`int_(0)^(pi)|1+2cosx|` dx is equal to :

Options

A. `(2pi)/(3)`
B. `pi`
C. `2`
D. `(pi)/(3) + 2sqrt(3)`

Answer

To solve the integral \( \int_{0}^{\pi} |1 + 2 \cos x| \, dx \), we need to analyze the expression inside the absolute value to determine where it is positive and where it is negative.

### Step 1: Determine where \( 1 + 2 \cos x \) is positive or negative
The expression \( 1 + 2 \cos x \) can change sign depending on the value of \( \cos x \). We need to find the points where \( 1 + 2 \cos x = 0 \).

\[ 1 + 2 \cos x = 0 \implies 2 \cos x = -1 \implies \cos x = -\frac{1}{2} \]

The values of \( x \) where \( \cos x = -\frac{1}{2} \) in the interval \( [0, \pi] \) are:

\[ x = \frac{2\pi}{3} \]

### Step 2: Split the integral at the point \( \frac{2\pi}{3} \)
Now we can split the integral into two parts:

\[ \int_{0}^{\pi} |1 + 2 \cos x| \, dx = \int_{0}^{\frac{2\pi}{3}} (1 + 2 \cos x) \, dx + \int_{\frac{2\pi}{3}}^{\pi} -(1 + 2 \cos x) \, dx \]

### Step 3: Evaluate the first integral \( \int_{0}^{\frac{2\pi}{3}} (1 + 2 \cos x) \, dx \)
Now we compute the first integral:

\[ \int_{0}^{\frac{2\pi}{3}} (1 + 2 \cos x) \, dx = \int_{0}^{\frac{2\pi}{3}} 1 \, dx + 2 \int_{0}^{\frac{2\pi}{3}} \cos x \, dx \]

Calculating each part:

1. \( \int_{0}^{\frac{2\pi}{3}} 1 \, dx = \left[ x \right]_{0}^{\frac{2\pi}{3}} = \frac{2\pi}{3} - 0 = \frac{2\pi}{3} \)
2. \( \int_{0}^{\frac{2\pi}{3}} \cos x \, dx = \left[ \sin x \right]_{0}^{\frac{2\pi}{3}} = \sin\left(\frac{2\pi}{3}\right) - \sin(0) = \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2} \)

Thus,

\[ 2 \int_{0}^{\frac{2\pi}{3}} \cos x \, dx = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]

Combining these results:

\[ \int_{0}^{\frac{2\pi}{3}} (1 + 2 \cos x) \, dx = \frac{2\pi}{3} + \sqrt{3} \]

### Step 4: Evaluate the second integral \( \int_{\frac{2\pi}{3}}^{\pi} -(1 + 2 \cos x) \, dx \)
Now we compute the second integral:

\[ \int_{\frac{2\pi}{3}}^{\pi} -(1 + 2 \cos x) \, dx = -\left( \int_{\frac{2\pi}{3}}^{\pi} 1 \, dx + 2 \int_{\frac{2\pi}{3}}^{\pi} \cos x \, dx \right) \]

Calculating each part:

1. \( \int_{\frac{2\pi}{3}}^{\pi} 1 \, dx = \left[ x \right]_{\frac{2\pi}{3}}^{\pi} = \pi - \frac{2\pi}{3} = \frac{\pi}{3} \)
2. \( \int_{\frac{2\pi}{3}}^{\pi} \cos x \, dx = \left[ \sin x \right]_{\frac{2\pi}{3}}^{\pi} = \sin(\pi) - \sin\left(\frac{2\pi}{3}\right) = 0 - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2} \)

Thus,

\[ 2 \int_{\frac{2\pi}{3}}^{\pi} \cos x \, dx = 2 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \]

Combining these results:

\[ \int_{\frac{2\pi}{3}}^{\pi} -(1 + 2 \cos x) \, dx = -\left( \frac{\pi}{3} - \sqrt{3} \right) = -\frac{\pi}{3} + \sqrt{3} \]

### Step 5: Combine both parts
Now we combine both parts of the integral:

\[ \int_{0}^{\pi} |1 + 2 \cos x| \, dx = \left( \frac{2\pi}{3} + \sqrt{3} \right) + \left( -\frac{\pi}{3} + \sqrt{3} \right) \]

Simplifying this:

\[ = \frac{2\pi}{3} - \frac{\pi}{3} + \sqrt{3} + \sqrt{3} = \frac{\pi}{3} + 2\sqrt{3} \]

### Final Answer
Thus, the value of the integral is:

\[ \int_{0}^{\pi} |1 + 2 \cos x| \, dx = \frac{\pi}{3} + 2\sqrt{3} \]