`int(sinx)/((sinx-cosx))dx=?`

Options

A. `(1)/(2)x-(1)/(2)log|sinx-cosx|+C`
B. `(1)/(2)x+(1)/(2)log|sinx-cosx|+C`
C. `log|sinx-cosx|+C`
D. none of these

Explanation

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin x}{\sin x - \cos x} \, dx \]

### Step 2: Multiply and Divide by 2
To simplify the integral, we can multiply the numerator and denominator by 2:
\[ I = \frac{1}{2} \int \frac{2\sin x}{\sin x - \cos x} \, dx \]

### Step 3: Split the Numerator
Next, we can express \( 2\sin x \) as \( (\sin x - \cos x) + (\sin x + \cos x) \):
\[ I = \frac{1}{2} \int \left( \frac{\sin x - \cos x}{\sin x - \cos x} + \frac{\sin x + \cos x}{\sin x - \cos x} \right) \, dx \]
This simplifies to:
\[ I = \frac{1}{2} \int \left( 1 + \frac{\sin x + \cos x}{\sin x - \cos x} \right) \, dx \]

### Step 4: Separate the Integral
Now we can separate the integral:
\[ I = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx \]

### Step 5: Solve the First Integral
The first integral is straightforward:
\[ \frac{1}{2} \int 1 \, dx = \frac{x}{2} \]

### Step 6: Substitute for the Second Integral
For the second integral, we can use the substitution \( t = \sin x - \cos x \). Then, we find \( dt = (\cos x + \sin x) \, dx \) or \( dx = \frac{dt}{\sin x + \cos x} \).

### Step 7: Rewrite the Second Integral
Substituting into the integral gives:
\[ \frac{1}{2} \int \frac{1}{t} \, dt \]

### Step 8: Solve the Second Integral
The integral of \( \frac{1}{t} \) is:
\[ \frac{1}{2} \ln |t| + C \]
Substituting back \( t = \sin x - \cos x \):
\[ \frac{1}{2} \ln |\sin x - \cos x| + C \]

### Step 9: Combine Results
Combining both parts, we have:
\[ I = \frac{x}{2} + \frac{1}{2} \ln |\sin x - \cos x| + C \]

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