`int sqrt(1+ cos 2x) dx = ?`

Options

A. `sqrt(2) cos x + C`
B. `sqrt(2) sinx + C`
C. `-sqrt(2) cos x + C`
D. `-sqrt(2) sin x + C`

Explanation

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

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

### Step 2: Substitute into the integral
Now, 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: Factor out the constant
Since \( \sqrt{2} \) is a constant, we can factor it out of the integral:
\[ \sqrt{2} \int \cos x \, dx \]

### Step 5: Integrate \( \cos x \)
The integral of \( \cos x \) is:
\[ \int \cos x \, dx = \sin x + C \]
where \( C \) is the constant of integration.

### Step 6: Combine the results
Putting it all together, we have:
\[ \sqrt{2} \int \cos x \, dx = \sqrt{2} (\sin x + C) = \sqrt{2} \sin x + C\sqrt{2} \]

### 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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