How to Integrate sqrt(cot x)

\begin{array}{l} \int \sqrt{c o t x} d x = \\ B y S u b s t i t u t i o n \\ l e t \Rightarrow u = \sqrt{\cot x} u^{2} = \cot x 2 u \\ d u = - \sec^{2} x d x = - (1 + \cot^{2} x) d x = - (1 + u^{4}) d x \\ d x = \frac{- 2 u}{1 + u^{4}} d u \\ \int \frac{- 2 u^{2}}{1 + u^{4}} d u \\ - \int \frac{(u^{2} + 1) + (u^{2} - 1)}{u^{4} + 1} d u = - \int \frac{u^{2} + 1}{u^{4} + 1} d u - \int \frac{u^{2} - 1}{u^{4} + 1} d u \\ - \int \frac{1 + \frac{1}{u^{2}}}{u^{2} + \frac{1}{u^{2}}} d u - \int \frac{1 - \frac{1}{u^{2}}}{u^{2} + \frac{1}{u^{2}}} d u = - \int \frac{1 + \frac{1}{u^{2}}}{{(u - \frac{1}{u})}^{2} + 2} d u - \int \frac{1 - \frac{1}{u^{2}}}{{(u + \frac{1}{u})}^{2} - 2} d u \\ b y s u b s t i t u t i o n \\ l e t \Rightarrow g = u - \frac{1}{u} d g = (1 + \frac{1}{u^{2}}) d u \\ a n d l e t \Rightarrow v = u + \frac{1}{u} d v = (1 - \frac{1}{u^{2}}) d u \\ - \int \frac{d g}{g^{2} + 2} - \int \frac{d v}{v^{2} - 2} = \frac{- 1}{\sqrt{2}} \tan^{- 1} \frac{g}{\sqrt{2}} - \frac{1}{2 \sqrt{2}} l n | \frac{v - \sqrt{2}}{v + \sqrt{2}} | + c \\ \frac{- 1}{\sqrt{2}} \tan^{- 1} \frac{u - \frac{1}{u}}{\sqrt{2}} - \frac{1}{2 \sqrt{2}} l n | \frac{u + \frac{1}{u} - \sqrt{2}}{u + \frac{1}{u} + \sqrt{2}} | + c \\ \frac{- 1}{\sqrt{2}} \tan^{- 1} \frac{\sqrt{\cot x} - \frac{1}{\sqrt{\cot x}}}{\sqrt{2}} - \frac{1}{2 \sqrt{2}} l n | \frac{\sqrt{\cot x} + \frac{1}{\sqrt{\cot x}} - \sqrt{2}}{\sqrt{\cot x} + \frac{1}{\sqrt{\cot x}} + \sqrt{2}} | + c \end{array}