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Step-by-Step Solution
Step 1: Write down the integral
We need to evaluate
$$
I = \int_{-2}^{2} \frac{\sin^2 x}{\left[ \frac{x}{\pi} \right] + \frac{1}{2}} \, dx,
$$
where $[\,\cdot\,]$ denotes the greatest integer (floor) function.
Step 2: Recognize symmetry and split the integral
Notice the integration limits go from $-2$ to $2$, and $\sin^2x$ is an even function, i.e. $\sin^2(-x) = \sin^2 x$. We use this fact to break the integral into two parts:
$$
I = \int_{-2}^{2} \frac{\sin^2 x}{\left[ \frac{x}{\pi} \right] + \frac{1}{2}} \, dx
= \int_{0}^{2} \frac{\sin^2 x}{\left[ \frac{x}{\pi} \right] + \frac{1}{2}} \, dx
+ \int_{0}^{2} \frac{\sin^2(-t)}{\left[ \frac{-t}{\pi} \right] + \frac{1}{2}} \, (-dt),
$$
where in the second term we substitute $x = -t$ so that when $x=-2 \rightarrow t=2$ and when $x=0 \rightarrow t=0.$ Changing the variable recasts the second integral with the same limits $0$ to $2$ but an extra negative sign, so:
$$
I = \int_{0}^{2} \frac{\sin^2 x}{\left[ \frac{x}{\pi} \right] + \frac{1}{2}} \, dx
+ \int_{0}^{2} \frac{\sin^2(-t)}{\left[ -\frac{t}{\pi} \right] + \frac{1}{2}} \, dt.
$$
Since $\sin^2(-t) = \sin^2 t$, we write
$$
I = \int_{0}^{2} \left[ \frac{\sin^2 x}{\left[ \frac{x}{\pi} \right] + \frac{1}{2}}
+ \frac{\sin^2 x}{\left[ -\frac{x}{\pi} \right] + \frac{1}{2}} \right] dx.
$$
Step 3: Use the property of the floor function
A crucial property for $x \neq n\pi$ (which never occurs in our interval $[-2,2]$ since $\pi \approx 3.14$ and there is no integer multiple of $\pi$ between $-2$ and $2$) is:
$$
\left\lfloor \frac{x}{\pi} \right\rfloor
+ \left\lfloor -\frac{x}{\pi} \right\rfloor = -1.
$$
Therefore, if we let
$$
A = \left\lfloor \frac{x}{\pi} \right\rfloor + \frac{1}{2},
\quad
B = \left\lfloor -\frac{x}{\pi} \right\rfloor + \frac{1}{2},
$$
then
$$
A + B = \left\lfloor \frac{x}{\pi} \right\rfloor + \left\lfloor -\frac{x}{\pi} \right\rfloor
+ 1 = -1 + 1 = 0.
$$
Step 4: Show that the integrand sums to zero
Notice that under the integral from $0$ to $2$,
$$
\frac{\sin^2 x}{A} + \frac{\sin^2 x}{B} = \sin^2 x \left( \frac{1}{A} + \frac{1}{B} \right).
$$
Since $A + B = 0$, we can rewrite
$$
\frac{1}{A} + \frac{1}{B}
= \frac{A + B}{AB}
= \frac{0}{AB}
= 0.
$$
Hence each pair of terms in the integrand adds up to zero for all $x$ in $[0,2]$.
Step 5: Conclude the value of the integral
Because the integrand’s overall sum is identically zero, the integral over the entire interval $[-2,2]$ must be zero:
$$
I = \int_{-2}^{2} \frac{\sin^2 x}{\left\lfloor \frac{x}{\pi} \right\rfloor + \frac{1}{2}} \, dx = 0.
$$
Final Answer
$\boxed{0}$
