Finding the limit of improper integral Got this limit to solve, and the solution apparently is $\sqrt{π^2-4} $.
$$\lim_{n\to\infty} \int_0^π {(x(x-π))^n\over 1+(x(x-π))^n } dx$$
I have no idea how to solve it. 
Thanks in advance.

Edit: I will assume thet the integrand includes $x(\pi-x)$ and not $x(x-\pi)$. Othewise the denominator vanishes at two points in $(0,1)$ and the integral is not convergent.
$x(\pi-x)=1$ has two solutions:
$$
x_\pm=\frac{\pi\pm\sqrt{\pi^2-4}}{2}.
$$
It follows that $0\le x(\pi-x)<1$ on $[0,x_-)\cup(x_+,\pi]$ and $x(x-\pi)>1$ on $(x_-,x_+)$. Then
$$
\lim_{n\to\infty}\frac{(x(\pi-x))^n}{1+(x(\pi-x))^n }=\begin{cases}
0 & x\in[0,x_-)\cup(x_+,\pi]\\
1 & x\in(x_-,x_+)\\
1/2 & x=x_\pm
\end{cases}
$$
Since the integral is uniformly bounded by $1$, we can interchange the limit and the integral:
$$
\lim_{n\to\infty}\int_0^\pi\frac{(x(\pi-x))^n}{1+(x(\pi-x))^n }\,dx=\int_{x_-}^{x_+}dx=x_+-x_-=\sqrt{\pi^2-4}.
$$
