TITLE: How do we find closed form of $\int_{0}^{1}{\sqrt{x^n\over 1-x^m}}\mathrm dx=F(n,m)?$
QUESTION [7 upvotes]: Consider this integral
Motivated by this question

$$\int_{0}^{1}{\sqrt{x^n\over 1-x^m}}\mathrm dx=F(n,m)\tag1$$
  Where $n\ge -1$ and $m\ge 1$

We note the following values for $F(n,m):$
$$F(-1,1)=\pi$$
$$F(1,1)={\pi\over 2}$$
$$F(1,3)={\pi\over 3}$$
Where $k $ is an integer, what are other values of $ n$ and $m$ that will give ${\pi\over k}$?
I guess to answer this question we have to find the closed form for $(1)$

REPLY [8 votes]: Hint. This is the Euler beta function in disguise, by the change of variable 
$$u=x^m, \quad x=u^{1/m}, \quad dx=\frac1m\cdot u^{1/m-1}du,
$$ one gets, for $n\ge-1,\, m>0,$

$$
\begin{align}
\int_{0}^{1}{\sqrt{x^n\over 1-x^m}}\:\mathrm dx&=\int_{0}^{1}(1-u)^{-\large\frac12}u^{\large\frac{n+2}{2m}-1}\:du=\frac{\sqrt{\pi }\:\Gamma \left(\frac{n+2}{2m}\right)}{m\: \Gamma \left(\frac{m+n+2}{2 m}\right)},\quad 
\end{align}
$$ 

then by properties of the gamma function one may simplify the preceding expression.