`int (cos(x+a))/(sin(x+b))dx`

Explanation

To solve the integral \( \int \frac{\cos(x + a)}{\sin(x + b)} \, dx \), we will follow these steps:

### Step 1: Rewrite the numerator using the cosine addition formula
Using the cosine addition formula, we can express \( \cos(x + a) \) as:
\[ \cos(x + a) = \cos x \cos a - \sin x \sin a \]
Thus, we rewrite the integral:
\[ \int \frac{\cos(x + a)}{\sin(x + b)} \, dx = \int \frac{\cos x \cos a - \sin x \sin a}{\sin(x + b)} \, dx \]

### Step 2: Split the integral
We can split the integral into two parts:
\[ \int \frac{\cos x \cos a}{\sin(x + b)} \, dx - \int \frac{\sin x \sin a}{\sin(x + b)} \, dx \]

### Step 3: Focus on the first integral
Let’s denote the first integral as \( I_1 \):
\[ I_1 = \int \frac{\cos x \cos a}{\sin(x + b)} \, dx \]
Using the substitution \( u = \sin(x + b) \), we have \( du = \cos(x + b) \, dx \). However, we need to express \( \cos x \) in terms of \( u \) or find another way to integrate.

### Step 4: Focus on the second integral
Now, let’s denote the second integral as \( I_2 \):
\[ I_2 = \int \frac{\sin x \sin a}{\sin(x + b)} \, dx \]
Using the same substitution \( u = \sin(x + b) \), we can express \( \sin x \) in terms of \( u \) and \( b \).

### Step 5: Combine the results
After evaluating both integrals, we can combine the results:
\[ \int \frac{\cos(x + a)}{\sin(x + b)} \, dx = I_1 - I_2 + C \]
where \( C \) is the constant of integration.

### Step 6: Final expression
The final expression will involve logarithmic functions based on the results of \( I_1 \) and \( I_2 \). The specific forms will depend on the evaluations of the integrals.