TITLE: Integral $\int\!\sqrt{\cot x}\,dx $
QUESTION [15 upvotes]: Find the integral
$$\int\!\sqrt{\cot x}\,dx $$
How can one solve this using substitution?
Can this be solved by complex methods?

REPLY [5 votes]: $$\cot x =t^2$$
$$dx=\frac{-2t}{1+t^4}{dt}$$
$$-\int\frac{2t^2}{1+t^4}{dt}$$
$$-\left[\int \frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2 +2}dt +\int \frac{1-\frac{1}{t^2}}{\left(t + \frac{1}{t}\right)^2 - 2}dt\right]$$
$$-\left(\frac{1}{\sqrt2}\tan^{-1}\left[\frac{\cot x -1}{\sqrt{2\cot x}}\right] +\frac{1}{2\sqrt2}\log\left|\frac{\cot x +1-\sqrt{2\cot x}}{\cot x +1+\sqrt{2\cot x}}\right|\right)$$