`int (sin x)/(sin x - cos x) dx = `

Options

A. `x + "log" (sin x - cos x)`
B. `1/2 [x + "log" (sin x - cos x)]`
C. `x - log "sin" (x + pi/4)`
D. `x + log cos (x + pi/4)`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin x}{\sin x - \cos x} \, dx \]
To simplify the expression, we can multiply and divide by 2:
\[ I = \int \frac{2\sin x}{2(\sin x - \cos x)} \, dx = \int \frac{2\sin x}{2\sin x - 2\cos x} \, dx \]

### Step 2: Split the Integral
Now, we can express the integrand as a sum of two fractions:
\[ I = \int \left(1 + \frac{\cos x + \sin x}{\sin x - \cos x}\right) \, dx \]
This gives us:
\[ I = \int 1 \, dx + \int \frac{\cos x + \sin x}{\sin x - \cos x} \, dx \]

### Step 3: Integrate the First Part
The first integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 4: Substitute for the Second Integral
Now, we focus on the second integral:
\[ \int \frac{\cos x + \sin x}{\sin x - \cos x} \, dx \]
We can use substitution. Let:
\[ t = \sin x - \cos x \]
Then, the derivative \( dt \) is:
\[ dt = (\cos x + \sin x) \, dx \]
Thus, we can rewrite the second integral in terms of \( t \):
\[ \int \frac{dt}{t} \]

### Step 5: Integrate the Second Part
The integral of \( \frac{1}{t} \) is:
\[ \int \frac{dt}{t} = \ln |t| + C \]
Substituting back for \( t \):
\[ \int \frac{\cos x + \sin x}{\sin x - \cos x} \, dx = \ln |\sin x - \cos x| + C \]

### Step 6: Combine the Results
Putting everything together, we have:
\[ I = x + \ln |\sin x - \cos x| + C \]

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