The value of `int_(0)^(2pi) |cos x -sin x|dx`is

Options

A. `(4)/(sqrt(2))`
B. `2sqrt(2)`
C. `(2)/(sqrt(2))`
D. `4sqrt(2)`

Answer

To solve the integral \( I = \int_{0}^{2\pi} |\cos x - \sin x| \, dx \), we will follow these steps:

### Step 1: Identify the intervals where \( \cos x - \sin x \) changes sign

To determine where \( |\cos x - \sin x| \) changes, we need to find the points where \( \cos x - \sin x = 0 \).

\[ \cos x = \sin x \]

This occurs when:

\[ \tan x = 1 \implies x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

Within the interval \( [0, 2\pi] \), the relevant points are:

\[ x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4} \]

### Step 2: Determine the sign of \( \cos x - \sin x \) in each interval

Now we evaluate the sign of \( \cos x - \sin x \) in the intervals:

1. **Interval \( [0, \frac{\pi}{4}] \)**:
- Choose \( x = 0 \):
\[ \cos(0) - \sin(0) = 1 - 0 = 1 \quad (\text{positive}) \]

2. **Interval \( [\frac{\pi}{4}, \frac{5\pi}{4}] \)**:
- Choose \( x = \pi \):
\[ \cos(\pi) - \sin(\pi) = -1 - 0 = -1 \quad (\text{negative}) \]

3. **Interval \( [\frac{5\pi}{4}, 2\pi] \)**:
- Choose \( x = \frac{3\pi}{2} \):
\[ \cos\left(\frac{3\pi}{2}\right) - \sin\left(\frac{3\pi}{2}\right) = 0 - (-1) = 1 \quad (\text{positive}) \]

### Step 3: Rewrite the integral based on the sign

Now we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx - \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{5\pi}{4}}^{2\pi} (\cos x - \sin x) \, dx \]

### Step 4: Calculate each integral

1. **Integral from \( 0 \) to \( \frac{\pi}{4} \)**:
\[ \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_{0}^{\frac{\pi}{4}} = \left( \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) \right) - (0 + 1) = \sqrt{2} - 1 \]

2. **Integral from \( \frac{\pi}{4} \) to \( \frac{5\pi}{4} \)**:
\[ \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} = \left( \sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right) \right) - \left( \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) \right) \]
\[ = \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) = -\sqrt{2} - \sqrt{2} = -2\sqrt{2} \]

3. **Integral from \( \frac{5\pi}{4} \) to \( 2\pi \)**:
\[ \int_{\frac{5\pi}{4}}^{2\pi} (\cos x - \sin x) \, dx = \left[ \sin x + \cos x \right]_{\frac{5\pi}{4}}^{2\pi} = \left( \sin(2\pi) + \cos(2\pi) \right) - \left( \sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right) \right) \]
\[ = (0 + 1) - \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = 1 + \sqrt{2} \]

### Step 5: Combine the results

Now we combine all the results:

\[ I = \left( \sqrt{2} - 1 \right) - (-2\sqrt{2}) + (1 + \sqrt{2}) = \sqrt{2} - 1 + 2\sqrt{2} + 1 + \sqrt{2} = 4\sqrt{2} \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{4\sqrt{2}} \]