To solve the integral \( I = \int_0^{\frac{\pi}{2}} \frac{\sin(nx)}{\sin(x)} \, dx \) where \( n = 2m + 1 \) and \( m \in \mathbb{N} \cup \{0\} \), we can follow these steps:
### Step 1: Substitute \( n \)
Given \( n = 2m + 1 \), we can start with the simplest case, \( m = 0 \), which gives \( n = 1 \).
### Step 2: Calculate the integral for \( n = 1 \)
Substituting \( n = 1 \):
\[
I = \int_0^{\frac{\pi}{2}} \frac{\sin(x)}{\sin(x)} \, dx = \int_0^{\frac{\pi}{2}} 1 \, dx
\]
Calculating this integral:
\[
I = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2}
\]
### Step 3: Check for \( n = 3 \)
Now, let's check for \( n = 3 \) (when \( m = 1 \)):
\[
I = \int_0^{\frac{\pi}{2}} \frac{\sin(3x)}{\sin(x)} \, dx
\]
Using the identity \( \sin(3x) = 3\sin(x) - 4\sin^3(x) \):
\[
I = \int_0^{\frac{\pi}{2}} \frac{3\sin(x) - 4\sin^3(x)}{\sin(x)} \, dx = \int_0^{\frac{\pi}{2}} (3 - 4\sin^2(x)) \, dx
\]
### Step 4: Simplify the integral
Now, we can separate the integral:
\[
I = \int_0^{\frac{\pi}{2}} 3 \, dx - 4 \int_0^{\frac{\pi}{2}} \sin^2(x) \, dx
\]
Calculating the first part:
\[
\int_0^{\frac{\pi}{2}} 3 \, dx = 3 \cdot \frac{\pi}{2} = \frac{3\pi}{2}
\]
### Step 5: Calculate \( \int_0^{\frac{\pi}{2}} \sin^2(x) \, dx \)
Using the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \):
\[
\int_0^{\frac{\pi}{2}} \sin^2(x) \, dx = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_0^{\frac{\pi}{2}} = \frac{1}{2} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{4}
\]
### Step 6: Substitute back into the integral
Now substituting back:
\[
I = \frac{3\pi}{2} - 4 \cdot \frac{\pi}{4} = \frac{3\pi}{2} - \pi = \frac{\pi}{2}
\]
### Conclusion
We have shown that for both \( n = 1 \) and \( n = 3 \), the integral evaluates to \( \frac{\pi}{2} \). By similar reasoning, this will hold for any odd integer \( n \) of the form \( 2m + 1 \).
Thus, the final answer is:
\[
\boxed{\frac{\pi}{2}}
\]
