This solution provides a step-by-step approach to finding the Laplace transform of a piecewise-defined function $f(t)$, which involves both exponential decay and trigonometric functions, combined with a unit step function $U(t - \pi)$.

Here's a breakdown of the approach taken:

1. Expression for $f(t)$: The function $f(t) = e^{-2t} \sin t - e^{-2t} \sin(t - \pi) U(t - \pi)$ is given. This suggests that for $t \geq \pi$, the function shifts and is modified by the unit step function $U(t - \pi)$.

2. Rewrite using trigonometric identities: The solution rewrites $\sin(t - \pi)$ using trigonometric identities, allowing for simplification: $\sin(t - \pi) = \sin t \cos \pi - \cos t \sin \pi = -\sin t$ Then, the function $f(t)$ is expressed in terms of $e^{-2t} \sin t$ and its shifted component.

3. Laplace Transform: The Laplace transform $F(s)$ is computed by transforming each term separately:

   • For $e^{-2t} \sin t$, we have a standard Laplace transform formula.
   • For the second term involving $U(t - \pi)$, the time-shifting property is used. This includes a factor of $e^{-\pi s}$ due to the shift by $\pi$.

The final form of $F(s)$ is: $F(s) = \frac{1}{(s+2)^2 + 1} + e^{-\pi s} \frac{1}{(s+2)^2 + 1}$

Would you like further details on any part of this solution or have any questions? Here are some related questions that may deepen your understanding:

1. How does the time-shifting property of the Laplace transform apply to unit step functions?
2. What are some alternative methods to solve Laplace transforms involving trigonometric functions?
3. How does the Laplace transform handle piecewise functions in general?
4. Could we generalize this approach to other trigonometric or exponential combinations?
5. What are some practical applications of this type of Laplace transform in engineering or physics?

Tip: Always check for trigonometric identities or simplifications in problems with trigonometric functions. This can significantly simplify the Laplace transform process.