1. **State the problem:** Evaluate the integral $$\int_0^{3\pi} \sin^5\left(\frac{x}{2}\right) dx.$$\n\n2. **Recall the formula and rules:** To integrate powers of sine, use the identity $$\sin^5 u = \sin^4 u \sin u = (\sin^2 u)^2 \sin u.$$ Also, use the power-reduction formula $$\sin^2 u = \frac{1 - \cos(2u)}{2}.$$\n\n3. **Rewrite the integral:** Let $$u = \frac{x}{2} \Rightarrow dx = 2 du.$$ The limits change from $$x=0 \Rightarrow u=0$$ to $$x=3\pi \Rightarrow u=\frac{3\pi}{2}.$$ So the integral becomes $$\int_0^{3\pi} \sin^5\left(\frac{x}{2}\right) dx = \int_0^{\frac{3\pi}{2}} \sin^5 u \cdot 2 du = 2 \int_0^{\frac{3\pi}{2}} \sin^5 u \, du.$$\n\n4. **Express $$\sin^5 u$$ using power reduction:**\n$$\sin^5 u = (\sin^2 u)^2 \sin u = \left(\frac{1 - \cos(2u)}{2}\right)^2 \sin u = \frac{(1 - \cos(2u))^2}{4} \sin u.$$\n\n5. **Substitute into the integral:**\n$$2 \int_0^{\frac{3\pi}{2}} \sin^5 u \, du = 2 \int_0^{\frac{3\pi}{2}} \frac{(1 - \cos(2u))^2}{4} \sin u \, du = \frac{1}{2} \int_0^{\frac{3\pi}{2}} (1 - \cos(2u))^2 \sin u \, du.$$\n\n6. **Expand the square:**\n$$(1 - \cos(2u))^2 = 1 - 2 \cos(2u) + \cos^2(2u).$$\n\n7. **Rewrite the integral:**\n$$\frac{1}{2} \int_0^{\frac{3\pi}{2}} \left(1 - 2 \cos(2u) + \cos^2(2u)\right) \sin u \, du = \frac{1}{2} \left[ \int_0^{\frac{3\pi}{2}} \sin u \, du - 2 \int_0^{\frac{3\pi}{2}} \cos(2u) \sin u \, du + \int_0^{\frac{3\pi}{2}} \cos^2(2u) \sin u \, du \right].$$\n\n8. **Evaluate each integral separately:**\n\n- First integral: $$\int_0^{\frac{3\pi}{2}} \sin u \, du = [-\cos u]_0^{\frac{3\pi}{2}} = -\cos\left(\frac{3\pi}{2}\right) + \cos 0 = -0 + 1 = 1.$$\n\n- Second integral: Use product-to-sum for $$\cos(2u) \sin u$$ or integration by parts. Using product-to-sum:\n$$\cos(2u) \sin u = \frac{1}{2} [\sin(3u) - \sin(u)].$$\nSo,\n$$\int_0^{\frac{3\pi}{2}} \cos(2u) \sin u \, du = \frac{1}{2} \int_0^{\frac{3\pi}{2}} (\sin(3u) - \sin u) du = \frac{1}{2} \left[ -\frac{\cos(3u)}{3} + \cos u \right]_0^{\frac{3\pi}{2}}.$$\nCalculate:\n$$-\frac{\cos(3 \cdot \frac{3\pi}{2})}{3} + \cos\left(\frac{3\pi}{2}\right) - \left(-\frac{\cos 0}{3} + \cos 0\right) = -\frac{\cos\left(\frac{9\pi}{2}\right)}{3} + 0 - \left(-\frac{1}{3} + 1\right).$$\nSince $$\cos\left(\frac{9\pi}{2}\right) = \cos\left(4\pi + \frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0,$$\nthis simplifies to\n$$0 - \left(-\frac{1}{3} + 1\right) = -(-\frac{1}{3} + 1) = \frac{1}{3} - 1 = -\frac{2}{3}.$$\nMultiply by $$\frac{1}{2}$$ gives $$-\frac{1}{3}.$$\n\n- Third integral: Use power-reduction for $$\cos^2(2u)$$:\n$$\cos^2(2u) = \frac{1 + \cos(4u)}{2}.$$\nSo,\n$$\int_0^{\frac{3\pi}{2}} \cos^2(2u) \sin u \, du = \frac{1}{2} \int_0^{\frac{3\pi}{2}} (1 + \cos(4u)) \sin u \, du = \frac{1}{2} \left[ \int_0^{\frac{3\pi}{2}} \sin u \, du + \int_0^{\frac{3\pi}{2}} \cos(4u) \sin u \, du \right].$$\nWe already know $$\int_0^{\frac{3\pi}{2}} \sin u \, du = 1.$$\nFor $$\int_0^{\frac{3\pi}{2}} \cos(4u) \sin u \, du,$$ use product-to-sum:\n$$\cos(4u) \sin u = \frac{1}{2} [\sin(5u) - \sin(3u)].$$\nThus,\n$$\int_0^{\frac{3\pi}{2}} \cos(4u) \sin u \, du = \frac{1}{2} \int_0^{\frac{3\pi}{2}} (\sin(5u) - \sin(3u)) du = \frac{1}{2} \left[ -\frac{\cos(5u)}{5} + \frac{\cos(3u)}{3} \right]_0^{\frac{3\pi}{2}}.$$\nCalculate:\n$$-\frac{\cos\left(5 \cdot \frac{3\pi}{2}\right)}{5} + \frac{\cos\left(3 \cdot \frac{3\pi}{2}\right)}{3} - \left(-\frac{\cos 0}{5} + \frac{\cos 0}{3}\right) = -\frac{\cos\left(\frac{15\pi}{2}\right)}{5} + \frac{\cos\left(\frac{9\pi}{2}\right)}{3} - \left(-\frac{1}{5} + \frac{1}{3}\right).$$\nSince $$\cos\left(\frac{15\pi}{2}\right) = \cos\left(7\pi + \frac{\pi}{2}\right) = 0$$ and $$\cos\left(\frac{9\pi}{2}\right) = 0,$$ this simplifies to\n$$0 + 0 - \left(-\frac{1}{5} + \frac{1}{3}\right) = -(-\frac{1}{5} + \frac{1}{3}) = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15}.$$\nMultiply by $$\frac{1}{2}$$ gives $$-\frac{1}{15}.$$\n\nSo the third integral is\n$$\frac{1}{2} (1 - \frac{1}{15}) = \frac{1}{2} \cdot \frac{14}{15} = \frac{7}{15}.$$\n\n9. **Combine all results:**\n$$\frac{1}{2} \left[1 - 2 \left(-\frac{1}{3}\right) + \frac{7}{15} \right] = \frac{1}{2} \left[1 + \frac{2}{3} + \frac{7}{15} \right].$$\nFind common denominator 15:\n$$1 = \frac{15}{15}, \quad \frac{2}{3} = \frac{10}{15}, \quad \frac{7}{15} = \frac{7}{15}.$$\nSum inside brackets:\n$$\frac{15}{15} + \frac{10}{15} + \frac{7}{15} = \frac{32}{15}.$$\nMultiply by $$\frac{1}{2}$$:\n$$\frac{1}{2} \cdot \frac{32}{15} = \frac{16}{15}.$$\n\n**Final answer:**\n$$\int_0^{3\pi} \sin^5\left(\frac{x}{2}\right) dx = \frac{16}{15}.$$