`intsqrt(1+sin 2x) dx`

Answer

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

### Step 1: Rewrite the integral
We start by letting:
\[ I = \int \sqrt{1 + \sin 2x} \, dx \]
Using the identity for \( \sin 2x \):
\[ \sin 2x = 2 \sin x \cos x \]
we can rewrite the integral as:
\[ I = \int \sqrt{1 + 2 \sin x \cos x} \, dx \]

### Step 2: Use the Pythagorean identity
We know that:
\[ 1 = \sin^2 x + \cos^2 x \]
Thus, we can express \( 1 + \sin 2x \) as:
\[ 1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2 \]
So, we rewrite the integral:
\[ I = \int \sqrt{(\sin x + \cos x)^2} \, dx \]

### Step 3: Simplify the square root
Since \( \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x| \), we can assume \( \sin x + \cos x \) is non-negative in the interval we are considering. Thus:
\[ I = \int (\sin x + \cos x) \, dx \]

### Step 4: Split the integral
Now we can split the integral:
\[ I = \int \sin x \, dx + \int \cos x \, dx \]

### Step 5: Integrate each term
We know:
\[ \int \sin x \, dx = -\cos x + C_1 \]
\[ \int \cos x \, dx = \sin x + C_2 \]
Combining these results, we have:
\[ I = -\cos x + \sin x + C \]

### Final Result
Thus, the final result for the integral is:
\[ I = \sin x - \cos x + C \]