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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

Grade:12

2 Answers

Arun
25750 Points
6 years ago
Dear student
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
Now you can solve this.
In case of any difficulty please feel free to ask
Arun (askIITians forum expert)
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
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

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