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

Answer

To solve the integral \( \int_{0}^{\pi} |1 + 2 \cos x| \, dx \), we will follow these steps:

### Step 1: Identify where the expression inside the modulus is zero
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 solutions to \( \cos x = -\frac{1}{2} \) in the interval \( [0, \pi] \) is:

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

### Step 2: Determine the sign of \( 1 + 2 \cos x \) in the intervals
We will analyze the sign of \( 1 + 2 \cos x \) in the intervals \( [0, \frac{2\pi}{3}] \) and \( [\frac{2\pi}{3}, \pi] \).

- For \( x \in [0, \frac{2\pi}{3}] \):
- Choose \( x = 0 \): \( 1 + 2 \cos(0) = 1 + 2(1) = 3 > 0 \)

- For \( x \in [\frac{2\pi}{3}, \pi] \):
- Choose \( x = \pi \): \( 1 + 2 \cos(\pi) = 1 + 2(-1) = -1 < 0 \)

Thus, we can conclude:
- \( 1 + 2 \cos x \geq 0 \) in \( [0, \frac{2\pi}{3}] \)
- \( 1 + 2 \cos x < 0 \) in \( [\frac{2\pi}{3}, \pi] \)

### Step 3: Rewrite the integral using the modulus
Now we can express the integral as:

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

### Step 4: Evaluate 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. \( \int_{0}^{\frac{2\pi}{3}} 1 \, dx = \left[ x \right]_{0}^{\frac{2\pi}{3}} = \frac{2\pi}{3} \)
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,
\[ \int_{0}^{\frac{2\pi}{3}} (1 + 2 \cos x) \, dx = \frac{2\pi}{3} + 2 \cdot \frac{\sqrt{3}}{2} = \frac{2\pi}{3} + \sqrt{3} \]

### Step 5: Evaluate the second integral
\[ \int_{\frac{2\pi}{3}}^{\pi} -(1 + 2 \cos x) \, dx = -\int_{\frac{2\pi}{3}}^{\pi} (1 + 2 \cos x) \, dx \]

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

### Step 6: Combine the results
Now we combine both integrals:
\[ \int_{0}^{\pi} |1 + 2 \cos x| \, dx = \left( \frac{2\pi}{3} + \sqrt{3} \right) + \left( -\frac{\pi}{3} + \sqrt{3} \right) \]

This simplifies to:
\[ \frac{2\pi}{3} - \frac{\pi}{3} + \sqrt{3} + \sqrt{3} = \frac{\pi}{3} + 2\sqrt{3} \]

### Final Result
Thus, the value of the integral is:
\[ \int_{0}^{\pi} |1 + 2 \cos x| \, dx = \frac{\pi}{3} + 2\sqrt{3} \]

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