1. **State the problem:** Evaluate the integral $$\int x^5 \cos(x^3) \, dx$$ using substitution and integration by parts. 2. **Substitution:** Let $$u = x^3$$. Then, $$du = 3x^2 \, dx$$ or $$dx = \frac{du}{3x^2}$$. 3. Rewrite the integral in terms of $$u$$: $$\int x^5 \cos(x^3) \, dx = \int x^5 \cos(u) \frac{du}{3x^2} = \frac{1}{3} \int x^{5-2} \cos(u) \, du = \frac{1}{3} \int x^3 \cos(u) \, du$$ 4. Since $$u = x^3$$, we have $$x^3 = u$$, so the integral becomes: $$\frac{1}{3} \int u \cos(u) \, du$$ 5. **Integration by parts:** Use the formula $$\int v \, dw = vw - \int w \, dv$$. Let $$v = u$$ and $$dw = \cos(u) \, du$$. Then, $$dv = du$$ and $$w = \sin(u)$$. 6. Apply integration by parts: $$\int u \cos(u) \, du = u \sin(u) - \int \sin(u) \, du$$ 7. Evaluate the remaining integral: $$\int \sin(u) \, du = -\cos(u)$$ 8. Substitute back: $$\int u \cos(u) \, du = u \sin(u) + \cos(u) + C$$ 9. Multiply by $$\frac{1}{3}$$: $$\frac{1}{3} \int u \cos(u) \, du = \frac{1}{3} (u \sin(u) + \cos(u)) + C$$ 10. Replace $$u = x^3$$: $$\boxed{\frac{1}{3} \left(x^3 \sin(x^3) + \cos(x^3)\right) + C}$$ This is the evaluated integral.