1. **State the problem:** We need to evaluate the integral $$\int x^2 (x^2 - 25)^{\frac{3}{2}} \, dx$$. 2. **Identify a substitution:** Let $$u = x^2 - 25$$. Then, $$du = 2x \, dx$$ or $$\frac{du}{2} = x \, dx$$. 3. **Rewrite the integral:** We have $$x^2 (x^2 - 25)^{\frac{3}{2}} dx = x^2 u^{\frac{3}{2}} dx$$. Since $$du = 2x dx$$, we can write $$x dx = \frac{du}{2}$$, but we still have an extra $$x$$ in $$x^2$$. Express $$x^2$$ in terms of $$u$$: $$x^2 = u + 25$$. 4. **Rewrite the integral in terms of $$u$$:** $$\int x^2 u^{\frac{3}{2}} dx = \int (u + 25) u^{\frac{3}{2}} \frac{du}{2x}$$. But we have a problem: the substitution introduces $$du = 2x dx$$, so $$dx = \frac{du}{2x}$$, and the integral still has $$x$$ in the denominator. 5. **Use trigonometric substitution instead:** Since $$x^2 - 25$$ appears, let $$x = 5 \sec \theta$$, so that $$x^2 - 25 = 25 \sec^2 \theta - 25 = 25 (\sec^2 \theta - 1) = 25 \tan^2 \theta$$. Then, $$\sqrt{x^2 - 25} = 5 \tan \theta$$. 6. **Rewrite the integral:** $$x^2 = 25 \sec^2 \theta$$, $$(x^2 - 25)^{\frac{3}{2}} = (25 \tan^2 \theta)^{\frac{3}{2}} = 25^{\frac{3}{2}} \tan^3 \theta = 125 \tan^3 \theta$$, and $$dx = 5 \sec \theta \tan \theta d\theta$$. 7. **Substitute all into the integral:** $$\int x^2 (x^2 - 25)^{\frac{3}{2}} dx = \int 25 \sec^2 \theta \cdot 125 \tan^3 \theta \cdot 5 \sec \theta \tan \theta d\theta$$ $$= 25 \times 125 \times 5 \int \sec^3 \theta \tan^4 \theta d\theta = 15625 \int \sec^3 \theta \tan^4 \theta d\theta$$. 8. **Simplify the integral:** Use $$\tan^4 \theta = (\tan^2 \theta)^2 = (\sec^2 \theta - 1)^2$$. So, $$\int \sec^3 \theta \tan^4 \theta d\theta = \int \sec^3 \theta (\sec^2 \theta - 1)^2 d\theta$$. 9. **Expand:** $$(\sec^2 \theta - 1)^2 = \sec^4 \theta - 2 \sec^2 \theta + 1$$, so the integral becomes $$\int \sec^3 \theta (\sec^4 \theta - 2 \sec^2 \theta + 1) d\theta = \int (\sec^7 \theta - 2 \sec^5 \theta + \sec^3 \theta) d\theta$$. 10. **Integrate each term:** The integrals of powers of secant are known or can be done by reduction formulas: - $$\int \sec^3 \theta d\theta = \frac{1}{2} (\sec \theta \tan \theta + \ln |\sec \theta + \tan \theta|) + C$$ - $$\int \sec^5 \theta d\theta = \frac{1}{4} (\sec^3 \theta \tan \theta + 3 \int \sec^3 \theta d\theta)$$ (using reduction formula) - $$\int \sec^7 \theta d\theta$$ can be reduced similarly. 11. **Use reduction formula for $$\int \sec^n \theta d\theta$$:** $$\int \sec^n \theta d\theta = \frac{\sec^{n-2} \theta \tan \theta}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2} \theta d\theta$$ for $$n > 1$$. Apply stepwise: - For $$n=7$$: $$\int \sec^7 \theta d\theta = \frac{\sec^5 \theta \tan \theta}{6} + \frac{5}{6} \int \sec^5 \theta d\theta$$ - For $$n=5$$: $$\int \sec^5 \theta d\theta = \frac{\sec^3 \theta \tan \theta}{4} + \frac{3}{4} \int \sec^3 \theta d\theta$$ 12. **Substitute back:** $$\int \sec^7 \theta d\theta = \frac{\sec^5 \theta \tan \theta}{6} + \frac{5}{6} \left( \frac{\sec^3 \theta \tan \theta}{4} + \frac{3}{4} \int \sec^3 \theta d\theta \right)$$ $$= \frac{\sec^5 \theta \tan \theta}{6} + \frac{5}{24} \sec^3 \theta \tan \theta + \frac{15}{24} \int \sec^3 \theta d\theta$$ 13. **Now, the original integral is:** $$\int (\sec^7 \theta - 2 \sec^5 \theta + \sec^3 \theta) d\theta = \int \sec^7 \theta d\theta - 2 \int \sec^5 \theta d\theta + \int \sec^3 \theta d\theta$$ Substitute the expressions: $$= \left( \frac{\sec^5 \theta \tan \theta}{6} + \frac{5}{24} \sec^3 \theta \tan \theta + \frac{15}{24} \int \sec^3 \theta d\theta \right) - 2 \left( \frac{\sec^3 \theta \tan \theta}{4} + \frac{3}{4} \int \sec^3 \theta d\theta \right) + \int \sec^3 \theta d\theta$$ $$= \frac{\sec^5 \theta \tan \theta}{6} + \frac{5}{24} \sec^3 \theta \tan \theta + \frac{15}{24} \int \sec^3 \theta d\theta - \frac{1}{2} \sec^3 \theta \tan \theta - \frac{3}{2} \int \sec^3 \theta d\theta + \int \sec^3 \theta d\theta$$ 14. **Combine like terms:** Coefficients of $$\int \sec^3 \theta d\theta$$: $$\frac{15}{24} - \frac{3}{2} + 1 = \frac{15}{24} - \frac{36}{24} + \frac{24}{24} = \frac{3}{24} = \frac{1}{8}$$ Coefficients of $$\sec^3 \theta \tan \theta$$: $$\frac{5}{24} - \frac{1}{2} = \frac{5}{24} - \frac{12}{24} = -\frac{7}{24}$$ So the integral is: $$\frac{\sec^5 \theta \tan \theta}{6} - \frac{7}{24} \sec^3 \theta \tan \theta + \frac{1}{8} \int \sec^3 \theta d\theta$$ 15. **Recall $$\int \sec^3 \theta d\theta$$:** $$\int \sec^3 \theta d\theta = \frac{1}{2} (\sec \theta \tan \theta + \ln |\sec \theta + \tan \theta|) + C$$ 16. **Substitute back:** $$\int \sec^7 \theta - 2 \sec^5 \theta + \sec^3 \theta d\theta = \frac{\sec^5 \theta \tan \theta}{6} - \frac{7}{24} \sec^3 \theta \tan \theta + \frac{1}{16} (\sec \theta \tan \theta + \ln |\sec \theta + \tan \theta|) + C$$ 17. **Recall the constant factor:** The original integral is $$15625 \times \text{above} + C$$ 18. **Rewrite in terms of $$x$$:** Recall $$x = 5 \sec \theta$$, so $$\sec \theta = \frac{x}{5}, \tan \theta = \frac{\sqrt{x^2 - 25}}{5}$$. Therefore, $$\sec^n \theta = \left( \frac{x}{5} \right)^n, \tan \theta = \frac{\sqrt{x^2 - 25}}{5}$$. 19. **Final answer:** $$\int x^2 (x^2 - 25)^{\frac{3}{2}} dx = 15625 \left[ \frac{\left( \frac{x}{5} \right)^5 \left( \frac{\sqrt{x^2 - 25}}{5} \right)}{6} - \frac{7}{24} \left( \frac{x}{5} \right)^3 \left( \frac{\sqrt{x^2 - 25}}{5} \right) + \frac{1}{16} \left( \frac{x}{5} \cdot \frac{\sqrt{x^2 - 25}}{5} + \ln \left| \frac{x}{5} + \frac{\sqrt{x^2 - 25}}{5} \right| \right) \right] + C$$ Simplify powers of 5: $$= 15625 \left[ \frac{x^5 \sqrt{x^2 - 25}}{6 \cdot 5^6} - \frac{7 x^3 \sqrt{x^2 - 25}}{24 \cdot 5^4} + \frac{1}{16} \left( \frac{x \sqrt{x^2 - 25}}{25} + \ln \left| \frac{x + \sqrt{x^2 - 25}}{5} \right| \right) \right] + C$$ Since $$15625 = 5^6$$, the first term simplifies to $$\frac{x^5 \sqrt{x^2 - 25}}{6}$$ Similarly for the second term: $$15625 \times \frac{7 x^3 \sqrt{x^2 - 25}}{24 \cdot 5^4} = \frac{7 \cdot 5^6 x^3 \sqrt{x^2 - 25}}{24 \cdot 5^4} = \frac{7 \cdot 5^2 x^3 \sqrt{x^2 - 25}}{24} = \frac{175 x^3 \sqrt{x^2 - 25}}{24}$$ And the third term: $$15625 \times \frac{1}{16} \times \frac{x \sqrt{x^2 - 25}}{25} = \frac{15625}{16 \times 25} x \sqrt{x^2 - 25} = \frac{625 x \sqrt{x^2 - 25}}{16}$$ And the logarithmic term: $$15625 \times \frac{1}{16} \ln \left| \frac{x + \sqrt{x^2 - 25}}{5} \right| = \frac{15625}{16} \ln \left| \frac{x + \sqrt{x^2 - 25}}{5} \right|$$ 20. **Therefore, the integral is:** $$\boxed{\frac{x^5 \sqrt{x^2 - 25}}{6} - \frac{175 x^3 \sqrt{x^2 - 25}}{24} + \frac{625 x \sqrt{x^2 - 25}}{16} + \frac{15625}{16} \ln \left| \frac{x + \sqrt{x^2 - 25}}{5} \right| + C}$$