If `f:R in R` be such that
f(x)`=sqrt(sin(cosx))+"In"(-2cos^(2) x+3 cos x-1)`, then `int_(x_(1))^(x_(2)) [cos x-(1)/(2)]dx` is equal to, where `x1.x2 in D` and [.] denotes the greatest integer function,

Options

A. 0
B. `(1)/(2)(x_(2)-x_(1))`
C. `x_(1)-x_(2)`
D. `(1)/(2)(x_(1)-x_(2))`

Explanation

To solve the problem, we need to evaluate the integral

\[ \int_{x_1}^{x_2} \left( \cos x - \frac{1}{2} \right) dx \]

where \( x_1 \) and \( x_2 \) are such that the expression inside the greatest integer function, \( \left\lfloor \cos x - \frac{1}{2} \right\rfloor \), is defined.

### Step 1: Analyze the function \( f(x) \)

Given:

\[ f(x) = \sqrt{\sin(\cos x)} + \ln(-2\cos^2 x + 3\cos x - 1) \]

We need to ensure that the logarithmic part is defined. The logarithm is defined when its argument is positive:

\[ -2\cos^2 x + 3\cos x - 1 > 0 \]

### Step 2: Solve the inequality

Rearranging the inequality gives:

\[ 2\cos^2 x - 3\cos x + 1 < 0 \]

Let \( y = \cos x \). The quadratic inequality becomes:

\[ 2y^2 - 3y + 1 < 0 \]

To find the roots, we use the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]

Thus, the roots are:

\[ y = 1 \quad \text{and} \quad y = \frac{1}{2} \]

### Step 3: Determine the intervals

The quadratic \( 2y^2 - 3y + 1 \) opens upwards (as the coefficient of \( y^2 \) is positive). Therefore, it is negative between the roots:

\[ \frac{1}{2} < y < 1 \]

This translates back to:

\[ \frac{1}{2} < \cos x < 1 \]

### Step 4: Identify the intervals for \( x \)

The condition \( \frac{1}{2} < \cos x < 1 \) corresponds to:

\[ 0 < x < \frac{\pi}{3} \quad \text{and} \quad \frac{5\pi}{3} < x < 2\pi \]

### Step 5: Evaluate the integral

Now we need to evaluate:

\[ \int_{x_1}^{x_2} \left( \cos x - \frac{1}{2} \right) dx \]

Given that \( \cos x \) varies between \( \frac{1}{2} \) and \( 1 \) in the intervals we identified, we can find the greatest integer function:

\[ \cos x - \frac{1}{2} > 0 \]

This means:

\[ \left\lfloor \cos x - \frac{1}{2} \right\rfloor = 0 \]

### Step 6: Conclusion

Thus, the integral simplifies to:

\[ \int_{x_1}^{x_2} 0 \, dx = 0 \]

The final answer is:

\[ \boxed{0} \]