Recently, I’ve started doing calculus and have ventured my way into the tedious integration of $\tan(x)$ to some specific root.
I’ve started wondering whether there is any generalisation of the integral of the nth root of $\tan(x)$, i.e.
$$\int\sqrt[n]{\tan x}\> dx $$ But I have no idea where to start myself. Any help will be much appreciated.
Apply the partial decomposition
\begin{align} \frac{nt^n}{t^{2n}+1}= \sum_{k=1}^{n} \frac{(-1)^{k+1} t\sin \theta_k}{t^2-2t\cos \theta_k +1}, \>\>\>\>\> \theta_k=\frac{(2k-1)\pi}{2n} \end{align} to integrate \begin{align} & \int\sqrt[n]{\tan x}\> dx=\int \frac{nt^n}{t^{2n}+1}dx\\ =& \sum_{k=0}^{n} {(-1)^{k+1}}\left( \sin \theta_k \ln\sqrt{t^2 - 2t\cos \theta_k + 1} +\cos \theta_k \tan^{-1}\frac{t- \cos \theta_k}{\sin \theta_k} \right) \end{align} where $t= \sqrt[n]{\tan x} $.