1. **Problem:** Evaluate the integral $$\int 3x^5 \sqrt{16 - x^2} \, dx$$. 2. **Formula and rules:** We will use substitution and algebraic manipulation. Recall that substitution is useful when the integral contains a function and its derivative. 3. **Step 1: Substitution.** Let $$u = 16 - x^2$$, then $$du = -2x \, dx$$ or $$-\frac{1}{2} du = x \, dx$$. 4. **Step 2: Express the integral in terms of $$u$$.** We have $$3x^5 \sqrt{16 - x^2} \, dx = 3x^5 \sqrt{u} \, dx$$. Note that $$x^5 = x^4 \cdot x = (x^2)^2 \cdot x$$. Since $$u = 16 - x^2$$, then $$x^2 = 16 - u$$. So, $$x^4 = (x^2)^2 = (16 - u)^2$$. Therefore, $$x^5 = (16 - u)^2 \cdot x$$. 5. **Step 3: Substitute into the integral:** $$\int 3x^5 \sqrt{u} \, dx = \int 3 (16 - u)^2 \cdot x \cdot \sqrt{u} \, dx$$. Using $$x \, dx = -\frac{1}{2} du$$, the integral becomes: $$\int 3 (16 - u)^2 \sqrt{u} \cdot \left(-\frac{1}{2} du\right) = -\frac{3}{2} \int (16 - u)^2 u^{1/2} \, du$$. 6. **Step 4: Expand and simplify the integrand:** Expand $$(16 - u)^2 = 256 - 32u + u^2$$. So the integral is: $$-\frac{3}{2} \int (256 - 32u + u^2) u^{1/2} \, du = -\frac{3}{2} \int (256 u^{1/2} - 32 u^{3/2} + u^{5/2}) \, du$$. 7. **Step 5: Integrate term by term:** $$\int u^{1/2} du = \frac{2}{3} u^{3/2}$$ $$\int u^{3/2} du = \frac{2}{5} u^{5/2}$$ $$\int u^{5/2} du = \frac{2}{7} u^{7/2}$$ Therefore, $$-\frac{3}{2} \left(256 \cdot \frac{2}{3} u^{3/2} - 32 \cdot \frac{2}{5} u^{5/2} + \frac{2}{7} u^{7/2} \right) + C$$ 8. **Step 6: Simplify coefficients:** $$-\frac{3}{2} \left( \frac{512}{3} u^{3/2} - \frac{64}{5} u^{5/2} + \frac{2}{7} u^{7/2} \right) + C$$ Multiply inside: $$= -\frac{3}{2} \cdot \frac{512}{3} u^{3/2} + \frac{3}{2} \cdot \frac{64}{5} u^{5/2} - \frac{3}{2} \cdot \frac{2}{7} u^{7/2} + C$$ $$= -256 u^{3/2} + \frac{96}{5} u^{5/2} - \frac{3}{7} u^{7/2} + C$$ 9. **Step 7: Substitute back $$u = 16 - x^2$$:** $$\boxed{ -256 (16 - x^2)^{3/2} + \frac{96}{5} (16 - x^2)^{5/2} - \frac{3}{7} (16 - x^2)^{7/2} + C }$$ This is the evaluated integral.