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?
$\begingroup$
$\endgroup$
3
asked Feb 22, 2016 at 15:57
-
1$\begingroup$ Not in terms of elementary functions. $\endgroup$– user65203Feb 22, 2016 at 16:00
-
3$\begingroup$ 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/… $\endgroup$– colormegoneFeb 22, 2016 at 16:06
-
1$\begingroup$ the solution containes the elliptic function $\endgroup$– Dr. Sonnhard GraubnerFeb 22, 2016 at 16:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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.
