1. **State the problem:** We want to evaluate the integral $$\int \frac{x^3}{\sqrt{x^2 - 9}} \, dx$$. 2. **Identify the substitution:** Since the integrand contains $\sqrt{x^2 - 9}$, a good substitution is to let $$x = 3\sec(\theta)$$ because $\sec^2(\theta) - 1 = \tan^2(\theta)$, which simplifies the square root. 3. **Compute $dx$ and substitute:** $$dx = 3\sec(\theta)\tan(\theta) \, d\theta$$ 4. **Rewrite the integral in terms of $\theta$:** $$\sqrt{x^2 - 9} = \sqrt{9\sec^2(\theta) - 9} = \sqrt{9(\sec^2(\theta) - 1)} = 3\tan(\theta)$$ The numerator: $$x^3 = (3\sec(\theta))^3 = 27\sec^3(\theta)$$ So the integral becomes: $$\int \frac{27\sec^3(\theta)}{3\tan(\theta)} \cdot 3\sec(\theta)\tan(\theta) \, d\theta = \int \frac{27\sec^3(\theta)}{3\tan(\theta)} \times 3\sec(\theta)\tan(\theta) \, d\theta$$ 5. **Simplify the integrand:** $$= \int \frac{27\sec^3(\theta)}{3\tan(\theta)} \times 3\sec(\theta)\tan(\theta) \, d\theta = \int 27\sec^4(\theta) \, d\theta$$ 6. **Integral to solve:** $$\int 27\sec^4(\theta) \, d\theta = 27 \int \sec^4(\theta) \, d\theta$$ 7. **Use reduction formula or rewrite:** Recall: $$\sec^4(\theta) = (\sec^2(\theta))^2 = (1 + \tan^2(\theta))^2 = 1 + 2\tan^2(\theta) + \tan^4(\theta)$$ Alternatively, use the reduction formula: $$\int \sec^n(\theta) \, d\theta = \frac{\sec^{n-2}(\theta) \tan(\theta)}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2}(\theta) \, d\theta$$ For $n=4$: $$\int \sec^4(\theta) \, d\theta = \frac{\sec^2(\theta) \tan(\theta)}{3} + \frac{2}{3} \int \sec^2(\theta) \, d\theta$$ 8. **Evaluate $\int \sec^2(\theta) \, d\theta$:** $$\int \sec^2(\theta) \, d\theta = \tan(\theta) + C$$ 9. **Substitute back:** $$\int \sec^4(\theta) \, d\theta = \frac{\sec^2(\theta) \tan(\theta)}{3} + \frac{2}{3} \tan(\theta) + C = \frac{\tan(\theta)}{3} (\sec^2(\theta) + 2) + C$$ 10. **Multiply by 27:** $$27 \int \sec^4(\theta) \, d\theta = 27 \times \frac{\tan(\theta)}{3} (\sec^2(\theta) + 2) + C = 9 \tan(\theta) (\sec^2(\theta) + 2) + C$$ 11. **Rewrite in terms of $x$:** Recall: $$x = 3\sec(\theta) \Rightarrow \sec(\theta) = \frac{x}{3}$$ $$\tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \sqrt{\frac{x^2}{9} - 1} = \frac{\sqrt{x^2 - 9}}{3}$$ So: $$9 \tan(\theta) (\sec^2(\theta) + 2) = 9 \times \frac{\sqrt{x^2 - 9}}{3} \left( \frac{x^2}{9} + 2 \right) = 3 \sqrt{x^2 - 9} \left( \frac{x^2}{9} + 2 \right)$$ Simplify inside the parentheses: $$\frac{x^2}{9} + 2 = \frac{x^2 + 18}{9}$$ Therefore: $$3 \sqrt{x^2 - 9} \times \frac{x^2 + 18}{9} = \frac{3}{9} (x^2 + 18) \sqrt{x^2 - 9} = \frac{1}{3} (x^2 + 18) \sqrt{x^2 - 9}$$ 12. **Final answer:** $$\int \frac{x^3}{\sqrt{x^2 - 9}} \, dx = \frac{1}{3} (x^2 + 18) \sqrt{x^2 - 9} + C$$