Prove that the integral
`int_(0)^(pi) (sin 2 k x)/(sin x) dx`
equals zero if k an integer.

Answer

To prove that the integral

\[ I = \int_{0}^{\pi} \frac{\sin(2kx)}{\sin(x)} \, dx \]

equals zero when \( k \) is an integer, we can use the property of definite integrals and the symmetry of the sine function.

### Step 1: Change of Variable
We will use the substitution \( x = \pi - t \). Then, \( dx = -dt \). The limits change as follows:
- When \( x = 0 \), \( t = \pi \)
- When \( x = \pi \), \( t = 0 \)

Thus, we can rewrite the integral as:

\[ I = \int_{\pi}^{0} \frac{\sin(2k(\pi - t))}{\sin(\pi - t)} (-dt) = \int_{0}^{\pi} \frac{\sin(2k(\pi - t))}{\sin(t)} dt \]

### Step 2: Simplifying the Integral
Using the sine function properties, we know that:

\[ \sin(2k(\pi - t)) = \sin(2k\pi - 2kt) = -\sin(2kt) \]

and

\[ \sin(\pi - t) = \sin(t) \]

Thus, we can rewrite the integral:

\[ I = \int_{0}^{\pi} \frac{-\sin(2kt)}{\sin(t)} dt = -\int_{0}^{\pi} \frac{\sin(2kt)}{\sin(t)} dt = -I \]

### Step 3: Adding the Two Expressions
Now we have:

\[ I = -I \]

Adding \( I \) to both sides gives:

\[ 2I = 0 \]

### Step 4: Conclusion
Thus, we find that:

\[ I = 0 \]

This proves that:

\[ \int_{0}^{\pi} \frac{\sin(2kx)}{\sin(x)} \, dx = 0 \]

for any integer \( k \).