On the integrals $\int_{-1}^0 \sqrt[2n+1]{x-\sqrt[2n+1]x} \mathrm dx$ Playing with integrals of type,
$$
I(n)=\int_{-1}^0 \sqrt[2n+1]{x-\sqrt[2n+1]x} \mathrm dx, 
$$
$$
n \in \mathbb{N}
$$
I got two interesting results for the limiting cases $n=1$ and $n \to \infty$:
$$
\lim_{n \to \infty} I(n) = 1
$$
The second result is perhaps more interesting,
$$
I(1)=\int_{-1}^0 \sqrt[3]{x-\sqrt[3]x} \mathrm dx \approx \frac {\pi}{\sqrt {27}} 
$$
The approximation is valid to $12$ places of decimal.
My question is straight, can we prove these results analytically? Any help would be appreciated.

For $n \in \mathbb{N}$ we have
\begin{align}
I (n) &= \int \limits_{-1}^0 \left[x - x^{\frac{1}{2n+1}}\right]^{\frac{1}{2n+1}} \mathrm{d} x \stackrel{x = -y}{=} \int \limits_0^1 \left[y^{\frac{1}{2n+1}} - y\right]^{\frac{1}{2n+1}} \mathrm{d} y = \int \limits_0^1 y^{\frac{1}{(2n+1)^2}}\left[1 - y^{\frac{2n}{2n+1}}\right]^{\frac{1}{2n+1}} \mathrm{d} y \\
&\hspace{-10pt}\stackrel{y = t^{\frac{2n+1}{2n}}}{=} \frac{2n+1}{2n} \int \limits_0^1 t^{\frac{n+1}{n (2n+1)}} (1-t)^{\frac{1}{2n+1}} \mathrm{d}t = \frac{2n+1}{2n} \operatorname{B}\left(\frac{n+1}{n(2n+1)}+1,\frac{1}{2n+1} + 1\right) \, .
\end{align}
Using $\Gamma(x+1) = x \Gamma(x)$ we can rewrite this result to find
$$ I (n) = \frac{\operatorname{B} \left(\frac{1}{n} - \frac{1}{2n+1}, \frac{1}{2n+1}\right)}{2 (2n+1)} = \frac{1}{2(2n+1)} \frac{\operatorname{\Gamma}\left(\frac{1}{n} - \frac{1}{2n+1}\right) \operatorname{\Gamma}\left(\frac{1}{2n+1}\right)}{\operatorname{\Gamma}\left(\frac{1}{n}\right)}$$
for $n \in \mathbb{N}$. In particular,
$$ I(1) = \frac{\operatorname{\Gamma}\left(\frac{2}{3}\right) \operatorname{\Gamma}\left(\frac{1}{3}\right)}{6} = \frac{\pi}{6 \sin \left(\frac{\pi}{3}\right)} = \frac{\pi}{3 \sqrt{3}} \,.$$
Moreover, we obtain
$$ \lim_{n \to \infty} I (n) \stackrel{\Gamma(x) \,  \stackrel{x \to 0}{\sim} \, \frac{1}{x}}{=} \lim_{n \to \infty} \frac{1}{2(2n+1)} \frac{\frac{n(2n+1)}{n+1} (2n+1)}{n} = \lim_{n \to \infty} \frac{2n+1}{2(n+1)} = 1 \, .$$ 
