to ask
Arun (askIITians forum expert)

Dear Student,
Please find below the solution to your problem.

I = cos-4/3 x + sin-4/3x
Divide both numerator and denominator by
cos4 x I = sec4 x/ tan8/3x I
= (1 + tan² x) sec² x/ tan8/3x
Put tan x = t then
Sec² x dx = dt I = ( 1+ t²) * dt/t8/3

Thanks and Regards