1. **State the problem:** We need to evaluate the definite integral $$\int_0^\pi 8 \sin^4 x \, dx$$. 2. **Recall the formula and rules:** To integrate powers of sine, use the power-reduction formula: $$\sin^4 x = \left(\sin^2 x\right)^2 = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{1}{4} (1 - 2\cos(2x) + \cos^2(2x))$$. Also, use the power-reduction for $$\cos^2(2x)$$: $$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$$. 3. **Rewrite the integrand:** $$8 \sin^4 x = 8 \times \frac{1}{4} (1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}) = 2 (1 - 2\cos(2x) + \frac{1}{2} + \frac{\cos(4x)}{2})$$ Simplify inside the parentheses: $$1 + \frac{1}{2} = \frac{3}{2}$$ So, $$8 \sin^4 x = 2 \left( \frac{3}{2} - 2\cos(2x) + \frac{\cos(4x)}{2} \right) = 3 - 4 \cos(2x) + \cos(4x)$$. 4. **Set up the integral:** $$\int_0^\pi 8 \sin^4 x \, dx = \int_0^\pi \left(3 - 4 \cos(2x) + \cos(4x)\right) dx$$. 5. **Integrate term-by-term:** - $$\int_0^\pi 3 \, dx = 3x \Big|_0^\pi = 3\pi$$ - $$\int_0^\pi -4 \cos(2x) \, dx = -4 \times \frac{\sin(2x)}{2} \Big|_0^\pi = -2 \sin(2x) \Big|_0^\pi = -2(0 - 0) = 0$$ - $$\int_0^\pi \cos(4x) \, dx = \frac{\sin(4x)}{4} \Big|_0^\pi = \frac{\sin(4\pi) - \sin(0)}{4} = 0$$ 6. **Sum the results:** $$3\pi + 0 + 0 = 3\pi$$. **Final answer:** $$\int_0^\pi 8 \sin^4 x \, dx = 3\pi$$.