Find the area between the x-axis and the curve ` y = sqrt(1+cos4x), 0 le x le pi`.

Answer

To find the area between the x-axis and the curve \( y = \sqrt{1 + \cos(4x)} \) for \( 0 \leq x \leq \pi \), we will follow these steps:

### Step 1: Simplify the Function
We start with the function:
\[ y = \sqrt{1 + \cos(4x)} \]
Using the double angle identity for cosine, we can rewrite \( \cos(4x) \):
\[ \cos(4x) = 2\cos^2(2x) - 1 \]
Substituting this back into the equation gives:
\[ y = \sqrt{1 + (2\cos^2(2x) - 1)} = \sqrt{2\cos^2(2x)} = \sqrt{2} |\cos(2x)| \]

### Step 2: Determine the Interval for Integration
Next, we need to analyze the function \( |\cos(2x)| \) over the interval \( [0, \pi] \). The function \( \cos(2x) \) oscillates between -1 and 1, and we need to find the points where it changes sign:
\[ \cos(2x) = 0 \Rightarrow 2x = \frac{\pi}{2} + n\pi \Rightarrow x = \frac{\pi}{4} + \frac{n\pi}{2} \]
In the interval \( [0, \pi] \), the relevant points are \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \).

### Step 3: Set Up the Integral
We will split the integral into two parts based on the sign of \( \cos(2x) \):
1. From \( 0 \) to \( \frac{\pi}{4} \), \( \cos(2x) \) is positive.
2. From \( \frac{\pi}{4} \) to \( \frac{3\pi}{4} \), \( \cos(2x) \) is negative.
3. From \( \frac{3\pi}{4} \) to \( \pi \), \( \cos(2x) \) is again positive.

Thus, the area \( A \) can be expressed as:
\[ A = \int_0^{\frac{\pi}{4}} \sqrt{2} \cos(2x) \, dx + \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} -\sqrt{2} \cos(2x) \, dx + \int_{\frac{3\pi}{4}}^{\pi} \sqrt{2} \cos(2x) \, dx \]

### Step 4: Calculate Each Integral
1. **First Integral**:
\[ \int_0^{\frac{\pi}{4}} \sqrt{2} \cos(2x) \, dx = \sqrt{2} \left[ \frac{\sin(2x)}{2} \right]_0^{\frac{\pi}{4}} = \sqrt{2} \left( \frac{\sin(\frac{\pi}{2})}{2} - 0 \right) = \frac{\sqrt{2}}{2} \]

2. **Second Integral**:
\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} -\sqrt{2} \cos(2x) \, dx = -\sqrt{2} \left[ \frac{\sin(2x)}{2} \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} = -\sqrt{2} \left( \frac{\sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2})}{2} \right) = -\sqrt{2} \left( \frac{-1 - 1}{2} \right) = \sqrt{2} \]

3. **Third Integral**:
\[ \int_{\frac{3\pi}{4}}^{\pi} \sqrt{2} \cos(2x) \, dx = \sqrt{2} \left[ \frac{\sin(2x)}{2} \right]_{\frac{3\pi}{4}}^{\pi} = \sqrt{2} \left( 0 - \frac{-1}{2} \right) = \frac{\sqrt{2}}{2} \]

### Step 5: Combine the Areas
Now, we sum the areas:
\[ A = \frac{\sqrt{2}}{2} + \sqrt{2} + \frac{\sqrt{2}}{2} = 2\sqrt{2} \]

### Final Answer
Thus, the total area between the x-axis and the curve \( y = \sqrt{1 + \cos(4x)} \) from \( 0 \) to \( \pi \) is:
\[ \boxed{2\sqrt{2}} \]