Studying $ u_n = \int_0^1 (\arctan x)^n \mathrm dx$ I would like to find an equivalent of:
$$ u_n = \int_0^1 (\arctan x)^n \mathrm dx$$
which might be: $$ u_n \sim \frac{\pi}{2n}  \left(\frac{\pi}{4} \right)^n$$
$$ 0\le u_n\le \left( \frac{\pi}{4} \right)^n$$
So $$ u_n \rightarrow 0$$
In order to get rid of $\arctan x$ I used the substitution
$$x=\tan \left(\frac{\pi t}{4n} \right) $$
which gives:
$$ u_n= \left(\frac{\pi}{4n} \right)^{n+1} \int_0^n t^n\left(1+\tan\left(\frac{\pi t}{4n} \right)^2 \right) \mathrm dt$$
But this integral is not easier to study!
Or: $$ t=(\arctan x)^n $$
$$ u_n = \frac{1}{n} \int_0^{(\pi/4)^n} t^{1/n}(1+\tan( t^{1/n})^2 ) \mathrm dt $$
How could I deal with this one?

Another (simpler) approach is to substitute $x = \tan{y}$ and get
$$u_n = \int_0^{\frac{\pi}{4}} dy \: y^n \, \sec^2{y}$$
Now we perform an analysis not unlike Laplace's Method: as $n \rightarrow \infty$, the contribution to the integral is dominated by the value of the integrand at $y=\pi/4$.  We may then say that
$$u_n \sim \sec^2{\frac{\pi}{4}} \int_0^{\frac{\pi}{4}} dy \: y^n = \frac{2}{n+1} \left ( \frac{\pi}{4} \right )^{n+1} (n \rightarrow \infty)  $$
The stated result follows.
