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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
25763 Points
3 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
10 months 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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