Ways to show that $\int_{0}^{1}((1-x^r)^{1/r}-x)^{2k}dx=\frac{1}{2k+1}$
We present 3 different solutions.
Solution 1 - slick substitution. We prove a more general statement:
Proposition. Let $R \in (0, \infty]$ and let $\varphi : [0, R] \to [0, R]$ satisfy the following conditions:
- $\varphi$ is continuous on $[0, R]$;
- $\varphi(0) = R$ and $\varphi(R) = 0$;
- $\varphi$ is bijective and $\varphi^{-1} = \varphi$.
Then for any integrable function $f$ on $[0, R]$, $$ \int_{0}^{R} f(|x-\varphi(x)|) \, \mathrm{d}x = \int_{0}^{R} f(x) \, \mathrm{d}x. $$
Proof. In case $\varphi$ is also continuously differentiable on $(0, R)$, by the substitution $y = \varphi(x)$, or equivalently, $x = \varphi(y)$,
$$ I := \int_{0}^{R} f( |x - \varphi(x)| ) \, \mathrm{d}x = -\int_{0}^{R} f( |\varphi(y) - y| ) \varphi'(y) \, \mathrm{d}y. $$
Summing two integrals,
\begin{align*} 2I &= \int_{0}^{R} f( |x - \varphi(x)| ) (1 - \varphi'(x)) \, \mathrm{d}x \\ &= \int_{-R}^{R} f( |u| ) \, \mathrm{d}u = 2\int_{0}^{R} f(u) \, \mathrm{d}u, \tag{$u = x - \varphi(x)$} \end{align*}
proving the claim when $\varphi$ is continuously differentiable. This proof can be easily adapted to general $\varphi$ by using Stieltjes integral. ■
Now plug $\varphi(x) = (1-x^r)^{1/r}$ with $R = 1$ and $f(x) = x^n$ for positive even integer $n$. Then
$$ \int_{0}^{1} \left( (1-x^r)^{1/r} - x \right)^n \, \mathrm{d}x = \int_{0}^{1} \left| x - (1-x^r)^{1/r} \right|^n \, \mathrm{d}x = \int_{0}^{1} x^n \, \mathrm{d}x = \frac{1}{n+1}. $$
Solution 2 - using beta function. Here is an alternative solution. Write $p = 1/r$. Then using the substitution $x = u^p$,
\begin{align*} \int_{0}^{1} \left( (1 - x^r)^{1/r} - x \right)^n \, \mathrm{d}x &= \int_{0}^{1} \left( (1 - u)^{p} - u^p \right)^{n} pu^{p-1} \, \mathrm{d}u \\ &= \sum_{k=0}^{n} (-1)^k \binom{n}{k} p \int_{0}^{1} (1-u)^{p(n-k)} u^{p(k+1)-1} \, \mathrm{d}u \\ &= \sum_{k=0}^{n} (-1)^k \binom{n}{k} p \cdot \frac{(p(n-k))!(p(k+1)-1)!}{(p(n+1))!} \end{align*}
Here, $s! = \Gamma(s+1)$. Now define $a_k = (pk)!/k!$. Then the above sum simplifies to
\begin{align*} \int_{0}^{1} \left( (1 - x^r)^{1/r} - x \right)^n \, \mathrm{d}x &= \frac{1}{(n+1)a_{n+1}} \sum_{k=0}^{n} (-1)^k a_{n-k}a_{k+1} \\ &= \frac{1}{(n+1)a_{n+1}} \left( a_0 a_{n+1} + \sum_{k=0}^{n-1} (-1)^k a_{n-k}a_{k+1} \right). \end{align*}
So it suffices to show that $\sum_{k=0}^{n-1} (-1)^k a_{n-k}a_{k+1} = 0$. But by the substitution $l = n-1-k$, we have
$$ \sum_{k=0}^{n-1} (-1)^k a_{n-k}a_{k+1} = - \sum_{l=0}^{n-1} (-1)^l a_{l+1}a_{n-l}. $$
(Here the parity of $n$ is used.) So the sum equals its negation, hence is zero as required.
Solution 3 - using multivariate calculus. Let $\mathcal{C}_r$ denote the curve defined by $x^r + y^r = 1$ in the first quadrant, oriented to the right. Then
$$ I(r) := \int_{0}^{1} \left( (1 -x^r)^{1/r} - x \right)^n \, \mathrm{d}x = \int_{\mathcal{C}_r} ( y - x )^n \, \mathrm{d}x. $$
Notice that if $0 < r < s$, then $\mathcal{C}_s$ lies above $\mathcal{C}_r$, and so, the curve $\mathcal{C}_r - \mathcal{C}_s$ bounds some region, which we denote by $\mathcal{D}$, counter-clockwise:
$\hspace{10em}$ 
Then by Green's theorem,
$$ I(r) - I(s) = \int_{\partial \mathcal{D}} ( y - x )^n \, \mathrm{d}x = - \iint_{\mathcal{D}} n (y - x)^{n-1} \, \mathrm{d}x\mathrm{d}y. $$
But since the region $\mathcal{D}$ is symmetric around $y = x$ and $n$ is even, interchanging the roles of $x$ and $y$ shows
$$ \iint_{\mathcal{D}} n (y - x)^{n-1} \, \mathrm{d}x\mathrm{d}y = \iint_{\mathcal{D}} n (x - y)^{n-1} \, \mathrm{d}x\mathrm{d}y = - \iint_{\mathcal{D}} n (y - x)^{n-1} \, \mathrm{d}x\mathrm{d}y. $$
Therefore $I(r) = I(s)$ for any $ r < s$, and in particular, letting $s \to \infty$ gives
$$ I(r) = \int _{0}^{1} (1 - x)^n \, \mathrm{d}x = \frac{1}{n+1}. $$
Here's your Beta integral $$S=\int_0^1x^{n-k}(1-x^r)^{k/r}dx$$ Setting $w=x^r$, we see that $$S=\frac1r\int_0^1w^{\frac{n+1-k-r}r}(1-w)^{k/r}dw$$ $$S=\frac1r\int_0^1w^{\frac{n+1-k}r-1}(1-w)^{\frac{k+r}r-1}dw$$ $$S=\frac1r\mathrm{B}\bigg(\frac{n+1-k}r,\frac{k+r}r\bigg)$$ So $$I(r,n)=\int_0^1[(1-x^r)^{1/r}-x]^ndx$$ $$I(r,n)=\frac1r\sum_{k=0}^{n}(-1)^{n-k}{n\choose k}\frac{\Gamma(\frac{n+1-k}r)\Gamma(\frac{k+r}r)}{\Gamma(1+\frac{n+1}r)}$$ $$I(r,n)=\frac1{r}\frac{\Gamma(n+1)}{\Gamma(1+\frac{n+1}r)}\sum_{k=0}^{n}(-1)^{n-k}\frac{\Gamma(\frac{n+1-k}r)\Gamma(\frac{k+r}r)}{\Gamma(k+1)\Gamma(n-k+1)}$$
Which is a closed form
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\left.\int_{0}^{1}\bracks{\pars{1 - x^{r}}^{1/r} - x}^{2k}\,\dd x\,\right\vert_{{\large r\ >\ 0} \atop {\large k\ \in\ \mathbb{N}_{\geq\ 0}}}} \\[5mm] \stackrel{x^{\large r}\ \mapsto\ x}{=}\,\,\,& \int_{0}^{1}\bracks{\pars{1 - x}^{1/r} - x^{1/r}}^{2k}\,{1 \over r}\, x^{1/r - 1}\,\dd x \\[5mm] \,\,\,\stackrel{x\ \mapsto\ x + 1/2}{=}\,\,\,& {1 \over r}\int_{-1/2}^{1/2}\bracks{\pars{{1 \over 2} - x}^{1/r} - \pars{{1 \over 2} + x}^{1/r}}^{2k}\, \pars{{1 \over 2} + x}^{1/r - 1}\,\dd x \\[8mm] = &\ {1 \over r}\int_{0}^{1/2}\bracks{\pars{{1 \over 2} - x}^{1/r} - \pars{{1 \over 2} + x\!}^{1/r}}^{2k}\times \\[2mm] &\ \phantom{{1 \over r}\int_{0}^{1/2}}\bracks{\pars{{1 \over 2} + x}^{1/r - 1} + \pars{{1 \over 2} - x}^{1/r - 1}\!}\!\dd x \\[8mm] = &\ -\int_{0}^{1/2}{1 \over 2k + 1}\,\partiald{}{x}\bracks{\pars{{1 \over 2} - x}^{1/r} - \pars{{1 \over 2} + x}^{1/r}}^{2k + 1}\,\dd x \\[5mm] = &\ \underbrace{\braces{-\bracks{\pars{{1 \over 2} - x}^{1/r} - \pars{{1 \over 2} + x}^{1/r}}^{2k + 1}}_{x\ =\ 0}^{x\ =\ 1/2}} _{\ds{=\ 1\ -\ 0\ =\ 1}}\,\,\,{1 \over 2k + 1} \\[5mm] = &\ \bbx{1 \over 2k + 1} \end{align}