What is `int_0^(2pi) sqrt(1+ sin""x/2 dx)` equal to

Options

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

Answer

To solve the integral \( \int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
We start with the expression under the integral:
\[ \sqrt{1 + \sin\frac{x}{2}} \]
We can use the identity \( \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \). However, in this case, we will use the half-angle identity for sine:
\[ \sin\frac{x}{2} = \frac{1 - \cos x}{2} \]
This gives us:
\[ 1 + \sin\frac{x}{2} = 1 + \frac{1 - \cos x}{2} = \frac{2 + 1 - \cos x}{2} = \frac{3 - \cos x}{2} \]
Thus, we can rewrite the integral as:
\[ \int_0^{2\pi} \sqrt{\frac{3 - \cos x}{2}} \, dx \]

### Step 2: Factor out constants
We can factor out the constant from the square root:
\[ \int_0^{2\pi} \sqrt{\frac{1}{2}} \sqrt{3 - \cos x} \, dx = \frac{1}{\sqrt{2}} \int_0^{2\pi} \sqrt{3 - \cos x} \, dx \]

### Step 3: Evaluate the integral
Now we need to evaluate the integral \( \int_0^{2\pi} \sqrt{3 - \cos x} \, dx \). This integral can be computed using symmetry and known results. The integral \( \int_0^{2\pi} \sqrt{a - b\cos x} \, dx \) has a known result:
\[ \int_0^{2\pi} \sqrt{a - b\cos x} \, dx = 2\pi \sqrt{a + b} \]
In our case, \( a = 3 \) and \( b = 1 \):
\[ \int_0^{2\pi} \sqrt{3 - \cos x} \, dx = 2\pi \sqrt{3 + 1} = 2\pi \sqrt{4} = 2\pi \cdot 2 = 4\pi \]

### Step 4: Combine results
Now substituting back into our expression:
\[ \frac{1}{\sqrt{2}} \int_0^{2\pi} \sqrt{3 - \cos x} \, dx = \frac{1}{\sqrt{2}} \cdot 4\pi = \frac{4\pi}{\sqrt{2}} = 2\sqrt{2}\pi \]

### Final Answer
Thus, the value of the integral \( \int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}} \, dx \) is:
\[ \boxed{2\sqrt{2}\pi} \]