1. **Stating the problem:** We want to find the integral $$\int 4 \sin^4 x \, dx$$ and match it with one of the given options. 2. **Rewrite the integral:** $$\int 4 \sin^4 x \, dx = 4 \int \sin^4 x \, dx$$ 3. **Use the power-reduction formula:** Recall that $$\sin^2 x = \frac{1 - \cos 2x}{2}$$ So, $$\sin^4 x = (\sin^2 x)^2 = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{1}{4}(1 - 2 \cos 2x + \cos^2 2x)$$ 4. **Substitute back into the integral:** $$4 \int \sin^4 x \, dx = 4 \int \frac{1}{4}(1 - 2 \cos 2x + \cos^2 2x) \, dx = \int (1 - 2 \cos 2x + \cos^2 2x) \, dx$$ 5. **Simplify the integral:** $$\int (1 - 2 \cos 2x + \cos^2 2x) \, dx = \int 1 \, dx - 2 \int \cos 2x \, dx + \int \cos^2 2x \, dx$$ 6. **Integrate the first two terms:** $$\int 1 \, dx = x$$ $$\int \cos 2x \, dx = \frac{\sin 2x}{2}$$ So, $$-2 \int \cos 2x \, dx = -2 \cdot \frac{\sin 2x}{2} = -\sin 2x$$ 7. **Handle the last integral using power-reduction again:** $$\cos^2 2x = \frac{1 + \cos 4x}{2}$$ So, $$\int \cos^2 2x \, dx = \int \frac{1 + \cos 4x}{2} \, dx = \frac{1}{2} \int 1 \, dx + \frac{1}{2} \int \cos 4x \, dx = \frac{x}{2} + \frac{1}{2} \cdot \frac{\sin 4x}{4} = \frac{x}{2} + \frac{\sin 4x}{8}$$ 8. **Combine all parts:** $$\int 4 \sin^4 x \, dx = x - \sin 2x + \frac{x}{2} + \frac{\sin 4x}{8} + C = \frac{3}{2} x - \sin 2x + \frac{1}{8} \sin 4x + C$$ 9. **Match with options:** This matches option (d): $$(3/2) x - \sin 2x + (1/8) \sin 4x$$ **Final answer:** $$\int 4 \sin^4 x \, dx = \frac{3}{2} x - \sin 2x + \frac{1}{8} \sin 4x + C$$