Evaluate :
`int(cos(x+a))/(cos(x-a))dx`

Answer

To evaluate the integral

\[ \int \frac{\cos(x + a)}{\cos(x - a)} \, dx, \]

we can follow these steps:

### Step 1: Substitution
Let \( t = x - a \). Then, we have:

\[ x = t + a \quad \text{and} \quad dx = dt. \]

### Step 2: Rewrite the Integral
Now, substituting \( x \) in the integral, we get:

\[ \int \frac{\cos((t + a) + a)}{\cos(t)} \, dt = \int \frac{\cos(t + 2a)}{\cos(t)} \, dt. \]

### Step 3: Use the Cosine Addition Formula
Using the cosine addition formula, \( \cos(t + 2a) = \cos(t)\cos(2a) - \sin(t)\sin(2a) \), we can rewrite the integral:

\[ \int \frac{\cos(t)\cos(2a) - \sin(t)\sin(2a)}{\cos(t)} \, dt. \]

### Step 4: Split the Integral
This can be split into two separate integrals:

\[ \int \cos(2a) \, dt - \int \sin(2a) \tan(t) \, dt. \]

### Step 5: Evaluate the First Integral
The first integral is straightforward:

\[ \int \cos(2a) \, dt = \cos(2a) t + C_1. \]

### Step 6: Evaluate the Second Integral
For the second integral, we have:

\[ -\sin(2a) \int \tan(t) \, dt. \]

The integral of \( \tan(t) \) is:

\[ \int \tan(t) \, dt = -\log|\cos(t)| + C_2. \]

Thus, we have:

\[ -\sin(2a)(-\log|\cos(t)|) = \sin(2a) \log|\cos(t)| + C_2. \]

### Step 7: Combine the Results
Combining both integrals, we get:

\[ \cos(2a) t + \sin(2a) \log|\cos(t)| + C, \]

where \( C = C_1 + C_2 \).

### Step 8: Substitute Back
Now, substituting back \( t = x - a \):

\[ \cos(2a)(x - a) + \sin(2a) \log|\cos(x - a)| + C. \]

### Final Answer
Thus, the final answer is:

\[ \int \frac{\cos(x + a)}{\cos(x - a)} \, dx = (x - a) \cos(2a) + \sin(2a) \log|\cos(x - a)| + C. \]