If `int(k)/(sin(x-a)cos(x-b))dx=log[(sin(x-a))/(cos(x-b))]+c`, then `k=`

Options

A. `sin(a-b)`
B. `cos(a-b)`
C. `tan(a-b)`
D. `sec(a-b)`

Answer

To find the value of \( k \) in the equation

\[ \int \frac{k}{\sin(x-a) \cos(x-b)} \, dx = \log\left(\frac{\sin(x-a)}{\cos(x-b)}\right) + c, \]

we will differentiate both sides with respect to \( x \).

### Step-by-Step Solution:

1. **Rearranging the Equation**:
We start with the equation:
\[ \int \frac{k}{\sin(x-a) \cos(x-b)} \, dx = \log(\sin(x-a)) - \log(\cos(x-b)) + c. \]

2. **Differentiating Both Sides**:
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx} \left( \int \frac{k}{\sin(x-a) \cos(x-b)} \, dx \right) = \frac{d}{dx} \left( \log(\sin(x-a)) - \log(\cos(x-b)) + c \right). \]
By the Fundamental Theorem of Calculus, the left-hand side simplifies to:
\[ \frac{k}{\sin(x-a) \cos(x-b)}. \]

3. **Differentiating the Right Side**:
Using the chain rule, we differentiate the right-hand side:
\[ \frac{d}{dx} \left( \log(\sin(x-a)) \right) = \frac{\cos(x-a)}{\sin(x-a)} \cdot \frac{d}{dx}(x-a) = \frac{\cos(x-a)}{\sin(x-a)}, \]
and
\[ \frac{d}{dx} \left( -\log(\cos(x-b)) \right) = -\frac{-\sin(x-b)}{\cos(x-b)} \cdot \frac{d}{dx}(x-b) = \frac{\sin(x-b)}{\cos(x-b)}. \]
Thus, the right-hand side becomes:
\[ \frac{\cos(x-a)}{\sin(x-a)} + \frac{\sin(x-b)}{\cos(x-b)}. \]

4. **Combining the Right Side**:
We can combine the two fractions:
\[ \frac{\cos(x-a) \cos(x-b) + \sin(x-b) \sin(x-a)}{\sin(x-a) \cos(x-b)}. \]

5. **Setting Both Sides Equal**:
Now we have:
\[ \frac{k}{\sin(x-a) \cos(x-b)} = \frac{\cos(x-a) \cos(x-b) + \sin(x-b) \sin(x-a)}{\sin(x-a) \cos(x-b)}. \]
By multiplying both sides by \( \sin(x-a) \cos(x-b) \), we get:
\[ k = \cos(x-a) \cos(x-b) + \sin(x-b) \sin(x-a). \]

6. **Using the Cosine Addition Formula**:
The right-hand side can be simplified using the cosine addition formula:
\[ k = \cos((x-a) - (x-b)) = \cos(b-a). \]

### Final Result:
Thus, we find that:
\[ k = \cos(a-b). \]