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.