1. The problem is to find the integral $$\int \sin^6(x) \, dx$$. 2. We start by expressing $$\sin^6(x)$$ in terms of powers of cosine using the identity $$\sin^2(x) = 1 - \cos^2(x)$$: $$\sin^6(x) = (\sin^2(x))^3 = (1 - \cos^2(x))^3$$. 3. Expand the cube: $$ (1 - \cos^2(x))^3 = 1 - 3\cos^2(x) + 3\cos^4(x) - \cos^6(x) $$. 4. Therefore, $$\int \sin^6(x) \, dx = \int (1 - 3\cos^2(x) + 3\cos^4(x) - \cos^6(x)) \, dx$$. 5. Next, express even powers of cosine in terms of cosines of multiples of x, using the power-reduction formulas: - $$\cos^2(x) = \frac{1 + \cos(2x)}{2}$$ - $$\cos^4(x) = (\cos^2(x))^2 = \left(\frac{1 + \cos(2x)}{2}\right)^2 = \frac{1}{4} + \frac{1}{2}\cos(2x) + \frac{1}{4}\cos^2(2x)$$ - For $$\cos^2(2x)$$, apply power reduction again: $$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$$ So, $$\cos^4(x) = \frac{1}{4} + \frac{1}{2}\cos(2x) + \frac{1}{8} + \frac{1}{8}\cos(4x) = \frac{3}{8} + \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)$$. 6. Similarly, for $$\cos^6(x) = (\cos^2(x))^3 = \left(\frac{1 + \cos(2x)}{2}\right)^3$$, use the binomial expansion: $$\left(\frac{1 + \cos(2x)}{2}\right)^3 = \frac{1}{8} (1 + 3\cos(2x) + 3\cos^2(2x) + \cos^3(2x))$$. 7. Use power-reduction on $$\cos^2(2x)$$ and use the triple-angle identity for $$\cos^3(2x)$$: - $$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$$ - $$\cos^3(2x) = \frac{3\cos(2x) + \cos(6x)}{4}$$ 8. Substitute to get: $$\cos^6(x) = \frac{1}{8} \left(1 + 3\cos(2x) + \frac{3}{2} + \frac{3}{2}\cos(4x) + \frac{3}{4}\cos(2x) + \frac{1}{4}\cos(6x) \right)$$ Combine like terms: $$\cos^6(x) = \frac{1}{8} \left(\frac{5}{2} + \frac{15}{4}\cos(2x) + \frac{3}{2}\cos(4x) + \frac{1}{4}\cos(6x)\right) = \frac{5}{16} + \frac{15}{32}\cos(2x) + \frac{3}{16}\cos(4x) + \frac{1}{32}\cos(6x)$$ 9. Substitute these expressions back into the integral: $$\int \sin^6(x) \, dx = \int \left[1 - 3 \left(\frac{1 + \cos(2x)}{2}\right) + 3 \left(\frac{3}{8} + \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)\right) - \left(\frac{5}{16} + \frac{15}{32}\cos(2x) + \frac{3}{16}\cos(4x) + \frac{1}{32}\cos(6x)\right) \right] dx$$ 10. Simplify inside the integral: $$1 - \frac{3}{2} - \frac{3}{2}\cos(2x) + \frac{9}{8} + \frac{3}{2}\cos(2x) + \frac{3}{8}\cos(4x) - \frac{5}{16} - \frac{15}{32}\cos(2x) - \frac{3}{16}\cos(4x) - \frac{1}{32}\cos(6x)$$ Group constants and cosines: - Constants: $$1 - \frac{3}{2} + \frac{9}{8} - \frac{5}{16} = \frac{35}{16}$$ - Coefficients of $$\cos(2x)$$: $$-\frac{3}{2} + \frac{3}{2} - \frac{15}{32} = -\frac{15}{32}$$ - Coefficients of $$\cos(4x)$$: $$\frac{3}{8} - \frac{3}{16} = \frac{3}{16}$$ - Coefficient of $$\cos(6x)$$: $$-\frac{1}{32}$$ 11. The integral becomes: $$\int \left(\frac{35}{16} - \frac{15}{32}\cos(2x) + \frac{3}{16}\cos(4x) - \frac{1}{32}\cos(6x)\right) \, dx$$ 12. Integrate term by term: - $$\int \frac{35}{16} \, dx = \frac{35}{16} x$$ - $$\int -\frac{15}{32}\cos(2x) \, dx = -\frac{15}{32} \frac{\sin(2x)}{2} = -\frac{15}{64} \sin(2x)$$ - $$\int \frac{3}{16} \cos(4x) \, dx = \frac{3}{16} \frac{\sin(4x)}{4} = \frac{3}{64} \sin(4x)$$ - $$\int -\frac{1}{32}\cos(6x) \, dx = -\frac{1}{32} \frac{\sin(6x)}{6} = -\frac{1}{192} \sin(6x)$$ 13. Combine all results and add the constant of integration $$C$$: $$\int \sin^6(x) \, dx = \frac{35}{16} x - \frac{15}{64} \sin(2x) + \frac{3}{64} \sin(4x) - \frac{1}{192} \sin(6x) + C$$.