Find $\int_0^{2 \pi} \arcsin(\sin(2x)) dx$ 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?
 A: Substitute $2x=t+2\pi$
$$\int_0^{2 \pi} \arcsin(\sin(2x)) dx
=\frac12\int_{-2\pi}^{2\pi} \arcsin(\sin t) dt=0$$
where the integral vanishes due to the odd integrand  $\arcsin(\sin t) $.
Note that the range of $\arcsin u$ is $[-\frac\pi2,\frac\pi2]$. So,  $\arcsin(\sin(2x))\ne 2x$ for all $x\in[0,2\pi]$ within the integration limits, which is the mistake you made.
A: The range of arcsine is $[-\pi/2, \pi/2]$.  This means the largest value that ever comes out of $\arcsin (\text{anything})$ is $\pi/2$.  The values of $2x$ over the interval $[0,2\pi]$ are quickly much bigger than $\pi/2$, so 
$$  2x \neq \arcsin \sin 2x  $$
over all of $x \in [0, 2\pi]$.  (This is an equality for the first little piece, $[0, \pi/4]$.)

So either we recognie the periodicity and symmetry to get the result $0$, or we need to attack this in pieces of the interval of integration.
$$  \arcsin \sin 2x = \begin{cases}
2x ,& 0 \leq x \leq \pi/4  \\
\pi - 2x ,& \pi/4 \leq x \leq 3\pi/4  \\
2x - 2 \pi ,& 3 \pi/4 \leq x \leq 5 \pi/4  \\
3\pi - 2x ,& 5 \pi/4 \leq x \leq 7 \pi/4  \\
2x - 4 \pi ,& 7 \pi/4 \leq x  \leq 2 \pi
\end{cases}$$
Where does that come from?  The equation 
$$  \sin r = \sin 2x  $$
has the complete solution
$$  r = 2x + 2\pi k \\
    \text{ or }  \\
    r = \pi - 2x + 2 \pi k
$$
for any integer $k$, under the condition that $-\pi/2 \leq r \leq \pi/2$.  With $k = 0$ we get the first two lines of the solution.  With $k=-1$, the next two lines, and so on.
A: $\text{arcsin}$ is a function. It takes numbers in the interval $[-1,1]$ and turns them into numbers in the interval $[-\pi/2, \pi/2]$. Given a number $y \in [-1,1]$, $\text{arcsin}(y)$ is the unique number $x$, between $-\pi/2$ and $\pi/2$, such that $\sin(x) = y$.
In particular, if $x$ is not a number between $-\pi/2$ and $\pi/2$, $\arcsin(\sin(x))$ cannot be $x$. By definition. For example, taking $x=\pi$. $\sin(\pi) = 0$, and the unique number $x$ in the interval $[-\pi/2, \pi/2]$ such that $\sin(x) = 0$ is $x=0$. So $\arcsin(\sin(\pi)) = 0$. Similarly, $\arcsin(\sin(3\pi/2)) = -\pi/2$. It's hopefully clear that if you know $x$, you can always figure out pretty easily what $\arcsin(\sin(x))$ is, but it's not exactly the same thing as $x$.
If you draw a picture, then you'll see that between $0$ and $2\pi$ your function looks like $2x$ for $x \in [0,\pi/4]$, like $\pi - 2x$ for $x \in [\pi/4, 3\pi/4]$, and so on. The area under that curve is clearly zero.
