If `int(1)/(1+sinx)dx=tan((x)/(2)+a)+b` then

Options

A. `a=-(pi)/(4),binR`
B. `a=(pi)/(4),binR`
C. `a=(5pi)/(4),b inR`
D. none of these

Answer

To solve the integral \( \int \frac{1}{1 + \sin x} \, dx \), we can use a trigonometric identity and substitution. Let's go through the solution step by step.

### Step 1: Use the Trigonometric Identity
We know that:
\[ \sin x = 2 \tan\left(\frac{x}{2}\right) \cdot \frac{1}{1 + \tan^2\left(\frac{x}{2}\right)} \]
This means:
\[ 1 + \sin x = 1 + 2 \tan\left(\frac{x}{2}\right) \cdot \frac{1}{1 + \tan^2\left(\frac{x}{2}\right)} = \frac{(1 + \tan^2\left(\frac{x}{2}\right)) + 2 \tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \]
Thus:
\[ 1 + \sin x = \frac{(1 + \tan\left(\frac{x}{2}\right))^2}{1 + \tan^2\left(\frac{x}{2}\right)} \]

### Step 2: Substitute
Let:
\[ t = \tan\left(\frac{x}{2}\right) \]
Then:
\[ dx = \frac{2}{1 + t^2} \, dt \]
Now substituting \( t \) into the integral:
\[ \int \frac{1}{1 + \sin x} \, dx = \int \frac{1 + t^2}{(1 + t)^2} \cdot \frac{2}{1 + t^2} \, dt = 2 \int \frac{1}{(1 + t)^2} \, dt \]

### Step 3: Integrate
Now we can integrate:
\[ 2 \int \frac{1}{(1 + t)^2} \, dt = 2 \left( -\frac{1}{1 + t} \right) + C = -\frac{2}{1 + t} + C \]

### Step 4: Substitute Back
Now substituting back \( t = \tan\left(\frac{x}{2}\right) \):
\[ -\frac{2}{1 + \tan\left(\frac{x}{2}\right)} + C \]

### Step 5: Final Form
Using the identity \( 1 + \tan\left(\frac{x}{2}\right) = \frac{1 + \sin x}{\cos x} \):
\[ -\frac{2 \cos\left(\frac{x}{2}\right)}{1 + \sin x} + C \]
Thus, we can express the integral as:
\[ \int \frac{1}{1 + \sin x} \, dx = -2 \tan\left(\frac{x}{2}\right) + C \]

### Conclusion
Comparing with the given form \( \tan\left(\frac{x}{2} + a\right) + b \), we can identify that:
- \( a = 0 \)
- \( b = -2 \)