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integral (sec^4/3 x . cosec^8/3 x) – please find the answer
Dear studentI = cos-4/3 x + sin-4/3xDivide both numerator and denominator by cos4 xI = sec4 x/ tan8/3xI = (1 + tan² x) sec² x/ tan8/3xPut tan x = t then Sec² x dx = dtI = ( 1+ t²) * dt/t8/3Now you can solve this.In case of any difficulty please feel free to askArun (askIITians forum expert)
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