The value of `int_0^pi(sin2kx)/(sinx)dx," where "k in I`, is

Options

A. `pi/2`
B. `pi`
C. `(3pi)/2`
D. `0`

Answer

To solve the integral \( I = \int_0^\pi \frac{\sin(2kx)}{\sin(x)} \, dx \), we will use a property of definite integrals.

### Step 1: Define the integral
Let:
\[ I = \int_0^\pi \frac{\sin(2kx)}{\sin(x)} \, dx \]

### Step 2: Use the property of definite integrals
We will use the property that states:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
Applying this to our integral, we substitute \( x \) with \( \pi - x \):
\[ I = \int_0^\pi \frac{\sin(2k(\pi - x))}{\sin(\pi - x)} \, dx \]

### Step 3: Simplify the integral
Using the identities \( \sin(\pi - x) = \sin(x) \) and \( \sin(2k(\pi - x)) = \sin(2k\pi - 2kx) = -\sin(2kx) \) (since \( \sin \) is an odd function):
\[ I = \int_0^\pi \frac{-\sin(2kx)}{\sin(x)} \, dx \]
This gives us:
\[ I = -\int_0^\pi \frac{\sin(2kx)}{\sin(x)} \, dx = -I \]

### Step 4: Solve for \( I \)
From the equation \( I = -I \), we can add \( I \) to both sides:
\[ 2I = 0 \]
Thus:
\[ I = 0 \]

### Final Answer
The value of the integral is:
\[ \int_0^\pi \frac{\sin(2kx)}{\sin(x)} \, dx = 0 \]