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.

Too long for a comment.
After ComplexYetTrivial's answer, we have
$$I_n=\frac{\Gamma \left(\frac{2 (n+1)}{2 n+1}\right) \Gamma \left(\frac{n+1}{n(2
   n+1)}\right)}{2 \Gamma \left(\frac{1}{n}\right)}$$ Expanding as series
$$I_n=1-\frac{1}{2 n}+\frac{12-\pi ^2}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ which is not "too bad" even for $n=1$; this would give $1-\frac{\pi ^2}{24}\approx 0.588766$ while $\frac{\pi}{3 \sqrt{3}}\approx 0.604600$. 
For a few values of $n$
$$\left(
\begin{array}{ccc}
n & \text{approximation} & \text{exact} \\
 1 & 0.588766 & 0.604600 \\
 2 & 0.772192 & 0.774848 \\
 3 & 0.843196 & 0.843965 \\
 4 & 0.880548 & 0.880851 \\
 5 & 0.903551 & 0.903695 \\
 6 & 0.919132 & 0.919210 \\
 7 & 0.930383 & 0.930429 \\
 8 & 0.938887 & 0.938916 \\
 9 & 0.945540 & 0.945560
\end{array}
\right)$$
