Choose the correct answer of the given question
`int sqrt((1-cosx)/2) dx`

Options

A. `-2cos(x/2)+c`
B. `e^(sqrtx) cosx+c`
C. 2sinx+c
D. `sqrt2 sinx+c`

Answer

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

### Step 1: Simplify the integrand
We start with the expression inside the square root:
\[ \sqrt{\frac{1 - \cos x}{2}} \]
Using the trigonometric identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \), we can rewrite the integrand:
\[ \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{2 \sin^2\left(\frac{x}{2}\right)}{2}} = \sqrt{\sin^2\left(\frac{x}{2}\right)} = \sin\left(\frac{x}{2}\right) \]

### Step 2: Rewrite the integral
Now we can rewrite the integral:
\[ \int \sqrt{\frac{1 - \cos x}{2}} \, dx = \int \sin\left(\frac{x}{2}\right) \, dx \]

### Step 3: Integrate
To integrate \( \sin\left(\frac{x}{2}\right) \), we use the formula for the integral of sine:
\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \]
In our case, \( k = \frac{1}{2} \):
\[ \int \sin\left(\frac{x}{2}\right) \, dx = -2 \cos\left(\frac{x}{2}\right) + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \sqrt{\frac{1 - \cos x}{2}} \, dx = -2 \cos\left(\frac{x}{2}\right) + C \]