$\displaystyle{\int_{0}^{3\pi}\left\lfloor\sin\left(x\right)\right\rfloor\,\mathrm{d}x = -\pi \approx -3.14159}$
Here, $\left\lfloor\cdots\right\rfloor$ means floor function
Why is it so ?. What I did is that I broke the limits from 0 to $\pi$, $\pi$ to $2\pi$ and $2\pi$ to $3\pi$.
In $0$ to $\pi$, and $x = \pi/2$, the value of $\sin\left(x\right)$ will be $1$. So, we should include that one in our final result, no ?.
From $\pi$ to $2\pi$, floor function of $\sin\left(x\right)$ becomes $-1$. Hence it comes as $-\left(2\pi - \pi\right) = -\pi$. Okay
Again from $2\pi$ to $3\pi$, at $x = 2\pi + \pi/2$, $\sin\left(x\right) = 1$. Hence we should include it.
So, the result should be $1 - \pi + 1 =2 - \pi$.
But Wolfram Alpha shows only -$\pi$. Why?
Notice that $0\leq\sin x<1$ for $2\pi\leq x\leq 3\pi$ except by one point, namely $x=\frac52\pi$, then
$$\int_{2\pi}^{3\pi}\lfloor\sin(x)\rfloor dx=\int_{2\pi}^{3\pi}0dx=0$$