We are tasked with finding the Laplace transform of $(t \cos(2t))^2$. Let's denote this function as $f(t) = (t \cos(2t))^2$.

Step 1: Simplify the Expression

We first simplify $f(t) = (t \cos(2t))^2$:

$f(t) = t^2 \cos^2(2t)$

Using the trigonometric identity:

$\cos^2(2t) = \frac{1 + \cos(4t)}{2}$

We can now rewrite $f(t)$ as:

$f(t) = t^2 \cdot \frac{1 + \cos(4t)}{2} = \frac{t^2}{2} + \frac{t^2 \cos(4t)}{2}$

Step 2: Apply the Laplace Transform

Now, apply the Laplace transform to both terms separately.

First Term: $\frac{t^2}{2}$

The Laplace transform of $t^n$ is:

$\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$

So, for $\frac{t^2}{2}$:

$\mathcal{L}\left\{\frac{t^2}{2}\right\} = \frac{1}{2} \cdot \frac{2!}{s^{2+1}} = \frac{1}{s^3}$

Second Term: $\frac{t^2 \cos(4t)}{2}$

The Laplace transform of $t^n \cos(at)$ is:

$\mathcal{L}\{t^n \cos(at)\} = \frac{s^2 - a^2}{(s^2 + a^2)^{n+1}}$

For $t^2 \cos(4t)$, with $n = 2$ and $a = 4$:

$\mathcal{L}\left\{\frac{t^2 \cos(4t)}{2}\right\} = \frac{1}{2} \cdot \frac{s^2 - 16}{(s^2 + 16)^3}$

Step 3: Combine the Results

Finally, we add the Laplace transforms of the two terms:

$\mathcal{L}\{(t \cos(2t))^2\} = \frac{1}{s^3} + \frac{1}{2} \cdot \frac{s^2 - 16}{(s^2 + 16)^3}$

This is the Laplace transform of $(t \cos(2t))^2$.

Would you like further details or clarifications on any of the steps?

Relative Questions:

1. What is the Laplace transform of $t^3 \sin(at)$?
2. How do you derive the Laplace transform of $\cos(at)$?
3. Can you explain the convolution theorem in Laplace transforms?
4. What is the inverse Laplace transform of $\frac{1}{s^3}$?
5. How does the shifting theorem apply to Laplace transforms?

Tip:

When dealing with products of $t$ and trigonometric functions, it's useful to break them down into simpler terms using trigonometric identities before applying the Laplace transform.