$$ s\left\lbrack\cos(\alpha t)\rbrack\right. $$

The expression you're dealing with is asking for the Laplace transform of $ \cos(\alpha t) $. Let's compute it step by step.

The Laplace transform of $ \cos(\alpha t) $ is given by the standard formula:

\[
\mathcal{L}\{\cos(\alpha t)\} = \frac{s}{s^2 + \alpha^2}
\]

Thus, the Laplace transform of $ \cos(\alpha t) $ is:

\[
\mathcal{L}\{\cos(\alpha t)\} = \frac{s}{s^2 + \alpha^2}
\]

Would you like further details or examples of this concept?

Here are 5 questions to expand on this:

1. What is the Laplace transform of $ \sin(\alpha t) $?
2. How does the Laplace transform handle exponential functions, like $ e^{at} \cos(\alpha t) $?
3. What is the inverse Laplace transform of $ \frac{s}{s^2 + \alpha^2} $?
4. How can we use Laplace transforms to solve differential equations?
5. What happens to the Laplace transform if the cosine function has a phase shift, like $ \cos(\alpha t + \phi) $?

**Tip**: The Laplace transform is particularly useful in solving linear differential equations, especially when initial conditions are provided.

Understand how to compute the Laplace transform of cos(αt) with a clear and concise explanation. This problem demonstrates the application of Laplace transforms to trigonometric functions and covers key mathematical concepts, including the standard formula for L{cos(αt)}. Suitable for high school and undergraduate students.

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