1. **State the problem:** Evaluate the integral $$\int \sqrt{9 - x^2} \, dx$$. 2. **Identify the formula and substitution:** This integral is of the form $$\int \sqrt{a^2 - x^2} \, dx$$ where $$a = 3$$. 3. **Use trigonometric substitution:** Let $$x = 3 \sin \theta$$, then $$dx = 3 \cos \theta \, d\theta$$. 4. **Rewrite the integral:** $$\int \sqrt{9 - x^2} \, dx = \int \sqrt{9 - 9 \sin^2 \theta} \cdot 3 \cos \theta \, d\theta = \int \sqrt{9(1 - \sin^2 \theta)} \cdot 3 \cos \theta \, d\theta$$ 5. **Simplify inside the square root:** $$\sqrt{9 \cos^2 \theta} = 3 \cos \theta$$ 6. **Substitute back:** $$\int 3 \cos \theta \cdot 3 \cos \theta \, d\theta = \int 9 \cos^2 \theta \, d\theta$$ 7. **Use the identity:** $$\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$$ 8. **Rewrite the integral:** $$\int 9 \cos^2 \theta \, d\theta = \int 9 \cdot \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{9}{2} \int (1 + \cos 2\theta) \, d\theta$$ 9. **Integrate:** $$\frac{9}{2} \left( \theta + \frac{\sin 2\theta}{2} \right) + C = \frac{9}{2} \theta + \frac{9}{4} \sin 2\theta + C$$ 10. **Back-substitute $$\theta$$:** Since $$x = 3 \sin \theta$$, then $$\sin \theta = \frac{x}{3}$$ and $$\theta = \arcsin \frac{x}{3}$$. 11. **Express $$\sin 2\theta$$:** $$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{x}{3} \cdot \sqrt{1 - \left(\frac{x}{3}\right)^2} = \frac{2x}{3} \sqrt{1 - \frac{x^2}{9}} = \frac{2x}{3} \cdot \frac{\sqrt{9 - x^2}}{3} = \frac{2x \sqrt{9 - x^2}}{9}$$ 12. **Substitute back into the integral:** $$\frac{9}{2} \arcsin \frac{x}{3} + \frac{9}{4} \cdot \frac{2x \sqrt{9 - x^2}}{9} + C = \frac{9}{2} \arcsin \frac{x}{3} + \frac{x \sqrt{9 - x^2}}{2} + C$$ **Final answer:** $$\int \sqrt{9 - x^2} \, dx = \frac{9}{2} \arcsin \frac{x}{3} + \frac{x \sqrt{9 - x^2}}{2} + C$$