function-and-relation-compositioninversetrigonometry
I can kind of understand the main direction (slope) of $y$ over the different $x$ intervals, but I can't figure out why the values of $y$ take on the shape of straight lines and not curves looking more like those of sin, cos…
EDIT: I understand that the derivative of Arccos(Sin(x)) gives 1 or -1 depending on the x interval, but it doesn't give me intuition into why that's the case.
Yes, that's the idea. The end of the bar is moving at a constant speed, so the $y$ component of its velocity (which is the rate of change of $\sin(\theta)$) is greatest in magnitude when all of that velocity is in the $y$ direction, and that happens when $\sin(\theta) = 0$.
When you talk about integrals, you could be talking about one of many distinct, but related topics. You could mean Riemann integrals, indefinite integrals, or many other types of integral out there.
You're probably talking about the indefinite integral when you say that the integral of $\sin$ is $-\cos$. The thing about the indefinite integral is that it is not really a single function. That's why you often display it with the little ${}+C$ at the end. An indefinite integral is more like a set of functions, which are all equal under vertical translations.
So, $\int f$ could be considered the set of functions which have a derivative $f$. There are many functions which have the derivative $\sin$, and they are all of the form $-\cos+C$, where $C$ is some real number. This is why it is said the integral of $\sin$ is $-\cos$.
But why does the integral then not represent the area? Well, it does. What you were saying was that $-\cos(x)$ should be the area under $\sin(x)$ up to $x$. The question is, up to $x$ from where? There is no lower bound! If you pick the lower bound $0$, then
$$-\cos(x)-(-\cos(0))=-\cos(x)+1$$
You get that $+1$ you were wondering about, and it does represent the area under $\sin(x)$ from $0$ to $x$.
Best Answer
Well, to think about the equation $$y=\arccos(\sin(x))$$ let's first look at an easier one, obtained by taking the cosine of both sides (noting that $\cos(\arccos(x))=x$ - that is $\arccos$ is a right-inverse of $\cos$): $$\cos(y)=\sin(x).$$ If we plot this, we get the following graph:
$\hskip1.5in$
which is just a series of diagonal lines going in two directions. This mystery unfolds when we rewrite $$\cos(y)=\cos\left(x+\frac{\pi}2\right)$$ and note that, from symmetries and periodicity, we have that $\cos(u)=\cos(v)$ whenever $u=2\pi k \pm v$ for some integer $k$. This gives us that there is a line $y=x+\frac{\pi}2$ included as well as the line $y=x-\frac{\pi}2$ and any shift of these horizontally (or vertically) by $2\pi$, giving rise to the lattice pattern.
However, the definition of $\arccos$ that you are using seems to mandate that its output falls in the interval $[0,\pi]$, hence $y$ must be in there. If we highlight the suitable range of $y$ in gray and trace the lines it covers in red, we recover the pattern you observed:
$\hskip1.5in$
which makes it clear that the alternation is merely a small part of a larger pattern, than can't be fully represented due to the fact that $\arccos$ is limited in its range.