1. **State the problem:** We need to evaluate the integral $$\int 5 \sin^4(x) \cos^2(x) \, dx.$$\n\n2. **Rewrite the integral:** Express powers of sine and cosine in terms of powers of sine or cosine to simplify. Here, we keep it as is for substitution.\n\n3. **Use substitution:** Let $$u = \sin(x)$$ so that $$du = \cos(x) \, dx.$$\n\n4. **Rewrite the integral in terms of $$u$$:** We have $$\sin^4(x) = u^4$$ and $$\cos^2(x) = (\cos(x))^2$$. Also, $$dx = \frac{du}{\cos(x)}$$, but this complicates things. Instead, rewrite $$\cos^2(x) = 1 - \sin^2(x) = 1 - u^2$$.\n\n5. **Rewrite the integral:** $$\int 5 u^4 (1 - u^2) \, dx.$$ But we still have $$dx$$, so better to express $$dx$$ in terms of $$du$$ and $$\cos(x)$$. Since $$du = \cos(x) dx$$, then $$dx = \frac{du}{\cos(x)}$$. Substituting back, the integral becomes complicated.\n\n6. **Alternative approach:** Use the identity $$\sin^2(x) = 1 - \cos^2(x)$$ to express everything in terms of $$\cos(x)$$. Let $$t = \cos(x)$$, then $$dt = -\sin(x) dx$$ or $$-dt = \sin(x) dx$$.\n\n7. **Rewrite powers:** $$\sin^4(x) = (1 - t^2)^2$$ and $$\cos^2(x) = t^2$$. Also, $$\sin(x) dx = -dt$$, so $$dx = \frac{-dt}{\sin(x)}$$. But this is complicated.\n\n8. **Better substitution:** Use $$\sin^4(x) \cos^2(x) = (\sin^2(x))^2 (\cos^2(x))$$. Use the identity $$\sin^2(x) = 1 - \cos^2(x)$$.\n\n9. **Rewrite integral:** $$\int 5 (1 - t^2)^2 t^2 \, dx$$. Since $$dt = -\sin(x) dx$$, and $$\sin(x) = \sqrt{1 - t^2}$$, then $$dx = \frac{dt}{-\sin(x)} = \frac{dt}{-\sqrt{1 - t^2}}$$, which is complicated.\n\n10. **Use power-reduction formulas:** Recall $$\sin^2(x) = \frac{1 - \cos(2x)}{2}$$ and $$\cos^2(x) = \frac{1 + \cos(2x)}{2}$$.\n\n11. **Rewrite integral:** $$5 \int \sin^4(x) \cos^2(x) dx = 5 \int (\sin^2(x))^2 \cos^2(x) dx = 5 \int \left(\frac{1 - \cos(2x)}{2}\right)^2 \left(\frac{1 + \cos(2x)}{2}\right) dx.$$\n\n12. **Simplify:** $$= 5 \int \frac{(1 - \cos(2x))^2 (1 + \cos(2x))}{8} dx = \frac{5}{8} \int (1 - \cos(2x))^2 (1 + \cos(2x)) dx.$$\n\n13. **Expand:** $$(1 - \cos(2x))^2 (1 + \cos(2x)) = (1 - 2\cos(2x) + \cos^2(2x))(1 + \cos(2x))$$\n\n14. **Multiply:** $$= (1)(1 + \cos(2x)) - 2\cos(2x)(1 + \cos(2x)) + \cos^2(2x)(1 + \cos(2x))$$\n$$= 1 + \cos(2x) - 2\cos(2x) - 2\cos^2(2x) + \cos^2(2x) + \cos^3(2x)$$\n$$= 1 - \cos(2x) - \cos^2(2x) + \cos^3(2x).$$\n\n15. **Integral becomes:** $$\frac{5}{8} \int \left(1 - \cos(2x) - \cos^2(2x) + \cos^3(2x)\right) dx.$$\n\n16. **Integrate term by term:**\n- $$\int 1 dx = x$$\n- $$\int \cos(2x) dx = \frac{\sin(2x)}{2}$$\n- For $$\int \cos^2(2x) dx$$, use power-reduction: $$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$$, so $$\int \cos^2(2x) dx = \int \frac{1 + \cos(4x)}{2} dx = \frac{x}{2} + \frac{\sin(4x)}{8}$$\n- For $$\int \cos^3(2x) dx$$, use identity: $$\cos^3(\theta) = \frac{3 \cos(\theta) + \cos(3\theta)}{4}$$, so $$\int \cos^3(2x) dx = \int \frac{3 \cos(2x) + \cos(6x)}{4} dx = \frac{3}{4} \int \cos(2x) dx + \frac{1}{4} \int \cos(6x) dx = \frac{3}{4} \cdot \frac{\sin(2x)}{2} + \frac{1}{4} \cdot \frac{\sin(6x)}{6} = \frac{3 \sin(2x)}{8} + \frac{\sin(6x)}{24}.$$\n\n17. **Putting it all together:**\n$$\frac{5}{8} \left(x - \frac{\sin(2x)}{2} - \left(\frac{x}{2} + \frac{\sin(4x)}{8}\right) + \left(\frac{3 \sin(2x)}{8} + \frac{\sin(6x)}{24}\right) \right) + C.$$\n\n18. **Simplify inside parentheses:**\n$$x - \frac{\sin(2x)}{2} - \frac{x}{2} - \frac{\sin(4x)}{8} + \frac{3 \sin(2x)}{8} + \frac{\sin(6x)}{24} = \frac{x}{2} - \frac{4 \sin(2x)}{8} + \frac{3 \sin(2x)}{8} - \frac{\sin(4x)}{8} + \frac{\sin(6x)}{24} = \frac{x}{2} - \frac{\sin(2x)}{8} - \frac{\sin(4x)}{8} + \frac{\sin(6x)}{24}.$$\n\n19. **Final answer:**\n$$\frac{5}{8} \left( \frac{x}{2} - \frac{\sin(2x)}{8} - \frac{\sin(4x)}{8} + \frac{\sin(6x)}{24} \right) + C = \frac{5x}{16} - \frac{5 \sin(2x)}{64} - \frac{5 \sin(4x)}{64} + \frac{5 \sin(6x)}{192} + C.$$