Can someone integrate the square root of tangent x for me?

Question

Please work this out don't just give me an answer.

[What is the integral of sqrt(tan(x))dx ?]

Answer 1

I don't know the exact integration
of this expression
substitute x=tan^(-1)u
dx/du=1/(1+u^2)
dx= du/(1+u^2)
new integral
= (√u)/(1+u^2) du
I get stuck similarly
on the integral of √secx dx

Answer 2

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