What is `int_(0)^(2pi) sqrt(1+ sin'x/2) dx` equal to ?

Options

A. `8`
B. `4`
C. `2`
D. `0`

Explanation

To solve the integral \( I = \int_{0}^{2\pi} \sqrt{1 + \sin \frac{x}{2}} \, dx \), we will follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the expression inside the square root. We know that:
\[ \sin \frac{x}{2} = \sin \left(\frac{2x}{4}\right) = 2 \sin \frac{x}{4} \cos \frac{x}{4} \]
Thus,
\[ 1 + \sin \frac{x}{2} = 1 + 2 \sin \frac{x}{4} \cos \frac{x}{4} \]

### Step 2: Use the Pythagorean identity
We can express 1 as:
\[ 1 = \sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} \]
So we have:
\[ 1 + \sin \frac{x}{2} = \sin^2 \frac{x}{4} + \cos^2 \frac{x}{4} + 2 \sin \frac{x}{4} \cos \frac{x}{4} \]
This can be rewritten as:
\[ 1 + \sin \frac{x}{2} = \left(\sin \frac{x}{4} + \cos \frac{x}{4}\right)^2 \]

### Step 3: Substitute back into the integral
Now substituting this back into the integral, we get:
\[ I = \int_{0}^{2\pi} \sqrt{\left(\sin \frac{x}{4} + \cos \frac{x}{4}\right)^2} \, dx \]
Since the square root and square cancel out, we have:
\[ I = \int_{0}^{2\pi} \left(\sin \frac{x}{4} + \cos \frac{x}{4}\right) \, dx \]

### Step 4: Integrate term by term
Now we can integrate term by term:
\[ I = \int_{0}^{2\pi} \sin \frac{x}{4} \, dx + \int_{0}^{2\pi} \cos \frac{x}{4} \, dx \]

#### Integrating \(\sin \frac{x}{4}\):
Using the substitution \( u = \frac{x}{4} \), \( dx = 4 \, du \):
\[ \int \sin \frac{x}{4} \, dx = -4 \cos \frac{x}{4} \bigg|_{0}^{2\pi} = -4 \left(\cos \frac{\pi}{2} - \cos 0\right) = -4(0 - 1) = 4 \]

#### Integrating \(\cos \frac{x}{4}\):
Using the same substitution:
\[ \int \cos \frac{x}{4} \, dx = 4 \sin \frac{x}{4} \bigg|_{0}^{2\pi} = 4 \left(\sin \frac{\pi}{2} - \sin 0\right) = 4(1 - 0) = 4 \]

### Step 5: Combine the results
Now, we combine the results of both integrals:
\[ I = 4 + 4 = 8 \]

Thus, the value of the integral is:
\[ \boxed{8} \]