# Integral of $\int \sin x\sqrt{\tan x}dx$

How to integrate

$$\int \sin x\sqrt{\tan x}dx$$

I put $\cos x=t^2$ but ended up with $(1-t^4)^{1/4}$. Can this integral be solved in terms of elementary functions?

• Not in terms of elementary functions.
– user65203
Feb 22 '16 at 16:00
• WA finding an anti-derivative expressed as a hypergeometric function is not a good sign... It's also not comforting when we already see this for $\ \int \ \sqrt{\tan x} \ \ dx \$ : math.stackexchange.com/questions/828640/… Feb 22 '16 at 16:06
• the solution containes the elliptic function Feb 22 '16 at 16:26

I put $\cos x=t^2,$ but ended up with $\Big(1-t^4\Big)^{1/4}$.

We know that $~\displaystyle\int_0^1\sqrt[m]{1-t^n}~dt~=~\int_0^1\sqrt[n]{1-t^m}~dt~=~{a+b\choose a}^{-1}~=~{a+b\choose b}^{-1},~$ where

$a=\dfrac1m~$ and $~b=\dfrac1n~$ $($or viceversa, it doesn't matter, since the entire expression is completely

symmetrical$),~$ is the beta function in disguise, as can be shown through a simple substitution.

Using the fact that $~\Gamma\bigg(\dfrac12\bigg)~=~\sqrt\pi~,~$ for $m=n=4$ we have $~\displaystyle\int_0^1\sqrt[4]{1-t^4}~dt~=~\dfrac{\Gamma^2\bigg(\dfrac14\bigg)}{8\sqrt\pi},~$

see $\Gamma$ function for more information. However, if you are absolutely certain that the indefinite

integral is what you're really after, then your only two solutions are to either express it in terms

of incomplete beta functions, or to expand the integrand into its own binomial series, and then

reverse the order of summation and integration, so as to obtain a hypergeometric function.