1. Problem 1: Evaluate the integral $$\int 12 \sin^6(x) \cos^3(x) \, dx$$ 2. To solve this, we use substitution and trigonometric identities. Notice the powers of sine and cosine. 3. Let’s rewrite the integral: $$\int 12 \sin^6(x) \cos^3(x) \, dx = 12 \int \sin^6(x) \cos^3(x) \, dx$$ 4. We can split $$\cos^3(x) = \cos^2(x) \cos(x)$$ and use the identity $$\cos^2(x) = 1 - \sin^2(x)$$: $$12 \int \sin^6(x) (1 - \sin^2(x)) \cos(x) \, dx$$ 5. Use substitution: let $$u = \sin(x)$$, so $$du = \cos(x) \, dx$$. 6. Substitute into the integral: $$12 \int u^6 (1 - u^2) \, du = 12 \int (u^6 - u^8) \, du$$ 7. Integrate term-by-term: $$12 \left( \frac{u^7}{7} - \frac{u^9}{9} \right) + C = 12 \left( \frac{\sin^7(x)}{7} - \frac{\sin^9(x)}{9} \right) + C$$ 8. Simplify the constant factor: $$= \frac{12}{7} \sin^7(x) - \frac{12}{9} \sin^9(x) + C = \frac{12}{7} \sin^7(x) - \frac{4}{3} \sin^9(x) + C$$ --- 9. Problem 2: Evaluate the integral $$\int t^7 \ln(t) \, dt$$ 10. Use integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ 11. Let $$u = \ln(t)$$ so $$du = \frac{1}{t} dt$$, and let $$dv = t^7 dt$$ so $$v = \frac{t^8}{8}$$. 12. Apply integration by parts: $$\int t^7 \ln(t) \, dt = \frac{t^8}{8} \ln(t) - \int \frac{t^8}{8} \cdot \frac{1}{t} dt = \frac{t^8}{8} \ln(t) - \frac{1}{8} \int t^7 dt$$ 13. Integrate $$\int t^7 dt$$: $$\int t^7 dt = \frac{t^8}{8} + C$$ 14. Substitute back: $$\frac{t^8}{8} \ln(t) - \frac{1}{8} \cdot \frac{t^8}{8} + C = \frac{t^8}{8} \ln(t) - \frac{t^8}{64} + C$$ --- Final answers: - For $$\int 12 \sin^6(x) \cos^3(x) \, dx$$: $$\boxed{\frac{12}{7} \sin^7(x) - \frac{4}{3} \sin^9(x) + C}$$ - For $$\int t^7 \ln(t) \, dt$$: $$\boxed{\frac{t^8}{8} \ln(t) - \frac{t^8}{64} + C}$$