माना `I = int sec^(6//5)x cosec ^(4//5) x dx`
`= int(sec^(2)x//dx)/(sin^(4//5)x) xx (sec^(4//5) x)/(sec^(4//5x)) dx`
अश और हर को `sec ^(4//5)` x से गुणा करने पर
`int (sec^(2)x dx)/(tan^(4//5)x)`
माना `tan x = t rArr sec^(2)xdx = dt`
`therefore" "I = int (dt)/(t^(4//5)) = int t^(4//5)dt = 5t^(1//5) 5(tan x)^(1//5)`