`int sqrt(1+sin2x)dx`

Answer

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

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

**Hint:** Remember that \( \sin 2x \) can be expressed in terms of \( \sin x \) and \( \cos x \).

### Step 2: Rewrite \( 1 \) using the Pythagorean identity
We can express \( 1 \) as \( \sin^2 x + \cos^2 x \):
\[ 1 + 2 \sin x \cos x = \sin^2 x + \cos^2 x + 2 \sin x \cos x \]

**Hint:** Use the identity \( \sin^2 x + \cos^2 x = 1 \) to combine terms.

### Step 3: Recognize the expression as a perfect square
The expression \( \sin^2 x + \cos^2 x + 2 \sin x \cos x \) can be factored as:
\[ (\sin x + \cos x)^2 \]
Thus, we can rewrite the integral:
\[ \int \sqrt{(\sin x + \cos x)^2} \, dx \]

**Hint:** Recognize that the square root of a square gives the absolute value, but since \( \sin x + \cos x \) is positive in the intervals we are considering, we can simplify it to:
\[ \int (\sin x + \cos x) \, dx \]

### Step 4: Integrate the expression
Now we can integrate:
\[ \int (\sin x + \cos x) \, dx = \int \sin x \, dx + \int \cos x \, dx \]
The integrals are:
\[ -\cos x + \sin x + C \]

**Hint:** Remember the basic integrals of sine and cosine.

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