Area of a super circle I have to find the area enclosed by $x^{2n}+y^{2n}=1, n \in \mathbb{Z}$ in terms of $F(\alpha)=\int_0^{\pi}\frac{dt}{(\sin{t})^{\alpha}}$
I tried the substitution $x=r\cos^{1/n}{\theta}, y=r\sin^{1/n}{\theta}$, but the Jacobian determinant is really horribly ugly, and I don't think this works. I ended up with $\frac{1}{2^{1/n}n} \int_0^{2\pi}\sin^{1/n}({2\theta})(\tan{\theta}+\cot{\theta})d\theta$. I don't see any other reasonable substitutions.
 A: The area $A$ of the supercircle is 
$$A=4\int_{0}^1 (1-x^{2n})^{1/2n}\,dx \tag 1$$
Enforcing the substitution $x^{2n}\to x$ in $(1)$ reveals
$$\begin{align}
A&=\frac2n \int_0^1 (1-x)^{1/2n} x^{1/2n - 1}\,dx\\\\
&=\frac2n B\left(\frac{1}{2n},1+\frac{1}{2n}\right) \tag 2\\\\
&=\frac2n \frac{\Gamma\left(1+\frac{1}{2n}\right)\Gamma\left(\frac{1}{2n}\right)}{\Gamma\left(1+\frac{1}{n}\right)} \tag 3\\\\
&=\frac2n \frac{1}{2n}\frac{\Gamma^2\left(\frac{1}{2n}\right)}{\frac1n \Gamma\left(\frac1n\right)} \tag 4\\\\
&= \frac{\Gamma^2\left(\frac{1}{2n}\right)}{n \Gamma\left(\frac1n\right)} \tag 5\\\\
&=\frac{2^{1-1/n}\sqrt{\pi}\,\Gamma\left(\frac{1}{2n}\right)}{n\Gamma\left(\frac12+\frac{1}{2n}\right)}\tag 6
\end{align}$$
In arriving at $(1)$, we used the standard integral representation of the Beta function, $B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \,dt$.
In going from $(2)$ to $(3)$ we used the relationship between the Beta and Gamma functions, $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$.
In going from $(3)$ to $(4)$, we used the functional equation for the Gamma function, $\Gamma(1+x)=x\Gamma(x)$.
In going from $(4)$ to $(5)$, we simplified the terms in $(4)$.
In going from $(5)$ to $(6)$, we used the duplication formula, $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi}\,\Gamma(2z)$
A: Here is another idea.
You can parametrize the curve $x^{2n}+y^{2n}=1$ as follows (if $n\neq 0$):
$$
\begin{cases}
x(t)=(\cos t)^\frac{1}{n}\\
y(t)=(\sin t)^\frac{1}{n}
\end{cases}
\quad t\in [0,2\pi]
$$
And then use Green's theorem to compute the area inside the curve:
$$
A=\oint_Cx\,dy = \int_0^{2\pi}x(t)y'(t)\,dt=\frac{1}{n}\int_0^{2\pi}
\frac{(\cos(t))^{1+\frac{1}{n}}}
     {(\sin(t))^{1-\frac{1}{n}}}
\,dt
$$
This being said, maybe there is a way to make $F(\alpha)$  appear in this expression?
