To solve the problem, we need to analyze the expression given for \( u_n \) and find the differences \( u_n - u_{n-1} \) for \( n = 2, 3, 4, \ldots \).
### Step-by-step Solution:
1. **Define the Expression**:
We have:
\[
u_n = \int_0^{\frac{\pi}{2}} \frac{\sin^2(nx)}{\sin x} \, dx
\]
2. **Find the Difference**:
We need to compute \( u_n - u_{n-1} \):
\[
u_n - u_{n-1} = \int_0^{\frac{\pi}{2}} \frac{\sin^2(nx)}{\sin x} \, dx - \int_0^{\frac{\pi}{2}} \frac{\sin^2((n-1)x)}{\sin x} \, dx
\]
This simplifies to:
\[
u_n - u_{n-1} = \int_0^{\frac{\pi}{2}} \frac{\sin^2(nx) - \sin^2((n-1)x)}{\sin x} \, dx
\]
3. **Use the Identity for Sine Squared**:
We can use the identity:
\[
\sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B)
\]
Let \( A = nx \) and \( B = (n-1)x \):
\[
\sin^2(nx) - \sin^2((n-1)x) = (\sin(nx) - \sin((n-1)x))(\sin(nx) + \sin((n-1)x))
\]
4. **Apply the Sine Difference Formula**:
We can use the sine difference formula:
\[
\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)
\]
Thus:
\[
\sin(nx) - \sin((n-1)x) = 2 \cos\left(\frac{(2n-1)x}{2}\right) \sin\left(\frac{x}{2}\right)
\]
5. **Substituting Back**:
Substitute back into the integral:
\[
u_n - u_{n-1} = \int_0^{\frac{\pi}{2}} \frac{2 \cos\left(\frac{(2n-1)x}{2}\right) \sin\left(\frac{x}{2}\right) \cdot (\sin(nx) + \sin((n-1)x))}{\sin x} \, dx
\]
6. **Evaluate the Integral**:
The integral can be evaluated, but we are primarily interested in the form of \( u_n - u_{n-1} \). The key point is that:
\[
u_n - u_{n-1} \text{ can be expressed as a function of } n
\]
7. **Check for Patterns**:
After evaluating for specific values of \( n \):
- \( u_2 - u_1 \)
- \( u_3 - u_2 \)
- \( u_4 - u_3 \)
We find that these differences yield terms of the form:
\[
\frac{1}{2n - 1}
\]
Specifically:
- \( u_2 - u_1 = \frac{1}{3} \)
- \( u_3 - u_2 = \frac{1}{5} \)
- \( u_4 - u_3 = \frac{1}{7} \)
8. **Conclusion**:
The sequence \( u_2 - u_1, u_3 - u_2, u_4 - u_3, \ldots \) is:
\[
\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \ldots
\]
The reciprocals of these terms are:
\[
3, 5, 7, \ldots
\]
which are in Arithmetic Progression (AP). Therefore, the original terms are in Harmonic Progression (HP).
### Final Answer:
The differences \( u_2 - u_1, u_3 - u_2, u_4 - u_3, \ldots \) are in HP.
