The value of the integral `overset(pi)underset(0)int (sin k x)/(sin x)dx` (k is an even integer) is equal to

Options

A. `pi//3`
B. `pi`
C. 0
D. None of these

Answer

To solve the integral \( I = \int_0^\pi \frac{\sin(kx)}{\sin(x)} \, dx \) where \( k \) is an even integer, we can use properties of definite integrals and the sine function.

### Step-by-Step Solution:

1. **Define the Integral:**
Let \( I = \int_0^\pi \frac{\sin(kx)}{\sin(x)} \, dx \).

2. **Use the Property of Definite Integrals:**
We can use the property of definite integrals:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]
Here, \( a = 0 \) and \( b = \pi \). So, we replace \( x \) with \( \pi - x \):
\[ I = \int_0^\pi \frac{\sin(k(\pi - x))}{\sin(\pi - x)} \, dx. \]

3. **Simplify the Integral:**
Using the sine function property \( \sin(k(\pi - x)) = \sin(k\pi - kx) = (-1)^k \sin(kx) \) and \( \sin(\pi - x) = \sin(x) \):
\[ I = \int_0^\pi \frac{(-1)^k \sin(kx)}{\sin(x)} \, dx. \]
Since \( k \) is an even integer, \( (-1)^k = 1 \):
\[ I = \int_0^\pi \frac{\sin(kx)}{\sin(x)} \, dx = I. \]

4. **Combine the Results:**
Now we have:
\[ I = \int_0^\pi \frac{\sin(kx)}{\sin(x)} \, dx = -I. \]
Adding \( I \) to both sides gives:
\[ 2I = 0. \]

5. **Solve for \( I \):**
Dividing by 2:
\[ I = 0. \]

### Conclusion:
The value of the integral \( \int_0^\pi \frac{\sin(kx)}{\sin(x)} \, dx \) where \( k \) is an even integer is \( 0 \).