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 follow these steps:

### Step 1: Simplify the expression inside the square root

We can use the identities for the sum of cosines and sines. The sum of cosines can be simplified as follows:

\[ \cos x + \cos 2x + \cos 3x = \cos x + \cos 3x + \cos 2x. \]

Using the formula for the sum of cosines:

\[ \cos a + \cos b = 2 \cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right), \]

we can first combine \(\cos x\) and \(\cos 3x\):

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

Thus,

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

### Step 2: Simplify the sine part

Similarly, for the sine terms:

\[ \sin x + \sin 2x + \sin 3x = \sin x + \sin 3x + \sin 2x. \]

Using the sine addition formula:

\[ \sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right), \]

we combine \(\sin x\) and \(\sin 3x\):

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

Thus,

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

### Step 3: Substitute back into the integral

Now substituting 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: Determine where \(2 \cos x + 1\) is positive or negative

We set \(2 \cos x + 1 = 0\):

\[ \cos x = -\frac{1}{2} \implies x = \frac{2\pi}{3}. \]

Thus, 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 5: Evaluate the integrals

1. **First Integral**:

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

2. **Second Integral**:

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

Since \(\sin \pi = 0\) and \(\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\):

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

### Step 6: Combine the results

Combining both parts:

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

### Final Result

Thus, the value of the integral is:

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