integral of $\arcsin(\sin(x))$ I'm having trouble with this integral
$$\int\arcsin(\sin x)\,\mathrm dx$$
The problem is with the intervals of definition for each function :/ if someone could dumb it down for me.   


*

*$\arcsin\colon [-1, 1] \to [-\pi/2, \pi/2]$.  

*$\sin:\mathbb{R}\to [-1, 1]$.


right?
But what about $\arcsin(\sin x)$ ?

My previous hint was a bit misleading - hence here is a (the comments are still appropriate)
CORRECTED HINT 
On $I= [-\pi/2,\pi/2]$ the sine function grows from $\sin(-\pi/2)=-1$ to $\sin(\pi/2)=1$, hence it is invertible on $I$. The inverse, $\arcsin$, is defined on $[-1,1]$ and has the property $\arcsin(\sin x) = x$ if $-\pi/2\le x\le \pi/2$.
Since $\sin$ is $2\pi$-peiodic we must have $\arcsin(\sin (x+2n\pi))=\arcsin(\sin(x))$ for all $x$ and integers $n$. 
Hence it is sufficient to understand $\arcsin(\sin x)$ for $\pi/2\le x\le 3\pi/2$, to achieve that use $\sin(x+\pi)=\sin (-x)$.
When you see the function there will be not problem to perform any integration.
