<s>[INST] 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?
 [/INST] $$
\begin{align}
&\int_0^1\arctan^n(x)\,\mathrm{d}x\\
&=\int_0^{\pi/4}x^n\sec^2(x)\,\mathrm{d}x\\
&=\int_0^{\pi/4}\left(\frac\pi4-x\right)^n\sec^2\left(\frac\pi4-x\right)\,\mathrm{d}x\\
&=\frac{(\pi/4)^{n+1}}{n}\int_0^ne^{-x}\color{#C00000}{e^x\left(1-\frac{x}{n}\right)^n}\color{#00A000}{\sec^2\left(\frac\pi4-\frac{\pi x}{4n}\right)}\,\mathrm{d}x\\
&=\frac{(\pi/4)^{n+1}}{n}\int_0^ne^{-x}\color{#C00000}{\small\left(1-\frac{x^2}{2n}-\frac{8x^3-3x^4}{24n^2}+O\left(\frac1{n^3}\right)\right)}\color{#00A000}{\small\left(2-\frac{\pi x}{n}+\frac{\pi^2x^2}{2n^2}+O\left(\frac1{n^3}\right)\right)}\,\mathrm{d}x\\
&=\frac{(\pi/4)^{n+1}}{n}{\small\left(2-\frac{\pi+2}{n}+\frac{2+3\pi+\pi^2}{n^2}+O\left(\frac1{n^3}\right)\right)}
\end{align}
$$
</s>