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.

This problem simplifies the expression \( L\{\sqrt{1 - \sin^2 t}\} \) by using trigonometric identities and finding its Laplace transform. The key concepts include the Pythagorean identity, the Laplace transform of cosine, and handling absolute values. Suitable for undergraduate students studying calculus or differential equations.

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