1. We need to evaluate $I=\int_{-\pi/2}^{\pi/2} [\,[x]-\sin x\,] \, dx,$ where $[t]$ denotes the greatest integer less than or equal to $t$.

2. Let $n=[x].$ Since $x\in\left[-\frac\pi2,\frac\pi2\right]$, we have $x\in[-1.57\ldots,1.57\ldots],$ so the possible values of $[x]$ are only $-2,-1,0,1.$ We now examine the integrand $[[x]-\sin x]=[n-\sin x].$ Because $n$ is an integer, we use the property $[n+a]=n+[a] \quad (n\in\mathbb Z).$ Hence, $[n-\sin x]=n+[-\sin x]=[x]+[-\sin x].$ So $I=\int_{-\pi/2}^{\pi/2} [x] \, dx+\int_{-\pi/2}^{\pi/2} [-\sin x] \, dx.$

3. First compute $I_1=\int_{-\pi/2}^{\pi/2} [x] \, dx.$ Break at integer points in the interval:

• On $\left[-\frac\pi2,-1\right)$, $[x]=-2$
• On $[-1,0)$, $[x]=-1$
• On $[0,1)$, $[x]=0$
• On $[1,\frac\pi2]$, $[x]=1$

Therefore, $I_1=\int_{-\pi/2}^{-1} (-2)\,dx+\int_{-1}^{0} (-1)\,dx+\int_{0}^{1}0\,dx+\int_{1}^{\pi/2}1\,dx.$ Now, $\int_{-\pi/2}^{-1} (-2)dx=-2\left(-1+\frac\pi2\right)=2-\pi,$ $\int_{-1}^{0} (-1)dx=-1,$ $\int_0^1 0\,dx=0,$ $\int_1^{\pi/2} 1\,dx=\frac\pi2-1.$ Thus, $I_1=(2-\pi)-1+0+\left(\frac\pi2-1\right)=-\frac\pi2.$

4. Now compute $I_2=\int_{-\pi/2}^{\pi/2}[-\sin x] \, dx.$ Since $\sin x\in[-1,1]$ on this interval, $-\sin x\in[-1,1]$.

Let us determine $[-\sin x]$:

• If $x\in\left[-\frac\pi2,0\right)$, then $\sin x<0$, so $-\sin x\in(0,1]$. Hence $[-\sin x]=0$ except at $x=-\frac\pi2$, where it equals $1$, which does not affect the integral.
• If $x=0$, then $[-\sin 0]=0$.
• If $x\in\left(0,\frac\pi2\right]$, then $\sin x>0$, so $-\sin x\in[-1,0)$. Hence $[-\sin x]=-1$ except at $x=\frac\pi2$, where it is $-1$ anyway.

So effectively,