Express int(dx/(sqrt(1-x^2))) in terms of its trigonometric functions.

The integral you provided is:

\[
\int \frac{dx}{\sqrt{1 - x^2}}
\]

This is a standard integral that results in an inverse trigonometric function. The result of this integral is:

\[
\arcsin(x) + C
\]

where \( C \) is the constant of integration.

### Explanation:
The function inside the integral, \( \frac{1}{\sqrt{1 - x^2}} \), is the derivative of the arcsine function, \( \arcsin(x) \). Hence, the integral evaluates directly to \( \arcsin(x) \).

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Here are 5 related questions you may find useful:

1. What is the integral of \( \frac{dx}{1 + x^2} \)?
2. How can we derive the derivative of \( \arcsin(x) \)?
3. What is the integral of \( \frac{dx}{(a^2 - x^2)^{3/2}} \)?
4. How do trigonometric substitution methods help in evaluating integrals like \( \int \frac{dx}{\sqrt{1 - x^2}} \)?
5. How do we compute \( \int \arcsin(x) \, dx \)?

**Tip**: When dealing with integrals involving expressions like \( \sqrt{1 - x^2} \), it's often helpful to recognize them as derivatives of inverse trigonometric functions.

Discover how to express the integral of dx / √(1 - x^2) using inverse trigonometric functions. This calculus problem results in the arcsin function and demonstrates key concepts in trigonometric integrals and substitution methods. Ideal for Grades 11-12.

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