The given integral is:

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

Solution:

This is a standard integral, and the best approach is to use trigonometric substitution.

Steps:

1. Substitute: Let $x = \sin\theta$, so $dx = \cos\theta \, d\theta$ and $\sqrt{1 - x^2} = \sqrt{1 - \sin^2\theta} = \cos\theta$.

   Substituting these into the integral: $\int \sqrt{1 - x^2} \, dx = \int \cos\theta \cdot \cos\theta \, d\theta = \int \cos^2\theta \, d\theta$

2. Simplify using trigonometric identity: Use the identity: $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$

   The integral becomes: $\int \cos^2\theta \, d\theta = \int \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int 1 \, d\theta + \frac{1}{2} \int \cos(2\theta) \, d\theta$

3. Solve each term:

   • For $\int 1 \, d\theta = \theta$
   • For $\int \cos(2\theta) \, d\theta = \frac{\sin(2\theta)}{2}$

   Combine these: $\int \cos^2\theta \, d\theta = \frac{1}{2} \theta + \frac{1}{4} \sin(2\theta) + C$

4. Back-substitute: Recall that $x = \sin\theta$, so $\theta = \arcsin(x)$ and $\sin(2\theta) = 2\sin\theta\cos\theta = 2x\sqrt{1 - x^2}$.

   Substituting back: $\int \sqrt{1 - x^2} \, dx = \frac{1}{2} \arcsin(x) + \frac{1}{4} \cdot 2x\sqrt{1 - x^2} + C$

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

Final Answer:

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

Would you like a detailed explanation of any step or have additional questions?

Here are 5 questions that can deepen your understanding:

1. What is the purpose of trigonometric substitution in integral calculus?
2. Why is $\cos^2\theta$ rewritten using the identity $\frac{1 + \cos(2\theta)}{2}$?
3. How does $\sin(2\theta) = 2\sin\theta\cos\theta$ simplify back-substitution?
4. What would happen if we used different substitution methods (e.g., $x = \cos\theta$)?
5. How does the constant of integration $C$ affect definite integrals?

Tip:

Trigonometric substitutions are particularly useful for integrals involving $\sqrt{1 - x^2}$, $\sqrt{1 + x^2}$, or $\sqrt{x^2 - 1}$.