Solve the following integration

Question

$$\int \sqrt{cot\theta} d\theta$$ 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?

Answer 1

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: $$\frac{1}{2\sqrt{2}}\left(\log(\sqrt{2}\, t^2 \pm 2t+\sqrt{2})-2\arctan(1\pm\sqrt{2}\,t)\right).\tag{2}$$

Answer 2

Consider the integral \begin{align} I = \int \sqrt{\cot(\theta)} \, d\theta. \end{align} Make the substitution $t = \sqrt{\cot(\theta)}$ to obtain the integral \begin{align} I = -2 \int \frac{t^{2} \, dt}{1+t^{4}}. \end{align} Now it can be seen that \begin{align} \frac{t^{2}}{1+t^{4}} &= \frac{t^{2}}{(1+i t^{2})(1- i t^{2})} \\ &= \frac{1}{2i} \left( \frac{1}{1-i t^{2}} - \frac{1}{1+ i t^{2}} \right) \end{align} which leads to \begin{align} I = i \int \left( \frac{1}{1-i t^{2}} - \frac{1}{1+ i t^{2}} \right) \, dt. \end{align} Since \begin{align} \int \frac{dt}{1+a t^{2}} = \frac{1}{\sqrt{a}} \, \tan^{-1}(\sqrt{a} t) \end{align} then the integral $I$ becomes \begin{align} I = i \left[ e^{\pi i/4} \tan^{-1}(e^{\pi i/4} t) - e^{- \pi i/4} \tan^{-1}(e^{- \pi i/4} t) \right]. \end{align} By the $\tan^{-1}(x)$ addition and eliminating the complex components the result becomes \begin{align} I = \frac{1}{\sqrt{2}} \left[ \tan^{-1}\left( \frac{\sqrt{2} \, t}{1- t^{2}} \right) - \tanh^{-1}\left( \frac{\sqrt{2} \, t}{1+ t^{2}} \right) \right]. \end{align} This leads to the result \begin{align} \int \sqrt{\cot(\theta)} \, d\theta = \frac{1}{\sqrt{2}} \left[ \tan^{-1}\left( \frac{\sqrt{2 \cot(\theta)}}{1-\cot(\theta)} \right) - \tanh^{-1}\left( \frac{\sqrt{2 \cot(\theta)}}{1+\cot(\theta)} \right) \right]. \end{align}