Integral X3 Sin B66746

1. **State the problem:** We need to evaluate the definite integral $$\int_0^{5\pi} x^3 \sin(x) \, dx$$. 2. **Recall the formula:** To integrate products of powers of $$x$$ and trigonometric functions, we use integration by parts repeatedly. The formula for integration by parts is: $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = x^3 \implies du = 3x^2 \, dx$$ $$dv = \sin(x) \, dx \implies v = -\cos(x)$$ 4. **Apply integration by parts:** $$\int x^3 \sin(x) \, dx = -x^3 \cos(x) + \int 3x^2 \cos(x) \, dx$$ 5. **Repeat integration by parts on $$\int 3x^2 \cos(x) \, dx$$:** Let $$u = 3x^2 \implies du = 6x \, dx$$ $$dv = \cos(x) \, dx \implies v = \sin(x)$$ Then $$\int 3x^2 \cos(x) \, dx = 3x^2 \sin(x) - \int 6x \sin(x) \, dx$$ 6. **Repeat integration by parts on $$\int 6x \sin(x) \, dx$$:** Let $$u = 6x \implies du = 6 \, dx$$ $$dv = \sin(x) \, dx \implies v = -\cos(x)$$ Then $$\int 6x \sin(x) \, dx = -6x \cos(x) + \int 6 \cos(x) \, dx$$ 7. **Integrate $$\int 6 \cos(x) \, dx$$:** $$\int 6 \cos(x) \, dx = 6 \sin(x) + C$$ 8. **Combine all parts:** $$\int 6x \sin(x) \, dx = -6x \cos(x) + 6 \sin(x) + C$$ $$\int 3x^2 \cos(x) \, dx = 3x^2 \sin(x) - \left(-6x \cos(x) + 6 \sin(x)\right) + C = 3x^2 \sin(x) + 6x \cos(x) - 6 \sin(x) + C$$ $$\int x^3 \sin(x) \, dx = -x^3 \cos(x) + 3x^2 \sin(x) + 6x \cos(x) - 6 \sin(x) + C$$ 9. **Evaluate definite integral from $$0$$ to $$5\pi$$:** $$\int_0^{5\pi} x^3 \sin(x) \, dx = \left[-x^3 \cos(x) + 3x^2 \sin(x) + 6x \cos(x) - 6 \sin(x)\right]_0^{5\pi}$$ 10. **Calculate at upper limit $$x=5\pi$$:** $$\cos(5\pi) = \cos(\pi) = -1$$ (since $$5\pi$$ is an odd multiple of $$\pi$$) $$\sin(5\pi) = 0$$ So, $$- (5\pi)^3 (-1) + 3 (5\pi)^2 (0) + 6 (5\pi) (-1) - 6 (0) = (5\pi)^3 - 6 (5\pi)$$ 11. **Calculate at lower limit $$x=0$$:** $$-0^3 \cos(0) + 3 \cdot 0^2 \sin(0) + 6 \cdot 0 \cos(0) - 6 \sin(0) = 0$$ 12. **Final result:** $$\int_0^{5\pi} x^3 \sin(x) \, dx = (5\pi)^3 - 6 (5\pi) = 125 \pi^3 - 30 \pi$$