The value of `int_(-2)^(2)(sin^(2)x)/([(x)/(pi)]+(1)/(2))dx` where [.] denotes greatest integer function , is

Options

A. 1
B. 0
C. `4 sin4`
D. None of these

Answer

To solve the integral

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

we will determine whether the function \( f(x) = \frac{\sin^2 x}{\left(\frac{x}{\pi}\right) + \frac{1}{2}} \) is odd or even.

### Step 1: Check if \( f(x) \) is odd or even

To check if \( f(x) \) is odd, we need to evaluate \( f(-x) \):

\[ f(-x) = \frac{\sin^2(-x)}{\left(\frac{-x}{\pi}\right) + \frac{1}{2}}. \]

Using the property of sine, we know that \( \sin(-x) = -\sin(x) \), thus:

\[ \sin^2(-x) = \sin^2(x). \]

Now substituting this into \( f(-x) \):

\[ f(-x) = \frac{\sin^2 x}{\left(-\frac{x}{\pi}\right) + \frac{1}{2}} = \frac{\sin^2 x}{-\frac{x}{\pi} + \frac{1}{2}}. \]

### Step 2: Simplify \( f(-x) \)

We can rewrite \( f(-x) \):

\[ f(-x) = \frac{\sin^2 x}{-\left(\frac{x}{\pi} - \frac{1}{2}\right)} = -\frac{\sin^2 x}{\left(\frac{x}{\pi} - \frac{1}{2}\right)}. \]

Now, we can express \( f(-x) \) in terms of \( f(x) \):

\[ f(-x) = -\frac{\sin^2 x}{\left(\frac{x}{\pi} - \frac{1}{2}\right)} = -f(x). \]

### Step 3: Conclusion about the function

Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 4: Evaluating the integral

For any odd function \( f(x) \), the integral over a symmetric interval \([-a, a]\) is zero:

\[ \int_{-a}^{a} f(x) \, dx = 0. \]

Thus, we have:

\[ I = \int_{-2}^{2} f(x) \, dx = 0. \]

### Final Answer

The value of the integral is

\[ \boxed{0}. \]