We are given the integral:

$\int_{0}^{\infty} e^{-2t} t^2 \sin(3t) \, dt$

Solution using the Laplace transform:

The Laplace transform of a function $f(t)$ is given by:

$\mathcal{L} \{ f(t) \} = \int_{0}^{\infty} e^{-st} f(t) \, dt$

We recognize the given integral as the Laplace transform of $f(t) = t^2 \sin(3t)$ evaluated at $s = 2$, so we first determine the general Laplace transform.

Step 1: Laplace Transform of $t^n \sin(bt)$

The known Laplace transform formula for $t^n \sin(bt)$ is:

$\mathcal{L} \{ t^n \sin(bt) \} = \frac{n! \cdot s}{(s^2 + b^2)^{n+1}}$

For $n = 2$, the formula becomes:

$\mathcal{L} \{ t^2 \sin(bt) \} = \frac{2! \cdot s}{(s^2 + b^2)^3} = \frac{2s}{(s^2 + b^2)^3}$

Step 2: Substituting values

Given $b = 3$ and evaluating at $s = 2$:

$\int_{0}^{\infty} e^{-2t} t^2 \sin(3t) \, dt = \frac{2(2)}{(2^2 + 3^2)^3}$

$= \frac{4}{(4 + 9)^3}$

$= \frac{4}{13^3}$

$= \frac{4}{2197}$

Final Answer:

$\int_{0}^{\infty} e^{-2t} t^2 \sin(3t) \, dt = \frac{4}{2197}$

Would you like a step-by-step breakdown of the solution or any clarification?

Here are five related questions you might find helpful:

1. What is the general formula for the Laplace transform of $t^n e^{-\alpha t}$?
2. How can the convolution theorem be applied in Laplace transforms?
3. What are the applications of Laplace transforms in engineering and physics?
4. How do we find the inverse Laplace transform of a given function?
5. How does the presence of exponential terms in the Laplace transform affect convergence?

Tip: Always check for known Laplace transform formulas to simplify complex integral evaluations.