I have to find the integral:

$$\int\_0^{2 \pi} \arcsin(\sin(2x)) dx$$

This is what I did:

$$\int\_0^{2 \pi} \arcsin(\sin(2x)) dx = \int\_0^{2 \pi} 2x dx = \bigg [ x^2 \bigg ]\_0^{2 \pi} = 4\pi^2$$

Now, I know this is incorrect, since the answer given in my textbook is $0$. What I want to know is why is it wrong? Can somebody tell me, in as much detail as possible so I can understand it once and for all, **why** is my answer wrong and how should I correct it? What is the intuition behind what I actually should be doing?