`int(1)/(sqrt(1+sinx))dx=`

Options

A. `(2)/(3)(1-sinx)^(3//2)+c`
B. `2[cos((x)/(2))-sin((x)/(2))]+c`
C. `-sqrt2log[tan((3pi)/(8)-(x)/(4))]+c`
D. `sqrt(x-cosx)+c`

Answer

To solve the integral \( I = \int \frac{1}{\sqrt{1 + \sin x}} \, dx \), we can use a trigonometric identity to simplify the expression. Here’s a step-by-step solution:

### Step 1: Use the Identity for \( \sin x \)
We can use the identity \( 1 + \sin x = 1 + 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \). This will help us rewrite the integral.

### Step 2: Rewrite the Integral
Using the identity, we can express the integral as:
\[ I = \int \frac{1}{\sqrt{1 + \sin x}} \, dx = \int \frac{1}{\sqrt{1 + 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}} \, dx \]

### Step 3: Use Half-Angle Substitution
Let \( t = \tan\left(\frac{x}{2}\right) \). Then, we have:
\[ \sin x = \frac{2t}{1+t^2} \quad \text{and} \quad dx = \frac{2}{1+t^2} \, dt \]
Substituting these into the integral gives:
\[ I = \int \frac{2}{\sqrt{1 + \frac{2t}{1+t^2}}} \cdot \frac{1}{1+t^2} \, dt \]

### Step 4: Simplify the Integral
Now simplify the expression under the square root:
\[ 1 + \frac{2t}{1+t^2} = \frac{(1+t^2) + 2t}{1+t^2} = \frac{(1+t)^2}{1+t^2} \]
Thus, we have:
\[ \sqrt{1 + \sin x} = \sqrt{\frac{(1+t)^2}{1+t^2}} = \frac{1+t}{\sqrt{1+t^2}} \]
Substituting this back into the integral gives:
\[ I = \int \frac{2 \sqrt{1+t^2}}{1+t} \cdot \frac{1}{1+t^2} \, dt \]

### Step 5: Further Simplification
This simplifies to:
\[ I = \int \frac{2}{1+t} \, dt \]
This integral can be solved easily.

### Step 6: Solve the Integral
Now, we can integrate:
\[ I = 2 \ln|1+t| + C \]
Substituting back \( t = \tan\left(\frac{x}{2}\right) \):
\[ I = 2 \ln\left|1 + \tan\left(\frac{x}{2}\right)\right| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{1}{\sqrt{1 + \sin x}} \, dx = 2 \ln\left|1 + \tan\left(\frac{x}{2}\right)\right| + C \]