Integration of \( \frac{1}{\cos(x-a)\cos(x-b)} \)
Question: Evaluate the integral:
\[
\int \frac{1}{\cos(x-a)\cos(x-b)} \, dx
\]
Step 1: Multiply and divide by \( \sin(a-b) \)
\[
I = \int \frac{1}{\cos(x-a)\cos(x-b)} \, dx
\]
\[ I = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\cos(x-a)\cos(x-b)} \, dx \]
\[ I = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\cos(x-a)\cos(x-b)} \, dx \]
Step 2: Use the identity
\[
\sin(a-b) = \sin[(x-a) – (x-b)]
\]
\[ = \sin(x-a)\cos(x-b) – \cos(x-a)\sin(x-b) \]
\[ = \sin(x-a)\cos(x-b) – \cos(x-a)\sin(x-b) \]
Step 3: Substitute in the numerator
\[
I = \frac{1}{\sin(a-b)}
\int \frac{\sin(x-a)\cos(x-b) – \cos(x-a)\sin(x-b)}
{\cos(x-a)\cos(x-b)} \, dx
\]
Step 4: Separate the terms
\[
I = \frac{1}{\sin(a-b)}
\int \left[
\frac{\sin(x-a)}{\cos(x-a)}
– \frac{\sin(x-b)}{\cos(x-b)}
\right] dx
\]
Step 5: Recognize the tangent function
\[
I = \frac{1}{\sin(a-b)}
\int \left[
\tan(x-a) – \tan(x-b)
\right] dx
\]
Step 6: Integrate each term
We know:
\[
\int \tan(u) \, du = -\ln|\cos(u)|
\]
\[ I = \frac{1}{\sin(a-b)} \left[ -\ln|\cos(x-a)| + \ln|\cos(x-b)| \right] + C \]
\[ I = \frac{1}{\sin(a-b)} \left[ -\ln|\cos(x-a)| + \ln|\cos(x-b)| \right] + C \]
Step 7: Combine logarithms
\[
I = \frac{1}{\sin(a-b)}
\ln \left| \frac{\cos(x-b)}{\cos(x-a)} \right| + C
\]
Final Answer:
\[
\boxed{
\int \frac{1}{\cos(x-a)\cos(x-b)} \, dx
= \frac{1}{\sin(a-b)}
\ln \left| \frac{\cos(x-b)}{\cos(x-a)} \right| + C
}
\]