evaluate this integral

First we will try to eliminate the absolute value symbol. It can be done by splitting the integral into several integrals, but I used another way..

Let $\displaystyle u=x-\pi , du=dx$

$\displaystyle => \int_0^{2\pi} \mathrm \lfloor|\sin x|+|\cos x|\rfloor\,\mathrm{d}x=\int_{-\pi}^{\pi} \mathrm \lfloor|\sin (u+\pi)|+|\cos (u+\pi)|\rfloor\,\mathrm{d}u$

$\displaystyle =\int_ {-\pi}^{\pi} \mathrm \lfloor|\sin (u)|+|\cos (u)|\rfloor\,\mathrm{d}u =2\int_ 0^\pi \mathrm \lfloor|\sin (u)|+|\cos (u)|\rfloor\,\mathrm{d}u$

Now, let $\displaystyle y=u-\frac{\pi}{2} , dy=du$

$\displaystyle => 2\int_ 0^\pi \mathrm \lfloor|\sin (u)|+|\cos (u)|\rfloor\,\mathrm{d}u=2\int_ {-\frac{\pi}{2}}^{\frac{\pi}{2}} \mathrm \lfloor|\sin (y+\frac{\pi}{2})|+|\cos (y+\frac{\pi}{2})|\rfloor\,\mathrm{d}y$

$\displaystyle = 2\int_ {-\frac{\pi}{2}}^{\frac{\pi}{2}} \mathrm \lfloor|\cos (y)|+|\sin (y)|\rfloor\,\mathrm{d}y=4\int_ 0^{\frac{\pi}{2}} \mathrm \lfloor|\cos (y)|+|\sin (y)|\rfloor\,\mathrm{d}y$

The new limits of integration indicate that the argument of $\cos x$ and $\sin x$ are in the first quadrant, hence they are positive.

Now to deal with the floor function, let $\displaystyle f(y)=\sin y +\cos y$ . Using calculus, f(y) increases on $]0,\frac{\pi}{4}[$ and decreases on $]\frac{\pi}{4},\frac{\pi}{2}[$

So, $\displaystyle f(0)<f(y)<f(\frac{\pi}{4}) for 0<x<\frac{\pi}{4} and f(\frac{\pi}{4})<f(y)<f(\frac{\pi}{2}) for \frac{\pi}{4}<x<\frac{\pi}{2}$ , hence we can bound $f$ between $1$ and $\sqrt{2}$ on $]0,\frac{\pi}{2}[$

$\displaystyle => 4\int_ 0^{\frac{\pi}{2}} \mathrm \lfloor\cos (y)+\sin (y)\rfloor\,\mathrm{d}y=4\int_ 0^{\frac{\pi}{2}} \mathrm{d}y= 2\pi$

Therefore our desired answer is $\boxed{m=2}$