`int (sin (x-a))/(cos (x+b))ax`

Answer

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

### Step 1: Rewrite the integrand
We start by rewriting the sine function using the angle subtraction formula:
\[ \sin(x-a) = \sin x \cos a - \cos x \sin a \]
Thus, we can express the integral as:
\[ I = \int \frac{\sin x \cos a - \cos x \sin a}{\cos(x+b)} \, dx \]

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

### Step 3: Factor out constants
We can factor out the constants from the integrals:
\[ I = \cos a \int \frac{\sin x}{\cos(x+b)} \, dx - \sin a \int \frac{\cos x}{\cos(x+b)} \, dx \]

### Step 4: Use substitution for the first integral
For the first integral, we can use the substitution \( u = x + b \), which gives \( du = dx \) and \( x = u - b \):
\[ \int \frac{\sin x}{\cos(x+b)} \, dx = \int \frac{\sin(u-b)}{\cos u} \, du \]
Using the angle subtraction formula again:
\[ \sin(u-b) = \sin u \cos b - \cos u \sin b \]
Thus, we have:
\[ \int \frac{\sin(u-b)}{\cos u} \, du = \int \frac{\sin u \cos b - \cos u \sin b}{\cos u} \, du = \cos b \int \tan u \, du - \sin b \int 1 \, du \]

### Step 5: Integrate
Now, we can integrate:
\[ \int \tan u \, du = -\log|\cos u| + C \]
\[ \int 1 \, du = u + C \]
So, substituting back:
\[ \int \frac{\sin x}{\cos(x+b)} \, dx = \cos b (-\log|\cos(x+b)|) - \sin b (x+b) + C \]

### Step 6: Substitute back into the original integral
Now substituting back into our expression for \( I \):
\[ I = \cos a \left( \cos b (-\log|\cos(x+b)|) - \sin b (x+b) \right) - \sin a (x+b) + C \]

### Final Result
Thus, the final result for the integral is:
\[ I = -\cos a \cos b \log|\cos(x+b)| - (\cos a \sin b + \sin a)(x+b) + C \]