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The given integral is:

\[
\int_0^2 \frac{ds}{\sqrt{4 - s^2}}
\]

This is a standard integral of the form:

\[
\int \frac{ds}{\sqrt{a^2 - s^2}} = \arcsin\left(\frac{s}{a}\right) + C
\]

In this case, \(a = 2\), so the integral becomes:

\[
\int_0^2 \frac{ds}{\sqrt{4 - s^2}} = \arcsin\left(\frac{s}{2}\right) \Bigg|_0^2
\]

Now, evaluate the definite integral:

1. At \(s = 2\):
   \[
   \arcsin\left(\frac{2}{2}\right) = \arcsin(1) = \frac{\pi}{2}
   \]

2. At \(s = 0\):
   \[
   \arcsin\left(\frac{0}{2}\right) = \arcsin(0) = 0
   \]

Thus, the result of the integral is:

\[
\frac{\pi}{2} - 0 = \frac{\pi}{2}
\]

So, the value of the given integral is:

\[
\frac{\pi}{2}
\]

Would you like more details or clarifications? Here are 5 related questions to deepen your understanding:

1. What is the general formula for integrating expressions like \(\frac{ds}{\sqrt{a^2 - s^2}}\)?
2. How does the arcsine function work in trigonometric integrals?
3. Why does the definite integral simplify to \(\frac{\pi}{2}\) in this case?
4. How can substitution help in solving integrals involving square roots?
5. What other trigonometric integrals are similar to this one?

**Tip**: For integrals involving expressions like \(\sqrt{a^2 - x^2}\), try using trigonometric substitutions, such as \(x = a\sin\theta\), to simplify the calculation.

Calculate the integral from 0 to 2 of ds over the square root of (4 - s^2).

Explore how to compute the integral of ds over the square root of (4 - s^2) using trigonometric substitution, specifically focusing on arcsin transformations. This technique is demonstrated with step-by-step solutions and is suitable for college-level students studying integral calculus.

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