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hi.
I = ∫sec2x√tanx .dx
let tan x = t^2
then, sec^2x dx = 2t dt
(1+ t^4) dx = 2t dt
dx = 2t / (1+ t^4) dt
& cos 2x = 2 cos^2 x - 1
= 2 ( 1/ √ 1+ t^4 ) ^2 -1
= 2 / 1+ t^4 -1
= 1- t^4 / 1+ t^4
so , sec 2x = 1+ t^4 / 1- t^4
put these values in I
I = ∫ { 1+ t^4 / 1- t^4 } . t . {2t / (1+ t^4) } dt
= ∫ 2 t^2 / 1- t^4 dt
which can easily be evaluated by partial fraction
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