Studying $ u_{n}=\frac{1}{n!}\int_0^1 (\arcsin x)^n \mathrm dx $ I would like to find a simple equivalent of:
$$ u_{n}=\frac{1}{n!}\int_0^1 (\arcsin x)^n \mathrm dx $$
We have:
$$ 0\leq u_{n}\leq \frac{1}{n!}\left(\frac{\pi}{2}\right)^n \rightarrow0$$
So $$ u_{n} \rightarrow 0$$
Clearly:
$$ u_{n} \sim \frac{1}{n!} \int_{\sin(1)}^1 (\arcsin x)^n \mathrm dx $$
But is there a simpler equivalent for $u_{n}$?
Using integration by part:
$$ \int_0^1 (\arcsin x)^n \mathrm dx = \left(\frac{\pi}{2}\right)^n - n\int_0^1 \frac{x(\arcsin x)^{n-1}}{\sqrt{1-x^2}} \mathrm dx$$
But the relation 
$$ u_{n} \sim \frac{1}{n!} \left(\frac{\pi}{2}\right)^n$$
seems to be wrong...
 A: The change of variable $x=\cos\left(\frac{\pi s}{2n}\right)$ yields
$$
u_n=\frac1{n!}\left(\frac\pi2\right)^{n+2}\frac1{n^2}v_n,
$$
with
$$
v_n=\int_0^n\left(1-\frac{s}n\right)^n\,\frac{2n}\pi \sin\left(\frac{\pi s}{2n}\right)\,\mathrm ds.
$$
When $n\to\infty$, $\left(1-\frac{s}n\right)^n\mathbf 1_{0\leqslant s\leqslant n}\to\mathrm e^{-s}$ and $\frac{2n}\pi \sin\left(\frac{\pi s}{2n}\right)\mathbf 1_{0\leqslant s\leqslant n}\to s$. Both convergences are monotonic hence $v_n\to\int\limits_0^\infty\mathrm e^{-s}\,s\,\mathrm ds=1$. Finally,
$$
u_n\sim\frac1{(n+2)!}\left(\frac\pi2\right)^{n+2}.
$$
A: This is not a complete answer, but an improved inequality. From
$$
\arcsin x\le \frac{\pi}{2}\,x
$$
you get
$$
u_n\le\frac{1}{(n+1)!}\Bigr(\frac{\pi}{2}\Bigl)^n.
$$
A: Substitute $x = \sin(t)$ to get
$$
u_n = \frac{1}{n!}\int_0^{\pi/2} t^n \cos(t) dt.
$$
The following equalities hold for all integers $m \geq 0$ (which can be checked by partial integration):
$$
\frac{1}{n!}\int_0^{\pi/2} t^n \left(\tfrac{\pi}{2} - t \right)^m dt = \frac{m!}{(n + m + 1)!}\left(\frac{\pi}{2}\right)^{n + m + 1}.
$$
Then it follows from the Taylor expansion of $\cos(t)$ around $t = \pi/2$
$$
\cos(t) = \sin \left(\tfrac{\pi}{2} - t \right) = \sum_{m=0}^{\infty}\frac{(-1)^m}{(2m + 1)!}\left(\tfrac{\pi}{2} - t\right)^{2m+1}
$$
that
$$
u_n = \sum_{m=0}^{\infty} \frac{(-1)^m}{(n+2m+2)!}\left(\frac{\pi}{2}\right)^{n + 2m + 2}.
$$
The growth in $n$ is determined by the first term ($m=0$) so
$$
u_n \sim \frac{1}{(n+2)!}\left(\frac{\pi}{2}\right)^{n + 2}. 
$$
