Evaluate $\displaystyle \lim _{n \to \infty} \int_{0}^1 \arcsin(\sin(nx)) dx$ Evaluate
$$\displaystyle \lim _{n \to \infty} \int_{0}^1 \arcsin(\sin(nx)) dx$$
I know $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2} \: \forall x \in [-1, 1]$
and also $\arcsin(\sin x) = x \: \forall x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Someone suggested me to restraint the integral on $[0, 2\pi]$ but I don't know how can I do that and why I should.

$$\int_{0}^{1}\arcsin(\sin(nx))\,dx = \frac{1}{n}\int_{0}^{n}\arcsin(\sin x)\,dx \tag{1}$$
and $\arcsin(\sin x)$ is a $2\pi$-periodic function corresponding to a triangle wave. Such function is continuous, bounded and with mean zero, hence the wanted limit is simply zero.
We may actually prove that
$$ \forall n\in\mathbb{N},\qquad \left|\int_{0}^{n}\arcsin(\sin x)\,dx\right|\leq\frac{\pi^2}{4} = \int_{0}^{\pi}\arcsin(\sin x)\,dx \tag{2}$$
and also expand $\arcsin(\sin x)$ as a Fourier sine series:
$$ \arcsin(\sin x) = \frac{4}{\pi}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}\sin((2n+1)x).\tag{3}$$
Uniform convergence is ensured by $\sum_{n\geq 0}\frac{1}{(2n+1)^2}<+\infty$.
