How to evaluate $\lim_{n \to \infty} \int_0^1 \arcsin (\sin(nx)) dx$? I would like to evaluate $\lim_{n \to \infty} \int_0^1 \arcsin (\sin(nx)) dx$.
I think the answer is 0, but can't prove it. The problem is difficult because of rapid oscillations.
 A: Make a substitution:
$$
\left|\int_0^1 \arcsin(\sin(n x))\,dx \right|
$$
$$
\stackrel{t=n x}{=} \frac{1}{n}\left| \int _0^n \arcsin(\sin(t))\,dt\right|
$$But now the integrand is periodic with period $2\pi$. In fact, the integral is maximized if we just integrate over $[0,\pi]$ (assuming $n\ge 4$, which can clearly be done):
$$
\le \frac{1}{n} \left|\int _0^{\pi} \arcsin(\sin(t))\,dt\right|
$$
$$
=\frac{1}{n}\cdot \frac{\pi^2}{4} \to 0
$$So the limit indeed converges to zero.
A: I wanted to show an approach using the closed form of the integral. Making the same substitution $t = nx$ as Integrand yields $$\frac{1}{n}\int_{0}^{n}\arcsin\left(\sin\left(t\right)\right)dt$$
This can be split into groups of $2\pi$ to get $$\frac{1}{n} \sum_{k=1}^{\left\lfloor \frac{n}{2\pi}\right\rfloor} \int_{2\pi (k-1)}^{2\pi k} \arcsin(\sin(t)) dt + \frac{1}{n} \int_{2\pi \left\lfloor \frac{n}{2\pi}\right\rfloor}^n \arcsin(\sin(t))dt$$
The integral $\int_x^{x+2\pi}\arcsin(\sin(t))dt$ is equal to $0$, so this simplifies to $$\frac{1}{n}\int_{2\pi \left\lfloor \frac{n}{2\pi}\right\rfloor}^n \arcsin(\sin(t))dt$$
This is equivalent to $$\frac{1}{n}\int_{0}^{n\pmod {2\pi}} \arcsin(\sin(t))dt$$
To show convergence to $0$, you could note that the maximum of $\int_0^x \arcsin(\sin(t)) dt$ is bounded for $x \in [0, 2\pi)$, and as $n \to \infty$, the limit must be $0$.
Based on where $p = n\pmod {2\pi}$ lies in $[0, 2\pi)$, there are a couple different values for the integral. If $0\le p \le \frac{\pi}{2}$, the integral is $\frac{p^2}{2n}$. If $\frac{\pi}{2} \le p \le \frac{3\pi}{2}$, the integral is $\frac{-\left(p-\pi\right)^{2}+\frac{\pi^{2}}{2}}{2n}$. Finally, if $\frac{3\pi}{2} \le p < 2\pi$, the integral is $\frac{\left(p-2\pi\right)^{2}}{2n}$.
