integral tan ^(1/2)x?

Question

integrate square root of tanx.

Answer 1

Use substitution:
If u^2 = tan x
then u = sqrt(tan x)
2u(du) = [(sec x)^2](dx)
2u(du) = (u^2 + 1)(dx)
[2u / (u^2 + 1)](du) = dx
Integrate [sqrt(tan x)](dx)
= Integrate u[2u / (u^2 + 1)](du)
= Integrate [2(u^2) / (u^2 + 1)](du)
= Integrate {2 – [2 / (u^2 + 1)]}(du)
= 2u – 2(arctan u) + C
= 2[sqrt(tan x)] – 2{arctan [sqrt(tan x)]} + C

Answer 2

The previous answer starts well but then there is a mistake because sec^2= 1+u^4 and not 1+u^2. Anyway one is reduced to partial fractions.

Answer 3

So you want to solve

Integral (sqrt(tan(x))dx

This is NOT an easy integral to solve. Are you sure you didn't mean anything else?