`int(1)/(sin(x-a)cos(x-b))dx` is equal to

Options

A. `(1)/(sin(a-b))log|(sin(x-a))/(cos(x-b))|+C`
B. `(1)/(cos(a-b))log|(sin(x-a))/(cos(x-b))|+C`
C. `(1)/(sin(a+b))log|(sin(x-a))/(cos(x-b))|+C`
D. `(1)/(cos(a+b))log|(sin(x-a))/(cos(x-b))|+C`

Answer

To solve the integral \( I = \int \frac{1}{\sin(x-a) \cos(x-b)} \, dx \), we can use a trigonometric identity and some algebraic manipulation. Here’s a step-by-step solution:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{\sin(x-a) \cos(x-b)} \, dx \]

### Step 2: Use a Trigonometric Identity
Recall that \(\sin(x-a) \cos(x-b)\) can be rewritten using the product-to-sum identities. However, in this case, we will manipulate the expression directly.

### Step 3: Multiply and Divide by \(\sin(x-b)\)
To facilitate integration, we can multiply and divide the integrand by \(\sin(x-b)\):
\[ I = \int \frac{\sin(x-b)}{\sin(x-a) \cos(x-b) \sin(x-b)} \, dx \]

### Step 4: Simplify the Expression
Now, we can rewrite the integral:
\[ I = \int \frac{\sin(x-b)}{\sin(x-a) \sin(x-b) \cos(x-b)} \, dx \]

### Step 5: Use Substitution
Next, we can use the substitution \( u = x - b \), which gives \( du = dx \):
\[ I = \int \frac{\sin(u)}{\sin(u + (a-b)) \cos(u)} \, du \]

### Step 6: Further Simplification
Using the identity for \(\sin(u + (a-b))\):
\[ \sin(u + (a-b)) = \sin(u) \cos(a-b) + \cos(u) \sin(a-b) \]
We can substitute this back into the integral:
\[ I = \int \frac{\sin(u)}{\sin(u) \cos(a-b) + \cos(u) \sin(a-b) \cos(u)} \, du \]

### Step 7: Final Integration
This integral can be simplified further, and we can solve it using standard integration techniques or tables. The final result will depend on the constants \(a\) and \(b\).

### Final Result
Thus, the integral can be expressed in terms of logarithmic functions or other elementary functions depending on the specific values of \(a\) and \(b\).