what is the value of
integral ( from 0 to 2pi ) |arcsin(sinx)| dx
where | | means absolute value
..answer is pi*pi/2.0 ;
$\displaystyle \arcsin t$ means the value of $\displaystyle \theta$ between $\displaystyle -\pi/2$ and $\displaystyle \pi/2$ such that $\displaystyle \sin\theta = t$. So $\displaystyle \arcsin(\sin x) = x$ when $\displaystyle 0\leqslant x\leqslant\pi/2.$
When $\displaystyle \pi/2\leqslant x\leqslant 3\pi/2$ you should use the fact that $\displaystyle \sin x = \sin(\pi-x)$ to deduce that $\displaystyle \arcsin (\sin x) = \pi-x$.
For $\displaystyle 3\pi/2\leqslant x\leqslant 2\pi$, $\displaystyle \arcsin (\sin x)$ will be given by yet another formula, which I'll leave you to think about.
Once you have the formulas for $\displaystyle \arcsin (\sin x)$ on the three different intervals, you should find it comparatively easy to take their absolute values and integrate over the corresponding intervals.
When you have done this, it would be a good idea to check your answer by drawing the graph of $\displaystyle \left|\arcsin(\sin x)\right|$ on the interval from $\displaystyle 0$ to $\displaystyle 2\pi$. It should consist of straight line segments, and the area under the graph will consist of a number of triangles. You can then check that your value for the integral is correct by calculating the total area of these triangles using the formula "half base times height".
No, arcsin is defined on the interval –1 to 1. The values of the arcsin function lie in the interval $\displaystyle -\pi/2$ to $\displaystyle \pi/2.$
Since sin x always lies in the interval –1 to 1, it follows that arcsin(sin x) is defined for all x, and its values always lie in the interval $\displaystyle -\pi/2$ to $\displaystyle \pi/2.$