Integral de $$$\cos{\left(2 x \right)} \cos{\left(4 x \right)} \cos{\left(6 x \right)}$$$

Halla $$$\int \cos{\left(2 x \right)} \cos{\left(4 x \right)} \cos{\left(6 x \right)}\, dx$$$.

Reescribe $$$\cos\left(2 x \right)\cos\left(4 x \right)$$$ utilizando la fórmula $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ con $$$\alpha=2 x$$$ y $$$\beta=4 x$$$:

$${\color{red}{\int{\cos{\left(2 x \right)} \cos{\left(4 x \right)} \cos{\left(6 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(6 x \right)}}{2}\right) \cos{\left(6 x \right)} d x}}}$$

Desarrolla la expresión:

$${\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(6 x \right)}}{2}\right) \cos{\left(6 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)} \cos{\left(6 x \right)}}{2} + \frac{\cos^{2}{\left(6 x \right)}}{2}\right)d x}}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(x \right)} = \cos{\left(2 x \right)} \cos{\left(6 x \right)} + \cos^{2}{\left(6 x \right)}$$$:

$${\color{red}{\int{\left(\frac{\cos{\left(2 x \right)} \cos{\left(6 x \right)}}{2} + \frac{\cos^{2}{\left(6 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} \cos{\left(6 x \right)} + \cos^{2}{\left(6 x \right)}\right)d x}}{2}\right)}}$$

Integra término a término:

$$\frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} \cos{\left(6 x \right)} + \cos^{2}{\left(6 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x} + \int{\cos^{2}{\left(6 x \right)} d x}\right)}}}{2}$$

Sea $$$u=6 x$$$.

Entonces $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{6}$$$.

Entonces,

$$\frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos^{2}{\left(6 x \right)} d x}}}}{2} = \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{6} d u}}}}{2}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{6}$$$ y $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$:

$$\frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{6} d u}}}}{2} = \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos^{2}{\left(u \right)} d u}}{6}\right)}}}{2}$$

Aplica la fórmula de reducción de potencia $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ con $$$\alpha= u $$$:

$$\frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}}{12} = \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{12}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \cos{\left(2 u \right)} + 1$$$:

$$\frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{12} = \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}}{12}$$

Integra término a término:

$$\frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{24} = \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{24}$$

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:

$$\frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{\int{\cos{\left(2 u \right)} d u}}{24} + \frac{{\color{red}{\int{1 d u}}}}{24} = \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{\int{\cos{\left(2 u \right)} d u}}{24} + \frac{{\color{red}{u}}}{24}$$

Sea $$$v=2 u$$$.

Entonces $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (los pasos pueden verse »), y obtenemos que $$$du = \frac{dv}{2}$$$.

La integral se convierte en

$$\frac{u}{24} + \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{24} = \frac{u}{24} + \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{24}$$

Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$\frac{u}{24} + \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{24} = \frac{u}{24} + \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{24}$$

La integral del coseno es $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$\frac{u}{24} + \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{48} = \frac{u}{24} + \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\sin{\left(v \right)}}}}{48}$$

Recordemos que $$$v=2 u$$$:

$$\frac{u}{24} + \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{v}} \right)}}{48} = \frac{u}{24} + \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{48}$$

Recordemos que $$$u=6 x$$$:

$$\frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{\sin{\left(2 {\color{red}{u}} \right)}}{48} + \frac{{\color{red}{u}}}{24} = \frac{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}{2} + \frac{\sin{\left(2 {\color{red}{\left(6 x\right)}} \right)}}{48} + \frac{{\color{red}{\left(6 x\right)}}}{24}$$

Reescribe el integrando utilizando la fórmula $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ con $$$\alpha=2 x$$$ y $$$\beta=6 x$$$:

$$\frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}}}{2} = \frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{\cos{\left(8 x \right)}}{2}\right)d x}}}}{2}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(x \right)} = \cos{\left(4 x \right)} + \cos{\left(8 x \right)}$$$:

$$\frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{\cos{\left(8 x \right)}}{2}\right)d x}}}}{2} = \frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(4 x \right)} + \cos{\left(8 x \right)}\right)d x}}{2}\right)}}}{2}$$

Integra término a término:

$$\frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\int{\left(\cos{\left(4 x \right)} + \cos{\left(8 x \right)}\right)d x}}}}{4} = \frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\left(\int{\cos{\left(4 x \right)} d x} + \int{\cos{\left(8 x \right)} d x}\right)}}}{4}$$

Sea $$$u=4 x$$$.

Entonces $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{4}$$$.

Por lo tanto,

$$\frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{4} + \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{4} = \frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{4}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{4}$$$ y $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{4} = \frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{4} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{4}$$

La integral del coseno es $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{16} = \frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{16}$$

Recordemos que $$$u=4 x$$$:

$$\frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{16} = \frac{x}{4} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\int{\cos{\left(8 x \right)} d x}}{4} + \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{16}$$

Sea $$$u=8 x$$$.

Entonces $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{8}$$$.

La integral se convierte en

$$\frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\int{\cos{\left(8 x \right)} d x}}}}{4} = \frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{4}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{8}$$$ y $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{4} = \frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{8}\right)}}}{4}$$

La integral del coseno es $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(12 x \right)}}{48} + \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$

Recordemos que $$$u=8 x$$$:

$$\frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(12 x \right)}}{48} + \frac{\sin{\left({\color{red}{\left(8 x\right)}} \right)}}{32}$$

Por lo tanto,

$$\int{\cos{\left(2 x \right)} \cos{\left(4 x \right)} \cos{\left(6 x \right)} d x} = \frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(8 x \right)}}{32} + \frac{\sin{\left(12 x \right)}}{48}$$

Añade la constante de integración:

$$\int{\cos{\left(2 x \right)} \cos{\left(4 x \right)} \cos{\left(6 x \right)} d x} = \frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(8 x \right)}}{32} + \frac{\sin{\left(12 x \right)}}{48}+C$$

$$$\int \cos{\left(2 x \right)} \cos{\left(4 x \right)} \cos{\left(6 x \right)}\, dx = \left(\frac{x}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{\sin{\left(8 x \right)}}{32} + \frac{\sin{\left(12 x \right)}}{48}\right) + C$$$A