# Evaluate$\int \sqrt{\tan x}dx$

My instructor told us that the integral of square root of tangent can be evaluated. How can one really do it?

Let $$A=\int\sqrt{\tan x}\,\mathrm dx$$ $$B=\int\sqrt{\cot x}\,\mathrm dx$$
Then
$A+B=\int\left(\sqrt{\tan x}+\sqrt{\cot x}\right) dx=\sqrt{2} \int\frac{\sin x+\cos x}{\sqrt{\sin2x}}\mathrm dx$
$=\sqrt2\int\frac{(\sin x-\cos x)'}{\sqrt{1-(\sin x-\cos x)^2}}\,\mathrm dx =\sqrt2\int\frac{1}{\sqrt{1-u^2}}\,\mathrm du$
$=\sqrt2\sin^{-1}u =\sqrt2\sin^{-1}(\sin x-\cos x).$
Also
$A-B=\int\left(\sqrt{\tan x}-\sqrt{\cot x}\right)\,\mathrm dx =\sqrt2\int\frac{\sin x-\cos x}{\sqrt{\sin2x}} \,\mathrm dx$
$=-\sqrt2\int\frac{(\sin x+\cos x)'}{\sqrt{(\sin x+\cos x)^2-1}}\,\mathrm dx =-\sqrt2\int\frac{\mathrm du}{\sqrt{u^2-1}}$
$=-\sqrt2\cosh^{-1}(\sin x+\cos x)$
Thus we get
$A=\int\sqrt{\tan x} dx=\frac{(A-B)+(A+B)}2$
$= \frac{\sqrt2}2(\sin^{-1}(\sin x-\cos x)-\cosh^{-1}(\sin x+\cos x)) + C.$
Note than you also get
$B=\int\sqrt{\cot x} dx=-\frac{(A-B)-(A+B)}2$
$=- \frac{\sqrt2}2(\sin^{-1}(\sin x-\cos x)-\cosh^{-1}(\sin x+\cos x)) + C.$