The value of the integral `int_(-2)^2 sin^2x/(-2[x/pi]+1/2)dx` (where [x] denotes the greatest integer less then or equal to x) is

Options

A. 4
B. `4-sin4`
C. `sin4`
D. 0

Answer

To solve the integral

\[ I = \int_{-2}^{2} \frac{\sin^2 x}{-2\left[\frac{x}{\pi}\right] + \frac{1}{2}} \, dx \]

where \([\cdot]\) denotes the greatest integer function, we will break down the integral into segments where the greatest integer function \(\left[\frac{x}{\pi}\right]\) remains constant.

### Step 1: Identify the intervals
The greatest integer function \(\left[\frac{x}{\pi}\right]\) changes at integer multiples of \(\pi\). For the limits \(-2\) to \(2\):

- For \(x \in [-2, -\pi) \approx [-2, -3.14)\), \(\left[\frac{x}{\pi}\right] = -1\)
- For \(x \in [-\pi, 0)\), \(\left[\frac{x}{\pi}\right] = -1\)
- For \(x \in [0, \pi)\), \(\left[\frac{x}{\pi}\right] = 0\)
- For \(x \in [\pi, 2)\), \(\left[\frac{x}{\pi}\right] = 1\)

Since \(\pi \approx 3.14\), we can simplify our intervals as follows:
- \([-2, 0)\)
- \([0, 2)\)

### Step 2: Break the integral
Now we can break the integral into two parts:

\[ I = \int_{-2}^{0} \frac{\sin^2 x}{-2(-1) + \frac{1}{2}} \, dx + \int_{0}^{2} \frac{\sin^2 x}{-2(0) + \frac{1}{2}} \, dx \]

### Step 3: Simplify the integrals
Calculating the denominators:

1. For \(x \in [-2, 0)\):
\[ -2(-1) + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2} \]
Thus, the first integral becomes:
\[ \int_{-2}^{0} \frac{\sin^2 x}{\frac{5}{2}} \, dx = \frac{2}{5} \int_{-2}^{0} \sin^2 x \, dx \]

2. For \(x \in [0, 2)\):
\[ -2(0) + \frac{1}{2} = \frac{1}{2} \]
Thus, the second integral becomes:
\[ \int_{0}^{2} \frac{\sin^2 x}{\frac{1}{2}} \, dx = 2 \int_{0}^{2} \sin^2 x \, dx \]

### Step 4: Combine the integrals
Now we can express \(I\) as:
\[ I = \frac{2}{5} \int_{-2}^{0} \sin^2 x \, dx + 2 \int_{0}^{2} \sin^2 x \, dx \]

### Step 5: Evaluate the integrals
Using the identity \(\sin^2 x = \frac{1 - \cos(2x)}{2}\):
\[ \int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C \]

Calculating the integrals:
1. For \(\int_{-2}^{0} \sin^2 x \, dx\):
\[ = \left[ \frac{x}{2} - \frac{\sin(2x)}{4} \right]_{-2}^{0} = \left(0 - 0\right) - \left(-1 - \frac{\sin(-4)}{4}\right) = 1 + \frac{\sin(4)}{4} \]

2. For \(\int_{0}^{2} \sin^2 x \, dx\):
\[ = \left[ \frac{x}{2} - \frac{\sin(2x)}{4} \right]_{0}^{2} = \left(1 - \frac{\sin(4)}{4}\right) - \left(0 - 0\right) = 1 - \frac{\sin(4)}{4} \]

### Step 6: Substitute back into \(I\)
Now substituting back:
\[ I = \frac{2}{5} \left(1 + \frac{\sin(4)}{4}\right) + 2 \left(1 - \frac{\sin(4)}{4}\right) \]

### Step 7: Simplify
\[ = \frac{2}{5} + \frac{\sin(4)}{10} + 2 - \frac{\sin(4)}{2} \]
Combine terms:
\[ = \frac{2}{5} + 2 - \frac{5\sin(4)}{10} \]
\[ = \frac{2 + 10 - 5\sin(4)}{5} = \frac{12 - 5\sin(4)}{5} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{12 - 5\sin(4)}{5} \]