Let f(x)=In`[cos|x|+(1)/(2)]` where [.] denotes the greatest integer function, then `int_(x_(1))^(x_(2)) lim_(nto oo)(({f(x)}^(n))/(x^(2)+tan^(2)x))dx` is equal to, where `x_(1),x_(2) in [(-pi)/(6),(pi)/(6)]`

Options

A. 1
B. 2
C. -1
D. 0

Explanation

To solve the given problem step by step, we will analyze the function \( f(x) = \ln\left[\cos|x| + \frac{1}{2}\right] \) and evaluate the integral

\[ \int_{x_1}^{x_2} \lim_{n \to \infty} \frac{(f(x))^n}{x^2 + \tan^2 x} \, dx \]

where \( x_1, x_2 \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \).

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

First, we need to evaluate \( f(x) \):

\[ f(x) = \ln\left[\cos|x| + \frac{1}{2}\right] \]

For \( x \) in the interval \( \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \), \( |x| = x \). Thus, we can simplify \( f(x) \):

- At \( x = 0 \), \( f(0) = \ln\left[\cos(0) + \frac{1}{2}\right] = \ln\left[1 + \frac{1}{2}\right] = \ln\left(\frac{3}{2}\right) \).
- At \( x = \pm \frac{\pi}{6} \), \( f\left(\pm \frac{\pi}{6}\right) = \ln\left[\cos\left(\frac{\pi}{6}\right) + \frac{1}{2}\right] = \ln\left[\frac{\sqrt{3}}{2} + \frac{1}{2}\right] = \ln\left[\frac{\sqrt{3} + 1}{2}\right] \).

Now, we need to find the range of \( \cos|x| \) in this interval.

- The maximum value of \( \cos|x| \) is \( 1 \) (at \( x = 0 \)).
- The minimum value of \( \cos|x| \) is \( \frac{\sqrt{3}}{2} \) (at \( x = \pm \frac{\pi}{6} \)).

Thus, we have:

\[ \frac{\sqrt{3}}{2} + \frac{1}{2} \leq \cos|x| + \frac{1}{2} \leq 1 + \frac{1}{2} \]

Calculating these values:

- \( \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \) (which is greater than \( 1 \)).
- \( 1 + \frac{1}{2} = \frac{3}{2} \).

### Step 2: Greatest Integer Function

Now, we apply the greatest integer function:

\[ \lfloor \cos|x| + \frac{1}{2} \rfloor \]

Since \( \frac{\sqrt{3} + 1}{2} \) is approximately \( 1.366 \) and \( \frac{3}{2} = 1.5 \), we find:

\[ \lfloor \cos|x| + \frac{1}{2} \rfloor = 1 \]

Thus,

\[ f(x) = \ln(1) = 0 \]

### Step 3: Evaluate the Limit and Integral

Now substituting \( f(x) \) into the integral:

\[ \lim_{n \to \infty} \frac{(f(x))^n}{x^2 + \tan^2 x} = \lim_{n \to \infty} \frac{0^n}{x^2 + \tan^2 x} = 0 \]

for all \( x \) in the interval \( \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \).

### Step 4: Final Integral Evaluation

Thus, the integral simplifies to:

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

### Conclusion

The value of the integral is:

\[ \boxed{0} \]