I don't like Wolfram Alpha's evaluation of an integral Wolfram Alpha says that $\int|\sin(x)| = -\cos(x)\mathrm{sgn}(\sin(x))$
It cannot be true, since $|\sin(x)|$ is always positive and its integral must grow indefinitely.
But when it comes to calculating the definite integral over intervals the page is right.
What is wrong here?
 A: You forgot the $+C$.
The $+C$ is even more important than usual here; $C$ is only ever required to be locally constant. While usually that means it actually is constant, the antiderivative that Wolfram Alpha computed is not defined at any of the points $x = n \pi$, where $n$ is an integer
While $C$ must take on the same value across intervals, your domain is made up of many disjoint intervals of the form $(n \pi, (n+1)\pi)$ (where $n$ ranges over all integers), and thus $C$ is allowed to have a different value on each of them.
Technically, the expression WolframAlpha gave is defined everywhere once you've decided what $\operatorname{sgn}(0)$ is, but your function is clearly not differentiable at those points, so it can only be said to be an antiderivative when we exclude the integer multiples of $\pi$.
We can do better than Wolfram Alpha: we can compute an antiderivative that is defined everywhere, by lining up the constants so that the function can be continuously extended to have a value at the integer multiples of $\pi$: the correct antiderivative is
$$ C + \begin{cases} \frac{2x}{\pi} - 1 & \sin(x) = 0
\\ 2 \left\lfloor \frac{x}{\pi} \right\rfloor - \cos(x) \operatorname{sgn}(\sin(x)) 
& \sin(x) \neq 0 \end{cases}$$
A: Indefinite integrals are functions which, after deriving them, give the original function. Thus, for example, the function $g(x) = x - 100000$ is an indefinite integral of $f(x)=1$, because $g'=f.$
The thing is that if $g$ is an indefinite integral of $f$, then $h(x) = g(x) + C$ is an indefinite integral of $f$ for every $C\in\mathbb R$. Even more, all indefinite integrals of $f$ have the form $h(x) = g(x) + C$.
Your second question is now fairly simple. You probably know that if $g'(x) = f(x)$, then $$\int_a^bf(x) dx = g(b) - g(a).$$
Now, if you take any indefinite integral of $f$, say the function $h(x) = g(x) + C$ (for a small enough $C$, this will be a negative function even if $f$ is positive), then $$h(b)-h(a) = g(b) + C - (g(a) + C) = g(b) - g(a).$$
