To solve the Laplace transform of the given function $f(t) = \cos^2(2t)$, we can approach the problem using standard properties of the Laplace transform.

Step-by-Step Solution:

1. Use a Trigonometric Identity: The expression $\cos^2(2t)$ can be simplified using the identity: $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ Substituting $x = 2t$, we get: $\cos^2(2t) = \frac{1 + \cos(4t)}{2}$

   So, $f(t)$ becomes: $f(t) = \frac{1 + \cos(4t)}{2}$

2. Break the Function into Simpler Terms: Now we have: $f(t) = \frac{1}{2} + \frac{\cos(4t)}{2}$

3. Apply the Laplace Transform: We now take the Laplace transform of each term separately:

   • The Laplace transform of a constant $a$ is: $\mathcal{L}\{a\} = \frac{a}{s}$ So, for $\frac{1}{2}$, the Laplace transform is: $\mathcal{L}\left\{\frac{1}{2}\right\} = \frac{1}{2s}$

   • The Laplace transform of $\cos(bt)$ is: $\mathcal{L}\{\cos(bt)\} = \frac{s}{s^2 + b^2}$ For $\frac{\cos(4t)}{2}$, we use $b = 4$, so: $\mathcal{L}\left\{\frac{\cos(4t)}{2}\right\} = \frac{1}{2} \cdot \frac{s}{s^2 + 16}$

4. Combine the Results: Adding the two terms together, the total Laplace transform is: $\mathcal{L}\{f(t)\} = \frac{1}{2s} + \frac{s}{2(s^2 + 16)}$

Final Answer:

$\mathcal{L}\{ \cos^2(2t) \} = \frac{1}{2s} + \frac{s}{2(s^2 + 16)}$

Would you like more details or further clarifications?

Here are some related questions for deeper understanding:

1. How does the Laplace transform handle functions involving higher powers of trigonometric functions?
2. What is the significance of the identity $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ in simplifying such problems?
3. How would you apply Laplace transform properties to other trigonometric identities, like $\sin^2(t)$?
4. What is the inverse Laplace transform of $\frac{s}{2(s^2 + 16)}$?
5. How does the Laplace transform of a sum of functions relate to the individual transforms?

Tip: When dealing with trigonometric functions in Laplace transforms, it's helpful to memorize common identities and transforms to simplify the process.