`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 \( \int_{0}^{2\pi} \lfloor |\sin x| + |\cos x| \rfloor \, dx \), where \( \lfloor . \rfloor \) denotes the greatest integer function, we will follow these steps:

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

We know that both \( |\sin x| \) and \( |\cos x| \) are non-negative and oscillate between 0 and 1. The maximum value of \( |\sin x| + |\cos x| \) occurs when both sine and cosine are at their maximum values.

### Step 2: Find the maximum and minimum values of \( |\sin x| + |\cos x| \)

Using the Cauchy-Schwarz inequality, we can establish that:
\[ |\sin x| + |\cos x| \leq \sqrt{(|\sin x|^2 + |\cos x|^2)(1^2 + 1^2)} = \sqrt{1 \cdot 2} = \sqrt{2} \]
Thus, we have:
\[ 1 \leq |\sin x| + |\cos x| \leq \sqrt{2} \]

### Step 3: Determine the range of \( \lfloor |\sin x| + |\cos x| \rfloor \)

Since \( \sqrt{2} \approx 1.414 \), we can conclude that:
\[ 1 \leq |\sin x| + |\cos x| < 2 \]
This implies that:
\[ \lfloor |\sin x| + |\cos x| \rfloor = 1 \]
for all \( x \) in the interval \( [0, 2\pi] \).

### Step 4: Set up the integral

Now we can substitute this result into the integral:
\[ \int_{0}^{2\pi} \lfloor |\sin x| + |\cos x| \rfloor \, dx = \int_{0}^{2\pi} 1 \, dx \]

### Step 5: Evaluate the integral

The integral simplifies to:
\[ \int_{0}^{2\pi} 1 \, dx = [x]_{0}^{2\pi} = 2\pi - 0 = 2\pi \]

### Conclusion

Thus, the value of the integral is:
\[ \int_{0}^{2\pi} \lfloor |\sin x| + |\cos x| \rfloor \, dx = 2\pi \]