`int(1)/(sin(x-a)cos(x-b))dx` is equal to

Options

A. `1/(cos(a-b))ln|(sin(x-a))/(cos(x-b))|+C`
B. `1/(cos(a-b))ln|(cos(x-a))/(sin(x-a))|+C`
C. `1/(sin(a-b))ln|(sin(x-a))/(cos(x-b))|+C`
D. `1/(sin(a+b))ln|(cos(x-a))/(sin(x-b))|+C`

Answer

To solve the integral \( I = \int \frac{1}{\sin(x-a) \cos(x-b)} \, dx \), we can follow these steps:

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

### Step 2: Use the Cosine Angle Difference Identity
We can express \(\cos(b-a)\) using the cosine angle difference identity:
\[ \cos(b-a) = \cos(x-a) \cos(x-b) + \sin(x-a) \sin(x-b) \]
Thus, we rewrite the integral:
\[ I = \int \frac{\cos(b-a)}{\sin(x-a) \cos(x-b)} \, dx = \int \frac{\cos(x-a) \cos(x-b) + \sin(x-a) \sin(x-b)}{\sin(x-a) \cos(x-b)} \, dx \]

### Step 3: Separate the Integral
Now we can separate the integral into two parts:
\[ I = \int \left( \frac{\cos(x-a)}{\sin(x-a)} + \frac{\sin(x-a)}{\cos(x-b)} \right) \, dx \]
This simplifies to:
\[ I = \int \cot(x-a) \, dx + \int \tan(x-b) \, dx \]

### Step 4: Integrate Each Part
Now we can integrate each part:
1. The integral of \(\cot(x-a)\) is:
\[ \int \cot(x-a) \, dx = \ln |\sin(x-a)| + C_1 \]
2. The integral of \(\tan(x-b)\) is:
\[ \int \tan(x-b) \, dx = -\ln |\cos(x-b)| + C_2 \]

### Step 5: Combine the Results
Combining these results, we have:
\[ I = \ln |\sin(x-a)| - \ln |\cos(x-b)| + C \]
Using the property of logarithms:
\[ \ln a - \ln b = \ln \left(\frac{a}{b}\right) \]
We can write:
\[ I = \ln \left|\frac{\sin(x-a)}{\cos(x-b)}\right| + C \]

### Step 6: Final Result
Finally, we can express the result in terms of \(\cos(b-a)\):
\[ I = \frac{1}{\cos(b-a)} \ln \left|\frac{\sin(x-a)}{\cos(x-b)}\right| + C \]

### Final Answer
Thus, the integral evaluates to:
\[ I = \frac{1}{\cos(b-a)} \ln \left|\frac{\sin(x-a)}{\cos(x-b)}\right| + C \]