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

Question

Arun · Answer

Dear student I = cos -4/3 x + sin -4/3 x Divide both numerator and denominator by cos4 x I = sec4 x/ tan 8/3 x I = (1 + tan&sup2; x) sec&sup2; x/ tan 8/3 x Put tan x = t then Sec&sup2; x dx = dt I = ( 1+ t&sup2;) * dt/t 8/3 Now you can solve this. In case of any difficulty please feel free to ask Arun (askIITians forum expert)

Rishi Sharma · Answer

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

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

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

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