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Grade: 12
        
integral (sec^4/3 x . cosec^8/3 x) – please find the answer
one year ago

Answers : (1)

Arun
23008 Points
							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)
one year ago
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