Solve the following integration $$\int \sqrt{cot\theta} d\theta$$
I tried to set $t=\sqrt{cot\theta},t^2=cot\theta$ and substitute into the original integration and get$$-\int\frac{2t^2}{1+t^4}dt$$, but then what can I do?

Consider that $$t^4+1 = (t^4+2t^2+1)-2t^2 = (t^2+1)^2-(\sqrt{2}\,t)^2 = (t^2+\sqrt{2}\,t+1)(t^2-\sqrt{2}\, t+1)$$
hence:
$$\frac{2t^2}{t^4+1}=\frac{t}{\sqrt{2}\, t^2 -2t+\sqrt{2}}-\frac{t}{\sqrt{2}\, t^2 +2t+\sqrt{2}}\tag{1}$$
and the integral of both terms in the RHS is given by:
$$\frac{1}{2\sqrt{2}}\left(\log(\sqrt{2}\, t^2 \pm 2t+\sqrt{2})-2\arctan(1\pm\sqrt{2}\,t)\right).\tag{2}$$
