`int ((1+ sinx))/((1- sinx)) dx=?`

Options

A. `2 sin x + 2 sec x + x + C`
B. `2 tan x + 2 sec x - x + C`
C. `tanx + sec x - x + C`
D. none of these

Explanation

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1 + \sin x}{1 - \sin x} \, dx \]
We can multiply and divide the integrand by \(1 + \sin x\):
\[ \int \frac{(1 + \sin x)(1 + \sin x)}{(1 - \sin x)(1 + \sin x)} \, dx \]

### Step 2: Simplify the Denominator
Using the difference of squares, we simplify the denominator:
\[ (1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x \]
Thus, the integral becomes:
\[ \int \frac{(1 + \sin x)^2}{\cos^2 x} \, dx \]

### Step 3: Expand the Numerator
Now we expand the numerator:
\[ (1 + \sin x)^2 = 1 + 2\sin x + \sin^2 x \]
So the integral can be rewritten as:
\[ \int \frac{1 + 2\sin x + \sin^2 x}{\cos^2 x} \, dx \]

### Step 4: Split the Integral
We can split the integral into three separate integrals:
\[ \int \frac{1}{\cos^2 x} \, dx + 2 \int \frac{\sin x}{\cos^2 x} \, dx + \int \frac{\sin^2 x}{\cos^2 x} \, dx \]

### Step 5: Integrate Each Term
1. The first integral:
\[ \int \frac{1}{\cos^2 x} \, dx = \tan x \]
2. The second integral:
\[ 2 \int \frac{\sin x}{\cos^2 x} \, dx = 2 \cdot (-\frac{1}{\cos x}) = -2 \sec x \]
3. The third integral:
\[ \int \frac{\sin^2 x}{\cos^2 x} \, dx = \int \tan^2 x \, dx = \tan x - x \]

### Step 6: Combine the Results
Combining all the results, we have:
\[ \tan x - 2 \sec x + \tan x - x + C \]
This simplifies to:
\[ 2 \tan x - 2 \sec x - x + C \]

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