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?
 A: To get a rough asymptotic answer, you can substitute $$x=1-\frac{y}{n}.$$  Then
$$
\frac{u_n}{(1/n) (\pi/4)^n}=\int_0^n \left(\frac{\arctan (1-\frac{y}{n})}{\pi/4}\right)^n \, dy.
$$
For fixed $y$, as $n\to\infty$,
$$
\frac{\arctan (1-\frac{y}{n})}{\pi/4}=1-\frac{2}{\pi n} y + O(\frac{1}{n^2})= \exp(-\frac{2y}{\pi n} + O(\frac{1}{n^2})),
$$
so, pointwise, the integrand converges to $e^{-2y/\pi}$.  Since we have the
bound
$$
\arctan (1-\frac{y}{n})\le \frac{\pi}{4} e^{-2y/\pi n}, \qquad 0\le y\le n, \ \ \ (*)
$$
we can apply the dominated convergence theorem to conclude that
$$
u_n \sim \frac{1}{n}(\frac{\pi}{4})^n \int_0^\infty e^{-2y/\pi} \, dy=\frac{\pi}{2n} (\frac{\pi}{4})^n.
$$
Also, from the integral and (*),
$$
0\le u_n\le \frac{\pi}{2n}(\frac{\pi}{4})^n, \qquad \rm for\  all\ \it n.
$$
A: $$
\begin{align}
&\int_0^1\arctan^n(x)\,\mathrm{d}x\\
&=\int_0^{\pi/4}x^n\sec^2(x)\,\mathrm{d}x\\
&=\int_0^{\pi/4}\left(\frac\pi4-x\right)^n\sec^2\left(\frac\pi4-x\right)\,\mathrm{d}x\\
&=\frac{(\pi/4)^{n+1}}{n}\int_0^ne^{-x}\color{#C00000}{e^x\left(1-\frac{x}{n}\right)^n}\color{#00A000}{\sec^2\left(\frac\pi4-\frac{\pi x}{4n}\right)}\,\mathrm{d}x\\
&=\frac{(\pi/4)^{n+1}}{n}\int_0^ne^{-x}\color{#C00000}{\small\left(1-\frac{x^2}{2n}-\frac{8x^3-3x^4}{24n^2}+O\left(\frac1{n^3}\right)\right)}\color{#00A000}{\small\left(2-\frac{\pi x}{n}+\frac{\pi^2x^2}{2n^2}+O\left(\frac1{n^3}\right)\right)}\,\mathrm{d}x\\
&=\frac{(\pi/4)^{n+1}}{n}{\small\left(2-\frac{\pi+2}{n}+\frac{2+3\pi+\pi^2}{n^2}+O\left(\frac1{n^3}\right)\right)}
\end{align}
$$
A: From your simplification,
$$
u_n = \left(\frac{\pi}{4n}\right)^{n+1} \int_0^n t^n \left(1 + \tan^2\left(\frac{\pi t}{4n}\right) \right) dt
$$
note that the integrand is non-negative and in fact on $[0,n]$, $0 \leq \tan(\pi t/4n) \leq 1$. Applying this we get
$$
\left(\frac{\pi}{4n}\right)^{n+1} \int_0^n t^n dt \leq u_n \leq \left(\frac{\pi}{4n}\right)^{n+1} \int_0^n 2t^n dt
$$
which gives
$$
\left(\frac{\pi}{4n}\right)^n \frac{\pi}{4(n+1)} \leq u_n \leq \left(\frac{\pi}{4n}\right)^n \frac{\pi}{2(n+1)}
$$
Hope that helps.
A: 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.
