`int(cosx)/(sqrt(1+sinx))dx` is equal to

Options

A. `sin.(x)/(2)-cos.(x)/(2)+C`
B. `sin.(x)/(2)+cos.(x)/(2)+C`
C. `2[sin.(x)/(2)-cos.(x)/(2)]+C`
D. `2[sin.(x)/(2)+cos.(x)/(2)]+C`

Explanation

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

### Step 1: Substitution
Let \( t = \sqrt{1 + \sin x} \). Then, squaring both sides gives:
\[ t^2 = 1 + \sin x \quad \Rightarrow \quad \sin x = t^2 - 1 \]

### Step 2: Differentiate to find dx
Now, we differentiate \( t^2 \):
\[ \frac{d}{dx}(t^2) = \frac{d}{dx}(1 + \sin x) \quad \Rightarrow \quad 2t \frac{dt}{dx} = \cos x \]
From this, we can express \( dx \):
\[ dx = \frac{2t}{\cos x} \, dt \]

### Step 3: Substitute in the integral
Now, we substitute \( \sin x \) and \( dx \) into the integral:
\[ \int \frac{\cos x}{\sqrt{1 + \sin x}} \, dx = \int \frac{\cos x}{t} \cdot \frac{2t}{\cos x} \, dt \]
The \( \cos x \) terms cancel out:
\[ = \int 2 \, dt \]

### Step 4: Integrate
Now we can easily integrate:
\[ \int 2 \, dt = 2t + C \]

### Step 5: Substitute back for t
Recall that \( t = \sqrt{1 + \sin x} \):
\[ = 2\sqrt{1 + \sin x} + C \]

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

---