If `x in[(4n+1)(pi)/(2),(4n+3)(pi)/(2)]` and `n in N`, then the value of `int_(0)^(x) [cos t] dt`, is

Options

A. `(2n-1)(pi)/(2)-x`
B. `(2n-1)(pi)/(pi)/(2)-x`
C. `(2n+1)(pi)/(2)-x`
D. `(2n+1)(pi)/(2)+x`

Explanation

To solve the problem, we need to evaluate the definite integral of the greatest integer function of the cosine function over the specified limits. Let's break it down step by step.

### Step 1: Understand the limits of integration
Given that \( x \) is in the interval \(\left[\frac{(4n+1)\pi}{2}, \frac{(4n+3)\pi}{2}\right]\), we can identify that \( x \) is between two specific points on the cosine wave.

### Step 2: Break down the integral
We can express the integral as:
\[ I = \int_0^x \lfloor \cos t \rfloor \, dt \]
Since \( \lfloor \cos t \rfloor \) takes values based on the cosine function, we need to analyze the behavior of \(\cos t\) within the limits of integration.

### Step 3: Determine the behavior of \(\cos t\)
The cosine function oscillates between -1 and 1. In the interval \(\left[0, \frac{(4n+3)\pi}{2}\right]\), the cosine function will take the following values:
- From \(0\) to \(\frac{\pi}{2}\), \(\cos t\) decreases from \(1\) to \(0\).
- From \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\), \(\cos t\) decreases from \(0\) to \(-1\).
- From \(\frac{3\pi}{2}\) to \(2\pi\), \(\cos t\) increases from \(-1\) to \(0\).

### Step 4: Split the integral based on the intervals
We can split the integral at the points where \(\cos t\) changes its value:
\[ I = \int_0^{\frac{\pi}{2}} \lfloor \cos t \rfloor \, dt + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \lfloor \cos t \rfloor \, dt + \int_{\frac{3\pi}{2}}^{x} \lfloor \cos t \rfloor \, dt \]

### Step 5: Evaluate each integral
1. **From \(0\) to \(\frac{\pi}{2}\)**:
\(\lfloor \cos t \rfloor = 1\) for \(t \in [0, \frac{\pi}{2})\).
\[ \int_0^{\frac{\pi}{2}} 1 \, dt = \frac{\pi}{2} \]

2. **From \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\)**:
\(\lfloor \cos t \rfloor = -1\) for \(t \in (\frac{\pi}{2}, \frac{3\pi}{2})\).
\[ \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} -1 \, dt = -\left(\frac{3\pi}{2} - \frac{\pi}{2}\right) = -\pi \]

3. **From \(\frac{3\pi}{2}\) to \(x\)**:
For \(t \in \left(\frac{3\pi}{2}, x\right)\), \(\lfloor \cos t \rfloor = -1\) since \(x\) is in the interval where \(\cos t\) is negative.
\[ \int_{\frac{3\pi}{2}}^{x} -1 \, dt = -\left(x - \frac{3\pi}{2}\right) = -x + \frac{3\pi}{2} \]

### Step 6: Combine the results
Now, we can combine all parts:
\[ I = \frac{\pi}{2} - \pi - x + \frac{3\pi}{2} \]
Simplifying this gives:
\[ I = -x + \frac{3\pi}{2} - \pi + \frac{\pi}{2} = -x + \frac{\pi}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{\pi}{2} - x \]