0
$\begingroup$

I have already solved the following integrals $\sqrt {\tan \left( x\right) }$ and $\sqrt[4] {\tan \left( x\right) }$ (the last one with some help) so I want to know if it's possible to have a solution for the general case.

$\endgroup$
1
  • $\begingroup$ I have not much idea about it, but I think it's not possible $\endgroup$
    – ashi
    Apr 12 '16 at 4:05
0
$\begingroup$

Let us consider $$I_n=\int \sqrt[2 n]{\tan (x)}\,dx$$ Change variable $$\sqrt[2 n]{\tan (x)}=u\implies x=\tan ^{-1}\left(u^{2 n}\right)\implies dx=\frac{2 n u^{2 n-1}}{u^{4 n}+1}\,du$$ All of this makes $$I_n=2n\int\frac{ u^{2 n}}{u^{4 n}+1}\,du$$ For a specific value of $n$,we could consider that $$u^{4n}+1=\prod_{i=1}^{4n}(u-r_i)$$ and use partial fraction decompositions to get something like $$\frac{ u^{2 n}}{u^{4 n}+1}=\sum_{i=1}^{4n}\frac {a_i}{u-r_ i}$$ and integrate each term which will result in a complex logarithm. Recombining them, we should arrive to a sum of logarithms and arctangents. I do not see how to get a general expression except using the Gaussian hypergeometric function which would give $$I_n=\frac{2 n }{2 n+1}u^{2 n+1}\, _2F_1\left(1,\frac{1}{4} \left(\frac{1}{n}+2\right);\frac{1}{4} \left(\frac{1}{n}+6\right);-u^{4 n}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.