how to integrate (tanx)^1/2

 ∫√(tan x) dx
Let tan x = t^2
=> sec^2 x dx = 2t dt
=> dx = [2t / (1 + t^4)]dt
=> Integral
= ∫ 2t^2 / (1 + t^4) dt
= ∫[(t^2 + 1) + (t^2 - 1)] / (1 + t^4) dt
= ∫(t^2 + 1) / (1 + t^4) dt + ∫(t^2 - 1) / (1 + t^4) dt
= ∫(1 + 1/t^2) / (t^2 + 1/t^2) dt + ∫(1 - 1/t^2) / (t^2 + 1/t^2) dt
= ∫(1 + 1/t^2)dt / [(t - 1/t)^2 + 2] + ∫(1 - 1/t^2)dt / [(t + 1/t)^2 -2]
Let t - 1/t = u for the first integral => (1 + 1/t^2)dt = du
and t + 1/t = v for the 2nd integral => (1 - 1/t^2)dt = dv
Integral
= ∫du/(u^2 + 2) + ∫dv/(v^2 - 2)
= (1/√2) arctan (u/√2) + (1/2√2) ln l(v -√2)/(v + √2)l + c
= (1/√2) arctan [(t^2 - 1)/t√2] + (1/2√2) ln l (t^2 + 1 - t√2) / t^2 + 1 + t√2) + c
= (1/√2) arctan [(tanx - 1)/(√2tan x)] + (1/2√2) ln l [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c

 Q . Integrate 

ANS :
 

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sir really i m very thankful to u i will prefer this method instead of using
other method .
sir i want to ask u a question how to integrate 1/1+tanxsquare not (tanx)^2 sir please if possible solve this questions thanks in advance