`int_(0)^(2pi)[|sin x|+|cos x|]dx`, where [.] denotes the greatest integer function, is equal to :

Options

A. `pi`
B. `2pi`
C. `pi//sqrt(2)`
D. `pi sqrt(2)`

Answer

To solve the integral \( \int_{0}^{2\pi} \left\lfloor |\sin x| + |\cos x| \right\rfloor \, dx \), we will follow these steps:

### Step 1: Analyze the function \( |\sin x| + |\cos x| \)

The function \( |\sin x| + |\cos x| \) varies between certain bounds. We know that both \( |\sin x| \) and \( |\cos x| \) reach their maximum value of 1. Therefore, we can establish the following inequalities:

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

### Step 2: Determine the range of \( |\sin x| + |\cos x| \)

The minimum value occurs when either \( \sin x \) or \( \cos x \) is zero, and the other is 1. The maximum value occurs when both \( \sin x \) and \( \cos x \) are equal, which happens at \( x = \frac{\pi}{4} + n\frac{\pi}{2} \) for integers \( n \).

Calculating the maximum:

\[ |\sin x| + |\cos x| = \sqrt{2} \quad \text{(at } x = \frac{\pi}{4}, \frac{5\pi}{4} \text{)} \]

Thus, we have:

\[ 1 \leq |\sin x| + |\cos x| < \sqrt{2} \approx 1.414 \]

### Step 3: Apply the greatest integer function

Since \( |\sin x| + |\cos x| \) is always greater than or equal to 1 and less than \( \sqrt{2} \), we can conclude:

\[ \left\lfloor |\sin x| + |\cos x| \right\rfloor = 1 \quad \text{for all } x \in [0, 2\pi] \]

### Step 4: Evaluate the integral

Now, we can evaluate the integral:

\[ \int_{0}^{2\pi} \left\lfloor |\sin x| + |\cos x| \right\rfloor \, dx = \int_{0}^{2\pi} 1 \, dx \]

Calculating this integral:

\[ \int_{0}^{2\pi} 1 \, dx = [x]_{0}^{2\pi} = 2\pi - 0 = 2\pi \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{2\pi} \]