`int(sqrt(1+cosx))dx` equals

Options

A. `2sqrt2sin(x)/(2)+c`
B. `-2sqrt2sin(x)/(2)+c`
C. `-2sqrt2cos(x)/(2)+c`
D. `2sqrtcos(x)/(2)+c`

Explanation

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

### Step 1: Rewrite the integrand using the half-angle identity
We know from trigonometry that:
\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the integral as:
\[ I = \int \sqrt{1 + \cos x} \, dx = \int \sqrt{2 \cos^2\left(\frac{x}{2}\right)} \, dx \]

### Step 2: Simplify the square root
The square root can be simplified:
\[ \sqrt{2 \cos^2\left(\frac{x}{2}\right)} = \sqrt{2} \cdot \cos\left(\frac{x}{2}\right) \]
Now the integral becomes:
\[ I = \int \sqrt{2} \cdot \cos\left(\frac{x}{2}\right) \, dx \]

### Step 3: Factor out the constant
We can factor out the constant \( \sqrt{2} \):
\[ I = \sqrt{2} \int \cos\left(\frac{x}{2}\right) \, dx \]

### Step 4: Integrate using the standard integral formula
The integral of \( \cos(ax) \) is given by:
\[ \int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C \]
In our case, \( a = \frac{1}{2} \), so we have:
\[ \int \cos\left(\frac{x}{2}\right) \, dx = \frac{\sin\left(\frac{x}{2}\right)}{\frac{1}{2}} = 2 \sin\left(\frac{x}{2}\right) \]

### Step 5: Substitute back into the integral
Substituting this back into our expression for \( I \):
\[ I = \sqrt{2} \cdot 2 \sin\left(\frac{x}{2}\right) + C \]
This simplifies to:
\[ I = 2\sqrt{2} \sin\left(\frac{x}{2}\right) + C \]

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

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