Proving $\int_0^1(1-x^n)^{\frac{1}{m}}dx=\int_0^1(1-x^m)^{\frac{1}{n}}dx$ Without Using Beta Function I'm interested in the following identity:
$$
\int_0^1(1-x^n)^{\frac{1}{m}}dx=\int_0^1(1-x^m)^{\frac{1}{n}}dx \ \ \forall n,m \in \mathbb{R}^+
$$
I already know for a fact that this is true, as I have already managed to prove it using the Beta Function, denoted $B(x,y)$ and the following three identities (found on the Wikipedia page for the beta function):
$$
B(x,y)=n\int_0^1 t^{nx-1}(1-t^n)^{y-1}dt \ \ \textrm{Re}(x) > 0, \textrm{Re}(y) > 0, n >0 \\
B(x,y)=B(y,x) \\
B(x+1,y)=B(x,y) \cdot \frac{x}{x+y}
$$
I'm interested in if there are more elementary solutions to the problem. I've tried playing around with substitutions but so far I haven't been able to prove they are equal using this method. I'm not overly familiar with the Beta Function so I'm looking for more simple solutions so I can have a better intuition as to why the identity is actually true. With that being said, I'd be happy for any solution you can think of (besides one involving the beta function) because I'm sure someone on the site will find it useful.
 A: Both sides are equal to the area of $\{(x,y)|x\ge0,y\ge0,x^m+y^n\le1)\}$ by using the geometrical meaning of integral.
A: With the substitution $\small\left\lbrace\begin{aligned}y&=x^{m}\\ \mathrm{d}x&=\frac{1}{m}y^{\frac{1}{m}-1}\,\mathrm{d}y\end{aligned}\right. $, we get : \begin{aligned} \int_{0}^{1}{\left(1-x^{m}\right)^{\frac{1}{n}}\,\mathrm{d}x}&=\frac{1}{m}\int_{0}^{1}{y^{\frac{1}{m}-1}\left(1-y\right)^{\frac{1}{n}}\,\mathrm{d}y}\\ &=\left[y^{\frac{1}{m}}\left(1-y\right)^{\frac{1}{n}}\right]_{0}^{1}+\frac{1}{n}\int_{0}^{1}{y^{\frac{1}{m}}\left(1-y\right)^{\frac{1}{n}-1}\,\mathrm{d}y}\\ &=\frac{1}{n}\int_{0}^{1}{y^{\frac{1}{m}}\left(1-y\right)^{\frac{1}{n}-1}\,\mathrm{d}y} \end{aligned}
Using the substitution $\small\left\lbrace\begin{aligned}u&=1-y\\ \mathrm{d}y&=-\,\mathrm{d}u\end{aligned}\right. $, we get : $$ \int_{0}^{1}{\left(1-x^{m}\right)^{\frac{1}{n}}\,\mathrm{d}x}=\frac{1}{n}\int_{0}^{1}{u^{\frac{1}{n}-1}\left(1-u\right)^{\frac{1}{m}}\,\mathrm{d}u} $$
Finally, using the substitution $\small\left\lbrace\begin{aligned}u&=t^{n}\\ \mathrm{d}t&=\frac{1}{n}u^{\frac{1}{n}-1}\,\mathrm{d}u\end{aligned}\right. $, we get : $$ \int_{0}^{1}{\left(1-x^{m}\right)^{\frac{1}{n}}\,\mathrm{d}x}=\int_{0}^{1}{\left(1-t^{n}\right)^{\frac{1}{m}}\,\mathrm{d}t} $$
