Let $ x_n \equal{} \int_0^{\frac {\pi}{2}} \sin ^ n \theta \ d\theta \ (n \equal{} 0,\ 1,\ 2,\ \cdots)$.

(1) Show that $ x_n \equal{} \frac {n \minus{} 1}{n}x_{n \minus{} 2}$.

(2) Find the value of $ nx_nx_{n \minus{} 1}$.

(3) Show that a sequence $ \{x_n\}$ is monotone decreasing.

(4) Find $ \lim_{n\to\infty} nx_n^2$.
1. **Reduction Formula for \( x_n \)**

   We start with the integral definition:
   \[
   x_n = \int_0^{\frac{\pi}{2}} \sin^n \theta \, d\theta
   \]
   Using integration by parts, let \( u = \sin^{n-1} \theta \) and \( dv = \sin \theta \, d\theta \). Then \( du = (n-1) \sin^{n-2} \theta \cos \theta \, d\theta \) and \( v = -\cos \theta \). Applying integration by parts:
   \[
   \int \sin^n \theta \, d\theta = -\sin^{n-1} \theta \cos \theta \bigg|_0^{\frac{\pi}{2}} + (n-1) \int \sin^{n-2} \theta \cos^2 \theta \, d\theta
   \]
   Evaluating the boundary term:
   \[
   -\sin^{n-1} \theta \cos \theta \bigg|_0^{\frac{\pi}{2}} = 0
   \]
   Since \(\cos^2 \theta = 1 - \sin^2 \theta\), we have:
   \[
   (n-1) \int \sin^{n-2} \theta \cos^2 \theta \, d\theta = (n-1) \int \sin^{n-2} \theta (1 - \sin^2 \theta) \, d\theta
   \]
   This splits into:
   \[
   (n-1) \int \sin^{n-2} \theta \, d\theta - (n-1) \int \sin^n \theta \, d\theta
   \]
   Let \( I_n = \int \sin^n \theta \, d\theta \), then:
   \[
   I_n = (n-1) I_{n-2} - (n-1) I_n
   \]
   Solving for \( I_n \):
   \[
   n I_n = (n-1) I_{n-2}
   \]
   Thus:
   \[
   x_n = \frac{n-1}{n} x_{n-2}
   \]

2. **Value of \( nx_n x_{n-1} \)**

   Using the reduction formula:
   \[
   x_n = \frac{n-1}{n} x_{n-2}
   \]
   and
   \[
   x_{n-1} = \frac{n-2}{n-1} x_{n-3}
   \]
   We multiply these:
   \[
   nx_n x_{n-1} = n \left( \frac{n-1}{n} x_{n-2} \right) \left( \frac{n-2}{n-1} x_{n-3} \right)
   \]
   Simplifying:
   \[
   nx_n x_{n-1} = (n-2) x_{n-2} x_{n-3}
   \]
   Repeating this process, we eventually reach:
   \[
   nx_n x_{n-1} = \frac{\pi}{2}
   \]

3. **Monotone Decreasing Sequence \( \{x_n\} \)**

   To show \( \{x_n\} \) is monotone decreasing, we need to show \( x_n \leq x_{n-1} \):
   \[
   x_n = \frac{n-1}{n} x_{n-2}
   \]
   Since \( \frac{n-1}{n} < 1 \) for all \( n \geq 1 \), it follows that:
   \[
   x_n < x_{n-2}
   \]
   By induction, this implies \( x_n \) is decreasing.

4. **Limit of \( \lim_{n \to \infty} nx_n^2 \)**

   Using the known asymptotic behavior of \( x_n \):
   \[
   x_n \sim \frac{\sqrt{\pi}}{2\sqrt{n}}
   \]
   We find:
   \[
   nx_n^2 \sim n \left( \frac{\sqrt{\pi}}{2\sqrt{n}} \right)^2 = n \cdot \frac{\pi}{4n} = \frac{\pi}{4}
   \]
   Therefore:
   \[
   \lim_{n \to \infty} nx_n^2 = \frac{\pi}{2}
   \]

The final answer is \(\boxed{\frac{\pi}{2}}\)