Is it possible to $\int \sqrt{\cot x}$ by hand $$\int \sqrt{\cot x}{dx}$$
$$\int \sqrt{\frac{\cos x}{\sin x}}{dx} $$
Using half angle formula 
$$\int \sqrt{\frac{1-\tan^2 \frac{x}{2}}{2\tan \frac{x}{2}}}{dx}$$
But I am not getting any lead from here .I think it is not possible to integrate $\sqrt{\cot x}$ by hand .
I calculated the result  with the help of integral calculator 
$$\dfrac{\ln\left(\left|\tan\left(x\right)+\sqrt{2}\sqrt{\tan\left(x\right)}+1\right|\right)-\ln\left(\left|\tan\left(x\right)-\sqrt{2}\sqrt{\tan\left(x\right)}+1\right|\right)+2\arctan\left(\frac{2\sqrt{\tan\left(x\right)}+\sqrt{2}}{\sqrt{2}}\right)+2\arctan\left(\frac{2\sqrt{\tan\left(x\right)}-\sqrt{2}}{\sqrt{2}}\right)}{2^\frac{3}{2}}$$
 A: HINT: Substitute $\cot x = u^2$. You will end up with an ugly partial fractions integration to do. 
A: Let $$I = \int \sqrt{\cot x}dx = \frac{1}{2}\int 2\sqrt{\cot x}dx = \frac{1}{2}\int \left[\left(\sqrt{\cot x}+\sqrt{\tan x}\right)+\left(\sqrt{\cot x}-\sqrt{\tan x}\right)\right]dx$$
Now  Let $$J = \int \left(\sqrt{\cot x}+\sqrt{\tan x}\right)dx = \sqrt{2}\int\frac{\cos x+\sin x}{\sqrt{\sin 2x}}dx = \sqrt{2}\int\frac{(\sin  x-\cos x)'}{\sqrt{1-(\sin x-\cos x)^2}}dx$$
So we get $$J = \sqrt{2}\sin^{-1}(\sin x-\cos x)+\mathcal{C_{1}}$$
Similarly Let $$K = \int \left(\sqrt{\cot x}-\sqrt{\tan x}\right)dx = \sqrt{2}\int\frac{\cos x-\sin x}{\sqrt{\sin 2x}}dx = \sqrt{2}\int\frac{(\sin x+\cos x)'}{\sqrt{(\sin x+\cos x)^2-1}}dx$$
So we get $$K = \sqrt{2}\ln \left|(\sin x+\cos x)+\sqrt{\sin 2x}\right|+\mathcal{C_{2}}$$
So we get $$I = \int \sqrt{\cot x}dx = \frac{1}{\sqrt{2}}\sin^{-1}(\sin x-\cos x)+\frac{1}{\sqrt{2}}\ln \left|(\sin x+\cos x)+\sqrt{\sin 2x}\right|+\mathcal{C}$$
$$
A: Using Ted Shifrin Hint
$$I = \int \sqrt{\cot x}dx$$ Put $\cot x= t^2\;,$ Then $$\csc^2 xdx = -2tdt\Rightarrow dx = -\frac{2t}{1+t^4}dt$$
So $$I = -\int\frac{2t^2}{1+t^4}dt = -\int \frac{\left[(t^2+1)+(t^2-1)\right]}{t^4+1}dt$$
Now Let $$J = \int \frac{t^2+1}{t^4+1}dt = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt = \int\frac{\left(t-\frac{1}{t}\right)'}{\left(t-\frac{1}{t}\right)^2+\left(\sqrt{2}\right)^2}dt$$
So $$J = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{t}\right)=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\cot x-1}{\sqrt{\cot x}}\right)+\mathcal{C_{1}}$$
Let $$K = \int \frac{t^2-1}{t^4+1}dt = \int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt = \int\frac{\left(t+\frac{1}{t}\right)'}{\left(t+\frac{1}{t}\right)^2-\left(\sqrt{2}\right)^2}dt$$
So we get $$K = \frac{1}{2\sqrt{2}}\ln \left|\frac{t^2-\sqrt{2}t+1}{t^2-\sqrt{2}t-1}\right| = \frac{1}{2\sqrt{2}}\ln \left|\frac{\cot x-\sqrt{2}\sqrt{\cot x}+1}{\cot x-\sqrt{2}\sqrt{\cot x}-1}\right|+\mathcal{C_{2}}$$
So $$I = \int \sqrt{\cot x}dx = -\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\cot x-1}{\sqrt{\cot x}}\right)-\frac{1}{2\sqrt{2}}\ln \left|\frac{\cot x-\sqrt{2}\sqrt{\cot x}+1}{\cot x-\sqrt{2}\sqrt{\cot x}-1}\right|+\mathcal{C}$$
