`intsqrt(1+cos 2x) dx`

Answer

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

### Step 1: Use the identity for \(\cos 2x\)
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
Since \(\sqrt{2\cos^2 x} = \sqrt{2} \cdot \sqrt{\cos^2 x} = \sqrt{2} \cdot |\cos x|\). However, since we are integrating, we can assume \(x\) is in a range where \(\cos x\) is non-negative (for example, \(0 \leq x \leq \frac{\pi}{2}\)), thus:
\[ \sqrt{2\cos^2 x} = \sqrt{2} \cos x \]

### Step 4: Integrate
Now we can integrate:
\[ \int \sqrt{2} \cos x \, dx = \sqrt{2} \int \cos x \, dx \]
The integral of \(\cos x\) is \(\sin x\), so:
\[ \sqrt{2} \int \cos x \, dx = \sqrt{2} \sin x + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \sqrt{1 + \cos 2x} \, dx = \sqrt{2} \sin x + C \]

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