`intsqrt(1+cos 2x) dx`

Answer

To solve the integral \(\int \sqrt{1 + \cos 2x} \, dx\), we can follow these steps:

### Step 1: Use the double angle identity for cosine
We know that:
\[ \cos 2x = 2 \cos^2 x - 1 \]
Thus, we can rewrite \(1 + \cos 2x\) as:
\[ 1 + \cos 2x = 1 + (2 \cos^2 x - 1) = 2 \cos^2 x \]

### Step 2: Substitute into the integral
Now we substitute this back into the integral:
\[ \int \sqrt{1 + \cos 2x} \, dx = \int \sqrt{2 \cos^2 x} \, dx \]

### Step 3: Simplify the square root
The square root can be simplified:
\[ \sqrt{2 \cos^2 x} = \sqrt{2} \cdot |\cos x| \]
Thus, the integral becomes:
\[ \int \sqrt{2} \cdot |\cos x| \, dx \]

### Step 4: Handle the absolute value
The absolute value \(|\cos x|\) can be split into two cases based on the sign of \(\cos x\):
1. When \(\cos x \geq 0\) (i.e., \(x\) in intervals where cosine is non-negative), we have \(|\cos x| = \cos x\).
2. When \(\cos x < 0\), we have \(|\cos x| = -\cos x\).

### Step 5: Integrate based on cases
1. **Case 1**: If \(\cos x \geq 0\):
\[ \int \sqrt{2} \cos x \, dx = \sqrt{2} \sin x + C \]

2. **Case 2**: If \(\cos x < 0\):
\[ \int \sqrt{2} (-\cos x) \, dx = -\sqrt{2} \sin x + C \]

### Final Result
Thus, the final result of the integral \(\int \sqrt{1 + \cos 2x} \, dx\) is:
\[ \int \sqrt{1 + \cos 2x} \, dx = \begin{cases} \sqrt{2} \sin x + C & \text{if } \cos x \geq 0 \\ -\sqrt{2} \sin x + C & \text{if } \cos x < 0 \end{cases} \]