To evaluate the integral

$\int_0^{\infty} e^{-3t} \sin(2t) \, dt,$

we can use the formula for the Laplace transform of $e^{at} \sin(bt)$. The general formula for the integral

$\int_0^{\infty} e^{-pt} \sin(qt) \, dt$

is given by:

$\frac{q}{p^2 + q^2}$

where $p > 0$.

In this case:

• $p = 3$
• $q = 2$

Substitute these values into the formula:

$\int_0^{\infty} e^{-3t} \sin(2t) \, dt = \frac{2}{3^2 + 2^2} = \frac{2}{9 + 4} = \frac{2}{13}$

So, the value of the integral is:

$\frac{2}{13}$

Would you like a detailed step-by-step explanation or have any questions?

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Here are 5 related questions to practice:

1. Evaluate $\int_0^{\infty} e^{-2t} \cos(3t) \, dt$.
2. Find the Laplace transform of $e^{-3t} \cos(4t)$.
3. Determine the integral $\int_0^{\infty} e^{-t} \sin(5t) \, dt$.
4. Calculate $\int_0^{\infty} e^{-4t} \sin(2t) \, dt$.
5. What is the value of $\int_0^{\infty} e^{-2t} \sin(t) \, dt$?

Tip: Understanding the Laplace transform formulas can greatly simplify the evaluation of many improper integrals involving exponentials and trigonometric functions.