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

Options

A. ` -2sqrt(1-sinx)+C`
B. `sin(x//2)+cos(x//2)+C`
C. `cos(x//2)-sin(x//2)+C`
D. `2sqrt(1-sinx)+C`

Explanation

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

### Step 1: Rewrite the integrand
We start by rewriting \( 1 + \sin x \) in a more convenient form. We can use the identity:
\[ 1 + \sin x = \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2 \]
This is derived from the Pythagorean identity and the double angle formulas.

### Step 2: Substitute into the integral
Now we can substitute this back into the integral:
\[ \int \sqrt{1 + \sin x} \, dx = \int \sqrt{\left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2} \, dx \]
Since the square root and the square cancel each other, we have:
\[ = \int \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right) \, dx \]

### Step 3: Integrate the expression
Next, we can integrate the expression:
\[ \int \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right) \, dx \]
We can split this into two separate integrals:
\[ = \int \sin \frac{x}{2} \, dx + \int \cos \frac{x}{2} \, dx \]

### Step 4: Solve each integral
1. For \( \int \sin \frac{x}{2} \, dx \):
- Use the substitution \( u = \frac{x}{2} \) which gives \( du = \frac{1}{2} dx \) or \( dx = 2 du \):
\[ \int \sin \frac{x}{2} \, dx = 2 \int \sin u \, du = -2 \cos u + C_1 = -2 \cos \frac{x}{2} + C_1 \]

2. For \( \int \cos \frac{x}{2} \, dx \):
- Again using the substitution \( u = \frac{x}{2} \):
\[ \int \cos \frac{x}{2} \, dx = 2 \int \cos u \, du = 2 \sin u + C_2 = 2 \sin \frac{x}{2} + C_2 \]

### Step 5: Combine the results
Now we combine the results of both integrals:
\[ \int \sqrt{1 + \sin x} \, dx = -2 \cos \frac{x}{2} + 2 \sin \frac{x}{2} + C \]
where \( C \) is the constant of integration.

### Final Answer
Thus, the final result is:
\[ \int \sqrt{1 + \sin x} \, dx = 2 \sin \frac{x}{2} - 2 \cos \frac{x}{2} + C \]