`int_(0)^(2pi)|sqrt(15) sin x + cos x|dx`

Answer

To solve the integral \( I = \int_{0}^{2\pi} |\sqrt{15} \sin x + \cos x| \, dx \), we will follow these steps:

### Step 1: Rewrite the expression inside the absolute value
We can express the integrand in a different form. Notice that:
\[ \sqrt{15} \sin x + \cos x = 4 \left( \frac{\sqrt{15}}{4} \sin x + \frac{1}{4} \cos x \right) \]
This can be rewritten using the sine addition formula:
\[ \sqrt{15} \sin x + \cos x = 4 \sin\left(x + \alpha\right) \]
where \( \alpha \) is an angle such that:
\[ \cos \alpha = \frac{1}{4}, \quad \sin \alpha = \frac{\sqrt{15}}{4} \]

### Step 2: Change of variable
Let \( t = x + \alpha \). Then, \( dx = dt \). The limits change as follows:
- When \( x = 0 \), \( t = \alpha \)
- When \( x = 2\pi \), \( t = 2\pi + \alpha \)

Thus, we can rewrite the integral:
\[ I = \int_{\alpha}^{2\pi + \alpha} |\sqrt{15} \sin(t - \alpha) + \cos(t - \alpha)| \, dt \]
Since the sine function is periodic, we can use the property of integrals over a full period:
\[ I = \int_{0}^{2\pi} |\sqrt{15} \sin t + \cos t| \, dt \]

### Step 3: Determine where the expression changes sign
To evaluate the integral, we need to find when \( \sqrt{15} \sin t + \cos t = 0 \):
\[ \sqrt{15} \sin t = -\cos t \implies \tan t = -\frac{1}{\sqrt{15}} \]
This gives us two points in the interval \( [0, 2\pi] \). Let's denote these points as \( t_1 \) and \( t_2 \).

### Step 4: Evaluate the integral in segments
We can split the integral at the points \( t_1 \) and \( t_2 \):
\[ I = \int_{0}^{t_1} -(\sqrt{15} \sin t + \cos t) \, dt + \int_{t_1}^{t_2} (\sqrt{15} \sin t + \cos t) \, dt + \int_{t_2}^{2\pi} -(\sqrt{15} \sin t + \cos t) \, dt \]

### Step 5: Calculate each integral
1. For \( \int_{0}^{t_1} -(\sqrt{15} \sin t + \cos t) \, dt \)
2. For \( \int_{t_1}^{t_2} (\sqrt{15} \sin t + \cos t) \, dt \)
3. For \( \int_{t_2}^{2\pi} -(\sqrt{15} \sin t + \cos t) \, dt \)

### Step 6: Combine results
After calculating the definite integrals, add the results together to find the value of \( I \).

### Final Result
The final value of the integral \( I \) will be \( 16 \).