Evaluate
`int_(0)^(pi)sqrt((cosx+cos2x+cos3x)^(2)+(sinx+sin2x+sin3x)^(2))dx`

Answer

To evaluate the integral

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

we can simplify the expression inside the square root.

### Step 1: Simplifying the Cosine Terms

First, we simplify \(\cos x + \cos 2x + \cos 3x\). We can group \(\cos x\) and \(\cos 3x\):

\[ \cos x + \cos 3x = 2 \cos\left(\frac{x + 3x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \cos(2x) \cos(x). \]

Thus, we have:

\[ \cos x + \cos 2x + \cos 3x = 2 \cos(2x) \cos(x) + \cos(2x) = \cos(2x)(2 \cos(x) + 1). \]

### Step 2: Simplifying the Sine Terms

Now, we simplify \(\sin x + \sin 2x + \sin 3x\):

\[ \sin x + \sin 3x = 2 \sin\left(\frac{x + 3x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \sin(2x) \cos(x). \]

Thus, we have:

\[ \sin x + \sin 2x + \sin 3x = 2 \sin(2x) \cos(x) + \sin(2x) = \sin(2x)(2 \cos(x) + 1). \]

### Step 3: Substituting Back into the Integral

Now substituting these back into the integral:

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

Factoring out \((2 \cos(x) + 1)^2\):

\[ I = \int_{0}^{\pi} |2 \cos(x) + 1| \sqrt{\cos^2(2x) + \sin^2(2x)} \, dx. \]

Since \(\sqrt{\cos^2(2x) + \sin^2(2x)} = 1\), we have:

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

### Step 4: Finding the Points of Change

Next, we need to determine where \(2 \cos(x) + 1 = 0\):

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

This occurs at \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\) (but only \(\frac{2\pi}{3}\) is in the interval \([0, \pi]\)).

### Step 5: Splitting the Integral

We split the integral at \(x = \frac{2\pi}{3}\):

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

### Step 6: Evaluating the Integrals

1. **First Integral**:

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

2. **Second Integral**:

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

This simplifies to:

\[ -\left(\pi - \sqrt{3} - \frac{2\pi}{3}\right) = -\left(\frac{3\pi}{3} - \frac{2\pi}{3} - \sqrt{3}\right) = -\left(\frac{\pi}{3} - \sqrt{3}\right). \]

### Step 7: Combining the Results

Combining both integrals gives:

\[ I = \left(\sqrt{3} + \frac{2\pi}{3}\right) + \left(-\frac{\pi}{3} + \sqrt{3}\right) = 2\sqrt{3} + \frac{2\pi}{3} - \frac{\pi}{3} = 2\sqrt{3} + \frac{\pi}{3}. \]

### Final Answer

Thus, the final answer is:

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