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

Options

A. `x sin(a-b)-cos(a-b)log|sin(x+b)|+c`
B. `x sin(a-b)+cos(a-b)log|sin(x+b)|+c`
C. `x sin(a+b)-cos(a+b)log|sin(x+b)|+c`
D. `x sin(a+b)+cos(a+b)log|sin(x+b)|+c`

Answer

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

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

### Step 2: Substitute the sine term
Next, we will express \( \sin(x+b) \) as:
\[ \sin(x+b) = \sin x \cos b + \cos x \sin b \]
Now, substituting this into our integral gives:
\[ I = \int \frac{\cos x \cos a + \sin x \sin a}{\sin x \cos b + \cos x \sin b} \, dx \]

### Step 3: Split the integral
We can split the integral into two parts:
\[ I = \int \frac{\cos x \cos a}{\sin x \cos b + \cos x \sin b} \, dx + \int \frac{\sin x \sin a}{\sin x \cos b + \cos x \sin b} \, dx \]

### Step 4: Use substitution
For the first integral, we can use the substitution \( u = \sin(x+b) \), which gives \( du = \cos(x+b) \, dx \). This substitution will help simplify the integral.

### Step 5: Evaluate the integrals
After performing the substitution, we can evaluate each integral separately:
1. The first integral will yield a logarithmic term.
2. The second integral will yield a linear term.

### Step 6: Combine results
Finally, we combine the results from both integrals to get the final answer.

### Final Answer
Thus, the integral evaluates to:
\[ I = x \sin(a+b) + \cos(a+b) \log|\sin(x+b)| + C \]