Closed form for $\int \left(1-x^{2/3}\right)^{3/2}\:dx$ Find a closed-form solution to
\begin{align}\int_0^1 \left(1-x^{2/3}\right)^{3/2}\:dx\tag{1},\end{align}
or even more generally, is there a methodology to solving integrals of the type
\begin{align}
\int_0^1\left(1-x^r\right)^{1/r}\:dx,\:\:r\in\left\{s\in\mathbb{Q}:0<s<1\right\}.\tag{2}
\end{align}
This integral arose when trying to calculate the area of a Superellipse.
Thank you for your time,
 A: Since $x\leq1$
 , we can try $x=\sin^{3}\left(y\right)$
  and then $dx=3\sin^{2}\left(y\right)\cos\left(y\right)dy$
 , whence$$\intop\left(1-x^{2/3}\right)^{3/2}dx=3\intop\cos^{4}\left(y\right)\sin^{2}\left(y\right)dy.$$
Maybe this last integral is easier to work.
A: Through the Euler's Beta function it is not difficult to show (like in this closed question) that for any $p,q>0$ we have:
$$\int_{0}^{1}(1-x^p)^{\frac{1}{q}}\,dx=\int_{0}^{1}(1-x^q)^{\frac{1}{p}}\,dx =\frac{\Gamma\left(1+\frac{1}{p}\right)\Gamma\left(1+\frac{1}{q}\right)}{\Gamma\left(1+\frac{1}{p}+\frac{1}{q}\right)} $$
hence by choosing $p=q=\frac{2}{3}$ we have:
$$ I = \frac{\Gamma\left(\frac{5}{2}\right)^2}{\Gamma(4)} = \color{red}{\frac{3\pi}{32}}$$
and if $p=q=r$
$$ I = \frac{\Gamma\left(1+\frac{1}{r}\right)^2}{\Gamma\left(1+\frac{2}{r}\right)}=\frac{r}{2}\cdot\frac{\Gamma\left(\frac{1}{r}\right)^2}{\Gamma\left(\frac{2}{r}\right)}=\frac{r\sqrt{\pi}}{2^{2/r}}\cdot\frac{\Gamma\left(\frac{1}{r}\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{r}\right)} $$
by Legendre's duplication formula.
A: The (most) general answer involves the hypergeometric function $_pF_q$:
\begin{equation}
\int(1-x^r)^{1/r} dx=x\;_2F_1(-1/r,1/r,1+1/r,x^r)
\end{equation}
In the particular case of (1) the indefinite integral is given by:
\begin{equation}
\int(1-x^{2/3})^{3/2} dx=\frac{1}{16}\left(\sqrt{1-x^{2/3}}(-3x^{1/3}+14x-8x^{5/3})+3\arcsin(x^{1/3})\right)
\end{equation}
Evaluation at $0$ and $1$ gives us:
\begin{equation}
\int_{0}^{1}(1-x^{2/3})^{3/2} dx=\frac{3\pi}{32}
\end{equation}
And more generally:
\begin{equation}
\int_{0}^{1}(1-x^r)^{1/r} dx=\frac{\Gamma(1+\frac{1}{r})^2}{\Gamma(\frac{2+r}{r})}
\end{equation}
Where $\Gamma(x)=(x-1)!$ is the extension of the factorial function.
More information about the hypergeometric and gamma function can be found on: http://mathworld.wolfram.com/HypergeometricFunction.html
http://mathworld.wolfram.com/GammaFunction.html
A: This integral may belong within a special category. See a general case in  my post here.
A: $\int_0^1(1-x^r)^\frac{1}{r}~dx$
$=\int_0^1(1-x)^\frac{1}{r}~d\left(x^\frac{1}{r}\right)$
$=\dfrac{1}{r}\int_0^1x^{\frac{1}{r}-1}(1-x)^\frac{1}{r}~dx$
$=\dfrac{1}{r}B\left(\dfrac{1}{r},\dfrac{1}{r}+1\right)$
