The given integral is:

$\int_0^2 \frac{1}{\sqrt{4 - s^2}} \, ds$

This is a standard integral that resembles the form of an inverse trigonometric function. Specifically, it is of the form:

$\int \frac{1}{\sqrt{a^2 - s^2}} \, ds = \arcsin\left(\frac{s}{a}\right) + C$

In our case, $a = 2$. So we apply the formula:

$\int_0^2 \frac{1}{\sqrt{4 - s^2}} \, ds = \arcsin\left(\frac{s}{2}\right) \Bigg|_0^2$

Now evaluate the limits:

1. At $s = 2$:

$\arcsin\left(\frac{2}{2}\right) = \arcsin(1) = \frac{\pi}{2}$

2. At $s = 0$:

$\arcsin\left(\frac{0}{2}\right) = \arcsin(0) = 0$

Therefore, the value of the integral is:

$\frac{\pi}{2} - 0 = \frac{\pi}{2}$

Final Answer:

$\int_0^2 \frac{1}{\sqrt{4 - s^2}} \, ds = \frac{\pi}{2}$

Would you like more details or further clarifications?

Here are 5 relative questions to expand your understanding:

1. What is the general form of integrals involving inverse trigonometric functions?
2. How do we recognize when to use the arcsin formula in integrals?
3. What is the geometric interpretation of the arcsine function?
4. Can you derive the arcsin integral formula from basic principles?
5. How does the bounds of the arcsine function affect its evaluation?

Tip: Always check for substitutions that transform integrals into recognizable forms involving inverse trigonometric functions.