TITLE: Ways to show that $\int_{0}^{1}((1-x^r)^{1/r}-x)^{2k}dx=\frac{1}{2k+1}$
QUESTION [26 upvotes]: Through some calculation, I found that for all $r>0$
$$
\begin{array}{rcl}
{\displaystyle\int_{0}^{1}\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{2}\,\mathrm{d}x} &  {\displaystyle =} &
{\displaystyle{1 \over 3}}
\\
{\displaystyle\int_{0}^{1}\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{4}\,\mathrm{d}x} &  {\displaystyle =} &
{\displaystyle{1 \over 5}}
\\
{\displaystyle\int_{0}^{1}\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{6}\,\mathrm{d}x} &  {\displaystyle =} &
{\displaystyle{1 \over 7}}
\end{array}
$$
It seems like for  $k \in \mathbb{N}$ 

$$
\int_{0}^{1}
\left[\left(1 - x^{r}\right)^{1/r} - x\right]^{2k}\mathrm{d}x = {1 \over 2k + 1}
$$

I want to prove this general form.
Someone suggested to make the substitution 
$$y=(1-x^r)^{1/r}$$
Let $n=2k$, and I rewrote the integral into 
$$\int_{0}^{1}((1-x^r)^{1/r}-x)^ndx=\int_{0}^{1}(y-x)^ndx$$
and tried to use the binomial formula:
$$(y-x)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}x^{n-k}y^{k}$$
The integral then becomes
$$\begin{align}
\int_{0}^{1}(y-x)^ndx&=\int_{0}^{1}\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}x^{n-k}y^{k}dx\\
&=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}\int_{0}^{1}x^{n-k}y^{k}dx\\
&=\sum _{k=0}^{n}{\binom {n}{k}}(-1)^{n-k}\int_{0}^{1}x^{n-k}(1-x^r)^{k/r}dx\\
\end{align}$$
Now I think I need to use Beta function:
$$B(x,y) = \frac{(x-1)!(y-1)!}{(x+y-1)!}= \int_{0}^{1}u^{x-1}(1-u)^{y-1}du=\sum_{n=0}^{\infty}\frac{{\binom{n-y}{n}}}{x+n}$$
Am I on the right track? Are here any easier ways to prove the general form?

REPLY [5 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}