To solve the integral \( \int \sqrt{1 + \sin x} \, dx \), we can follow these steps:
### Step 1: Rewrite the integrand
We can use the identity \( 1 + \sin x = 1 + 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \). This allows us to express \( 1 + \sin x \) in a different form:
\[
1 + \sin x = \cos^2\left(\frac{x}{2}\right) + \sin^2\left(\frac{x}{2}\right) + 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) = \left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right)^2
\]
### Step 2: Substitute the expression
Now we can rewrite the integral:
\[
\int \sqrt{1 + \sin x} \, dx = \int \sqrt{\left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right)^2} \, dx
\]
This simplifies to:
\[
\int \left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right) \, dx
\]
### Step 3: Integrate the expression
Next, we can integrate the expression:
\[
\int \left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right) \, dx = \int \sin\left(\frac{x}{2}\right) \, dx + \int \cos\left(\frac{x}{2}\right) \, dx
\]
Using the substitution \( u = \frac{x}{2} \), we have \( dx = 2 \, du \). Thus:
\[
\int \sin\left(\frac{x}{2}\right) \, dx = 2 \int \sin(u) \, du = -2 \cos(u) + C = -2 \cos\left(\frac{x}{2}\right) + C
\]
\[
\int \cos\left(\frac{x}{2}\right) \, dx = 2 \int \cos(u) \, du = 2 \sin(u) + C = 2 \sin\left(\frac{x}{2}\right) + C
\]
### Step 4: Combine the results
Combining both integrals, we get:
\[
\int \sqrt{1 + \sin x} \, dx = -2 \cos\left(\frac{x}{2}\right) + 2 \sin\left(\frac{x}{2}\right) + C
\]
### Final Answer
Thus, the final result of the integral is:
\[
\int \sqrt{1 + \sin x} \, dx = 2 \left( \sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right) \right) + C
\]
