The value of the integral $$\int_{-2}^{2} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} dx$$ (where $$[x]$$ denotes the greatest integer less than or equal to x) is

We have to evaluate the definite integral

$$I=\int_{-2}^{2}\frac{\sin^{2}x}{\left[\dfrac{x}{\pi}\right]+\dfrac12}\,dx,$$

where $$[\,t\,]$$ denotes the greatest integer less than or equal to $$t$$ (the “floor” function).

First we study the denominator. For each $$x$$ lying between $$-2$$ and $$2$$ we look at the value of $$\dfrac{x}{\pi}$$:

$$-2\le x<0\quad\Longrightarrow\quad -\dfrac{2}{\pi}\le\dfrac{x}{\pi}<0 \quad\Longrightarrow\quad -0.637\ldots\le\dfrac{x}{\pi}<0.$$

Because every number in this interval is negative but lies strictly greater than $$-1$$, its greatest-integer part is $$-1$$:

$$\left[\dfrac{x}{\pi}\right]=-1\quad\text{when}\quad -2<x<0.$$

Similarly, for the non-negative half

$$0\le x\le 2\quad\Longrightarrow\quad 0\le\dfrac{x}{\pi}\le\dfrac{2}{\pi}\approx0.637\ldots,$$

and all these numbers have greatest-integer part $$0$$:

$$\left[\dfrac{x}{\pi}\right]=0\quad\text{when}\quad 0\le x\le 2.$$

Consequently the whole denominator takes only two constant values:

For $$x\in(-2,0):\qquad\left[\dfrac{x}{\pi}\right]+\dfrac12=-1+\dfrac12=-\dfrac12.$$

For $$x\in[0,2]:\qquad\left[\dfrac{x}{\pi}\right]+\dfrac12=0+\dfrac12=\dfrac12.$$

Using this piecewise constant denominator, we split the integral:

$$\begin{aligned} I&=\int_{-2}^{0}\frac{\sin^{2}x}{-\dfrac12}\,dx+\int_{0}^{2}\frac{\sin^{2}x}{\dfrac12}\,dx\\[4pt] &=-2\int_{-2}^{0}\sin^{2}x\,dx+2\int_{0}^{2}\sin^{2}x\,dx. \end{aligned}$$

Now recall the basic property that $$\sin^{2}x$$ is an even function, i.e.

$$\sin^{2}(-x)=\sin^{2}x\quad\text{for all }x.$$

For an even integrand the areas on symmetric intervals are equal, so we have

$$\int_{-2}^{0}\sin^{2}x\,dx=\int_{0}^{2}\sin^{2}x\,dx.$$ Let us denote this common value by $$A$$, i.e.

$$A=\int_{0}^{2}\sin^{2}x\,dx.$$

Substituting $$A$$ into the expression for $$I$$, we obtain

$$I=-2A+2A=0.$$

Hence, the correct answer is Option A.