Question

How to integrate square root of tan x?

Answer 1

Question #12765

How to integrate $\int \sqrt{tanx} dx$ .

Solution.

Make substitution $y = \sqrt{tanx}$ , then our integral obviously becomes $\int \frac{2y^2}{1 + y^4} dy = \int \frac{1 + 1/y^2}{y^2 + 1/y^2} dy + \int \frac{1 - 1/y^2}{y^2 + 1/y^2} dy = 1/\sqrt{2} \tan^{-1}\left(\frac{y^2 - 1}{\sqrt{2}y}\right) + \frac{1}{2\sqrt{2}} \log \left| \frac{y^2 - \sqrt{2}y + 1}{y^2 + \sqrt{2}y + 1} \right|$ .

Substitution $y = \tan x$ leads us to the desired result.