1. Planteamos el problema: calcular las integrales dadas de potencias de funciones trigonométricas. 2. Para resolver integrales de potencias de senos y cosenos, usamos identidades trigonométricas y fórmulas de reducción. 3. a) $\int \sin^2(5x) \, dx$ Usamos la identidad $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$: $$\int \sin^2(5x) \, dx = \int \frac{1 - \cos(10x)}{2} \, dx = \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(10x) \, dx$$ Integrando: $$= \frac{x}{2} - \frac{1}{2} \cdot \frac{\sin(10x)}{10} + C = \frac{x}{2} - \frac{\sin(10x)}{20} + C$$ 4. b) $\int \cos^7(4x) \, dx$ Separamos $\cos^7(4x) = \cos^6(4x) \cos(4x)$ y usamos $\cos^2(\theta) = 1 - \sin^2(\theta)$: $$\int \cos^7(4x) \, dx = \int (\cos^2(4x))^3 \cos(4x) \, dx = \int (1 - \sin^2(4x))^3 \cos(4x) \, dx$$ Sea $u = \sin(4x)$, entonces $du = 4 \cos(4x) dx$ o $\cos(4x) dx = \frac{du}{4}$: $$= \int (1 - u^2)^3 \frac{du}{4} = \frac{1}{4} \int (1 - u^2)^3 du$$ Expandimos: $$(1 - u^2)^3 = 1 - 3u^2 + 3u^4 - u^6$$ Integrando término a término: $$\frac{1}{4} \left( u - u^3 + \frac{3u^5}{5} - \frac{u^7}{7} \right) + C$$ Volvemos a $x$: $$= \frac{1}{4} \left( \sin(4x) - \sin^3(4x) + \frac{3 \sin^5(4x)}{5} - \frac{\sin^7(4x)}{7} \right) + C$$ 5. c) $\int \sin^3(2x) \cos^4(2x) \, dx$ Escribimos $\sin^3(2x) = \sin(2x) \sin^2(2x) = \sin(2x)(1 - \cos^2(2x))$: $$\int \sin(2x)(1 - \cos^2(2x)) \cos^4(2x) \, dx = \int \sin(2x)(\cos^4(2x) - \cos^6(2x)) \, dx$$ Sea $u = \cos(2x)$, entonces $du = -2 \sin(2x) dx$ o $\sin(2x) dx = -\frac{du}{2}$: $$= \int (u^4 - u^6) \sin(2x) dx = -\frac{1}{2} \int (u^4 - u^6) du = -\frac{1}{2} \left( \frac{u^5}{5} - \frac{u^7}{7} \right) + C$$ Volvemos a $x$: $$= -\frac{1}{2} \left( \frac{\cos^5(2x)}{5} - \frac{\cos^7(2x)}{7} \right) + C = -\frac{\cos^5(2x)}{10} + \frac{\cos^7(2x)}{14} + C$$ 6. d) $\int \sin^2(6x) \cos^4(6x) \, dx$ Usamos $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$ y $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$: $$\sin^2(6x) = \frac{1 - \cos(12x)}{2}, \quad \cos^4(6x) = (\cos^2(6x))^2 = \left( \frac{1 + \cos(12x)}{2} \right)^2$$ Entonces: $$\int \sin^2(6x) \cos^4(6x) dx = \int \frac{1 - \cos(12x)}{2} \cdot \frac{(1 + \cos(12x))^2}{4} dx = \frac{1}{8} \int (1 - \cos(12x))(1 + 2\cos(12x) + \cos^2(12x)) dx$$ Expandimos: $$= \frac{1}{8} \int (1 + 2\cos(12x) + \cos^2(12x) - \cos(12x) - 2\cos^2(12x) - \cos^3(12x)) dx$$ Simplificamos términos: $$= \frac{1}{8} \int (1 + \cos(12x) - \cos^2(12x) - \cos^3(12x)) dx$$ Usamos $\cos^2(12x) = \frac{1 + \cos(24x)}{2}$ para integrar: $$\int \cos^2(12x) dx = \int \frac{1 + \cos(24x)}{2} dx = \frac{x}{2} + \frac{\sin(24x)}{48} + C$$ La integral de $\cos^3(12x)$ se puede resolver usando reducción o fórmula estándar, pero para brevedad: $$\int \cos^3(ax) dx = \frac{\sin(ax)}{a} - \frac{\sin^3(ax)}{3a} + C$$ Aplicando con $a=12$: $$\int \cos^3(12x) dx = \frac{\sin(12x)}{12} - \frac{\sin^3(12x)}{36} + C$$ Finalmente: $$\int \sin^2(6x) \cos^4(6x) dx = \frac{1}{8} \left( x + \frac{\sin(12x)}{12} - \left( \frac{x}{2} + \frac{\sin(24x)}{48} \right) - \left( \frac{\sin(12x)}{12} - \frac{\sin^3(12x)}{36} \right) \right) + C$$ Simplificando: $$= \frac{1}{8} \left( \frac{x}{2} - \frac{\sin(24x)}{48} + \frac{\sin^3(12x)}{36} \right) + C = \frac{x}{16} - \frac{\sin(24x)}{384} + \frac{\sin^3(12x)}{288} + C$$ 7. e) $\int \sin(8x) \cos(3x) \, dx$ Usamos la fórmula producto a suma: $$\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$$ Entonces: $$\int \sin(8x) \cos(3x) dx = \frac{1}{2} \int [\sin(11x) + \sin(5x)] dx = \frac{1}{2} \left( -\frac{\cos(11x)}{11} - \frac{\cos(5x)}{5} \right) + C$$ 8. f) $\int \sin(5x) \cos(3x) \, dx$ Similar al anterior: $$= \frac{1}{2} \int [\sin(8x) + \sin(2x)] dx = \frac{1}{2} \left( -\frac{\cos(8x)}{8} - \frac{\cos(2x)}{2} \right) + C$$ **Respuestas finales:** $a) \frac{x}{2} - \frac{\sin(10x)}{20} + C$ $b) \frac{1}{4} \left( \sin(4x) - \sin^3(4x) + \frac{3 \sin^5(4x)}{5} - \frac{\sin^7(4x)}{7} \right) + C$ $c) -\frac{\cos^5(2x)}{10} + \frac{\cos^7(2x)}{14} + C$ $d) \frac{x}{16} - \frac{\sin(24x)}{384} + \frac{\sin^3(12x)}{288} + C$ $e) -\frac{\cos(11x)}{22} - \frac{\cos(5x)}{10} + C$ $f) -\frac{\cos(8x)}{16} - \frac{\cos(2x)}{4} + C$