We need to evaluate the integral:

$I = \int \frac{dx}{\sqrt{4 - x^2}}$

Step 1: Recognizing the Standard Form

This integral matches the standard form:

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C$

where $a = 2$.

Step 2: Substituting Values

Comparing with the standard form, we set $a = 2$, so the result is:

$I = \arcsin\left(\frac{x}{2}\right) + C$

Final Answer:

$\arcsin\left(\frac{x}{2}\right) + C$

Would you like a detailed breakdown of why this substitution works?

Related Questions:

1. How do you derive the standard integral formula for $\int \frac{dx}{\sqrt{a^2 - x^2}}$?
2. What is the geometric interpretation of this integral?
3. How would this integral change if the denominator were $\sqrt{x^2 - 4}$ instead?
4. How can trigonometric substitution be used to evaluate this integral?
5. What happens if we replace 4 with another constant $k$ in the square root?

Tip: Recognizing standard integral forms can save time in solving complex integrals!