`int (dx)/(sin(x-a)sin(x-b)`

Options

A. `sin(b-a)log |frac{sin(x-b)}{sin(x-a)}|+c`
B. `cosec(b-a)log|frac{sin(x-a)}{sin(x-b)}|+c`
C. `cosec(b-a)log |frac {sin(x-b)}{sin(x-a)}|+c`
D. `sin(b-a)log|frac{sin(x-a)}{sin(x-b)}+c`

Answer

To solve the integral

\[ I = \int \frac{dx}{\sin(x-a) \sin(x-b)}, \]

we will follow a series of steps to simplify and evaluate the integral.

### Step 1: Rewrite the Integral
We start with the integral:

\[ I = \int \frac{dx}{\sin(x-a) \sin(x-b)}. \]

### Step 2: Multiply and Divide by \(\sin(b-a)\)
To facilitate the integration, we multiply and divide the integrand by \(\sin(b-a)\):

\[ I = \int \frac{\sin(b-a) \, dx}{\sin(b-a) \sin(x-a) \sin(x-b)}. \]

### Step 3: Use the Sine Difference Identity
We can use the identity for sine:

\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right). \]

Here, we rewrite the numerator:

\[ \sin(x-b) - \sin(x-a) = 2 \cos\left(\frac{(x-a)+(x-b)}{2}\right) \sin\left(\frac{(x-a)-(x-b)}{2}\right). \]

### Step 4: Substitute in the Integral
Now substituting this back into the integral, we have:

\[ I = \int \frac{\sin(b-a) \, dx}{\sin(b-a) \left(\sin(x-b) - \sin(x-a)\right)}. \]

### Step 5: Simplify the Integral
This simplifies to:

\[ I = \int \frac{dx}{\sin(x-b) - \sin(x-a)}. \]

### Step 6: Break Down the Integral
Using the identity, we can express the integral as:

\[ I = \int \frac{dx}{2 \cos\left(\frac{(x-a)+(x-b)}{2}\right) \sin\left(\frac{(x-a)-(x-b)}{2}\right)}. \]

### Step 7: Evaluate the Integral
Now we can split the integral into two parts:

\[ I = \frac{1}{\sin(b-a)} \left( \int \cot(x-b) \, dx - \int \cot(x-a) \, dx \right). \]

### Step 8: Integrate Cotangent
The integral of \(\cot x\) is:

\[ \int \cot x \, dx = \ln |\sin x| + C. \]

Thus, we have:

\[ I = \frac{1}{\sin(b-a)} \left( \ln |\sin(x-b)| - \ln |\sin(x-a)| + C \right). \]

### Step 9: Combine the Logarithms
Using the property of logarithms, we can combine the two logarithmic terms:

\[ I = \frac{1}{\sin(b-a)} \ln \left| \frac{\sin(x-b)}{\sin(x-a)} \right| + C. \]

### Final Result
Thus, the final result for the integral is:

\[ I = \frac{1}{\sin(b-a)} \ln \left| \frac{\sin(x-b)}{\sin(x-a)} \right| + C. \]