Evaluate: `int1/(cos(x-a)cos(x-b))\ dx`

Answer

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

### Step-by-Step Solution:

1. **Multiply and Divide by a Trigonometric Constant**:
We start by multiplying and dividing the integrand by \( \frac{1}{\sin(a-b)} \):
\[ \int \frac{1}{\cos(x-a) \cos(x-b)} \, dx = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\cos(x-a) \cos(x-b)} \, dx \]

**Hint**: This step helps in transforming the integral into a more manageable form using trigonometric identities.

2. **Use the Sine Difference Identity**:
We can express \( \sin(a-b) \) in terms of \( x \):
\[ \sin(a-b) = \sin(x-b) \cos(x-a) - \cos(x-b) \sin(x-a) \]
Substituting this into the integral gives:
\[ \frac{1}{\sin(a-b)} \int \frac{\sin(x-b) \cos(x-a) - \cos(x-b) \sin(x-a)}{\cos(x-a) \cos(x-b)} \, dx \]

**Hint**: Recognizing the sine difference identity helps in rewriting the integral in a more useful form.

3. **Separate the Integral**:
Now we can separate the integral into two parts:
\[ \frac{1}{\sin(a-b)} \left( \int \tan(x-b) \, dx - \int \tan(x-a) \, dx \right) \]

**Hint**: Separating the integral allows us to integrate each term individually.

4. **Integrate Each Term**:
The integral of \( \tan(x) \) is:
\[ \int \tan(x) \, dx = -\ln|\cos(x)| + C \]
Thus, we have:
\[ \int \tan(x-b) \, dx = -\ln|\cos(x-b)| + C_1 \]
\[ \int \tan(x-a) \, dx = -\ln|\cos(x-a)| + C_2 \]

**Hint**: Remember the integral of \( \tan(x) \) is crucial for solving this problem.

5. **Combine the Results**:
Substituting back, we get:
\[ \frac{1}{\sin(a-b)} \left( -\ln|\cos(x-b)| + \ln|\cos(x-a)| \right) + C \]
This simplifies to:
\[ \frac{1}{\sin(a-b)} \ln \left| \frac{\cos(x-a)}{\cos(x-b)} \right| + C \]

**Hint**: Combining logarithmic terms is a useful algebraic manipulation.

### Final Answer:
Thus, the final result of the integral is:
\[ \int \frac{1}{\cos(x-a) \cos(x-b)} \, dx = \frac{1}{\sin(a-b)} \ln \left| \frac{\cos(x-a)}{\cos(x-b)} \right| + C \]