The expression $L\{\sqrt{1 - \sin^2 t}\}$ simplifies as follows:

We start by recognizing that:

$\sqrt{1 - \sin^2 t} = \sqrt{\cos^2 t} = |\cos t|$

Since the cosine function is positive for $t \in [0, \pi]$ and negative for $t \in [\pi, 2\pi]$, we write $|\cos t|$ to account for the absolute value. For simplicity, if we assume the function is evaluated where $\cos t \geq 0$ (i.e., $t \in [0, \pi]$), we can just take:

$\sqrt{1 - \sin^2 t} = \cos t$

Now, we need to find the Laplace transform $L\{\cos t\}$.

The Laplace transform of $\cos(at)$ is a standard result and is given by:

$L\{\cos(at)\} = \frac{s}{s^2 + a^2}$

In this case, $a = 1$, so:

$L\{\cos t\} = \frac{s}{s^2 + 1}$

Thus, the Laplace transform of $\sqrt{1 - \sin^2 t}$ is:

$L\{|\cos t|\} = \frac{s}{s^2 + 1}$

Do you want more details or have any questions?

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Here are 5 related questions:

1. What is the Laplace transform of $\sin(at)$?
2. How does the Laplace transform handle absolute values?
3. What is the inverse Laplace transform of $\frac{s}{s^2 + 1}$?
4. How does the Laplace transform behave with trigonometric functions?
5. How would the Laplace transform change for $\cos(bt)$ where $b \neq 1$?

Tip: The Laplace transform is often useful for solving differential equations, especially those with initial conditions.