The integral you provided is:

$$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx$$

This is a standard integral that can be solved using a trigonometric substitution. The appropriate substitution is:

$$x = a \sin(\theta)$$

Steps to solve:

1. Substitute $x = a \sin(\theta)$, then $dx = a \cos(\theta) \, d\theta$.

2. The integrand becomes:

   $$\int \frac{1}{\sqrt{a^2 - a^2 \sin^2(\theta)}} \cdot a \cos(\theta) \, d\theta$$

3. Simplify the square root:

   $$\sqrt{a^2 - a^2 \sin^2(\theta)} = \sqrt{a^2 (1 - \sin^2(\theta))} = \sqrt{a^2 \cos^2(\theta)} = a \cos(\theta)$$

   So the integrand simplifies to:

   $$\int \frac{a \cos(\theta)}{a \cos(\theta)} \, d\theta = \int 1 \, d\theta$$

4. Integrate with respect to $\theta$:

   $$\theta + C$$

5. Convert back to $x$ using $\theta = \arcsin\left(\frac{x}{a}\right)$:

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

Final Answer:

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

Would you like any more details on this solution, or do you have any questions?

Here are five related questions you might find interesting:

1. How would you solve the integral $\int \frac{x}{\sqrt{a^2 - x^2}} \, dx$?
2. What is the integral of $\int \frac{1}{\sqrt{a^2 + x^2}} \, dx$?
3. How does the substitution $x = a \sin(\theta)$ work geometrically in this context?
4. What is the significance of the constant $C$ in indefinite integrals?
5. Can you derive the integral using a different substitution, like $x = a \cos(\theta)$?

Tip: When dealing with integrals involving square roots like this, trigonometric substitutions are often useful because they simplify the radical expression.