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 integral \( \int_0^x \lfloor \cos t \rfloor \, dt \) where \( x \) is in the interval \( \left[\frac{(4n+1)\pi}{2}, \frac{(4n+3)\pi}{2}\right] \) and \( n \in \mathbb{N} \), we will follow these steps:

### Step 1: Understand the behavior of \( \cos t \)
The cosine function oscillates between -1 and 1. Specifically, in the intervals of interest:
- 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 and then back to 0.
- From \( \frac{3\pi}{2} \) to \( 2\pi \), \( \cos t \) increases from -1 to 1.

### Step 2: Determine the value of \( \lfloor \cos t \rfloor \)
In the interval \( \left[\frac{(4n+1)\pi}{2}, \frac{(4n+3)\pi}{2}\right] \):
- At \( t = \frac{(4n+1)\pi}{2} \), \( \cos t = 0 \).
- At \( t = \frac{(4n+3)\pi}{2} \), \( \cos t = 0 \).
- Between these points, \( \cos t \) takes negative values, specifically from \( 0 \) to \( -1 \).

Thus, \( \lfloor \cos t \rfloor = -1 \) for \( t \) in the interval \( \left(\frac{(4n+1)\pi}{2}, \frac{(4n+3)\pi}{2}\right) \).

### Step 3: Set up the integral
Now we can set up the integral:
\[ \int_0^x \lfloor \cos t \rfloor \, dt = \int_0^{\frac{(4n+1)\pi}{2}} \lfloor \cos t \rfloor \, dt + \int_{\frac{(4n+1)\pi}{2}}^x \lfloor \cos t \rfloor \, dt \]

### Step 4: Evaluate the first part of the integral
For \( t \) from \( 0 \) to \( \frac{(4n+1)\pi}{2} \):
- \( \lfloor \cos t \rfloor = 0 \) for \( t \in [0, \frac{\pi}{2}) \).
- Thus, \( \int_0^{\frac{(4n+1)\pi}{2}} \lfloor \cos t \rfloor \, dt = 0 \).

### Step 5: Evaluate the second part of the integral
For \( t \) from \( \frac{(4n+1)\pi}{2} \) to \( x \):
\[ \int_{\frac{(4n+1)\pi}{2}}^x \lfloor \cos t \rfloor \, dt = \int_{\frac{(4n+1)\pi}{2}}^x (-1) \, dt = -\left(t \bigg|_{\frac{(4n+1)\pi}{2}}^x\right) = -\left(x - \frac{(4n+1)\pi}{2}\right) \]
Thus,
\[ \int_{\frac{(4n+1)\pi}{2}}^x \lfloor \cos t \rfloor \, dt = -x + \frac{(4n+1)\pi}{2} \]

### Step 6: Combine the results
Combining both parts, we have:
\[ \int_0^x \lfloor \cos t \rfloor \, dt = 0 - x + \frac{(4n+1)\pi}{2} = \frac{(4n+1)\pi}{2} - x \]

### Final Result
Thus, the value of the integral is:
\[ \int_0^x \lfloor \cos t \rfloor \, dt = \frac{(4n+1)\pi}{2} - x \]