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 need to determine where the expression inside the modulus changes sign.

### Step 1: Find the points where \( \cos x = \sin x \)

We start by solving the equation \( \cos x = \sin x \). 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 points where \( \cos x = \sin x \) are:

- \( x = \frac{\pi}{4} \)
- \( x = \frac{5\pi}{4} \)

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

We will evaluate the sign of \( \cos x - \sin x \) in the intervals \( [0, \frac{\pi}{4}] \), \( [\frac{\pi}{4}, \frac{5\pi}{4}] \), and \( [\frac{5\pi}{4}, 2\pi] \).

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

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

3. **Interval \( [\frac{5\pi}{4}, 2\pi] \)**:
- Choose \( x = \frac{3\pi}{2} \): \( \cos(\frac{3\pi}{2}) - \sin(\frac{3\pi}{2}) = 0 - (-1) = 1 > 0 \) (positive)
- Thus, \( |\cos x - \sin x| = \cos x - \sin x \)

### Step 3: Set up the integral with the determined expressions

Now we can write the integral as:

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

### Step 4: Evaluate each integral

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

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

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

### Step 5: Combine all parts

Now we combine all the results:

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

Thus, the value of the integral is:

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