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

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

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

$${\color{red}{\int{\sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{2}\right) \cos{\left(2 x \right)} d x}}}$$

Desarrolla la expresión:

$${\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{2}\right) \cos{\left(2 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(x \right)} \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(3 x \right)} \cos{\left(2 x \right)}}{2}\right)d x}}}$$

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 $$$f{\left(x \right)} = \sin{\left(x \right)} \cos{\left(2 x \right)} + \sin{\left(3 x \right)} \cos{\left(2 x \right)}$$$:

$${\color{red}{\int{\left(\frac{\sin{\left(x \right)} \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(3 x \right)} \cos{\left(2 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(x \right)} \cos{\left(2 x \right)} + \sin{\left(3 x \right)} \cos{\left(2 x \right)}\right)d x}}{2}\right)}}$$

Integra término a término:

$$\frac{{\color{red}{\int{\left(\sin{\left(x \right)} \cos{\left(2 x \right)} + \sin{\left(3 x \right)} \cos{\left(2 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(x \right)} \cos{\left(2 x \right)} d x} + \int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}\right)}}}{2}$$

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

$$\frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(x \right)} \cos{\left(2 x \right)} d x}}}}{2} = \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{2}\right)d x}}}}{2}$$

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 $$$f{\left(x \right)} = - \sin{\left(x \right)} + \sin{\left(3 x \right)}$$$:

$$\frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{2}\right)d x}}}}{2} = \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(- \sin{\left(x \right)} + \sin{\left(3 x \right)}\right)d x}}{2}\right)}}}{2}$$

Integra término a término:

$$\frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- \sin{\left(x \right)} + \sin{\left(3 x \right)}\right)d x}}}}{4} = \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \int{\sin{\left(x \right)} d x} + \int{\sin{\left(3 x \right)} d x}\right)}}}{4}$$

La integral del seno es $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

$$\frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{\int{\sin{\left(3 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin{\left(x \right)} d x}}}}{4} = \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{\int{\sin{\left(3 x \right)} d x}}{4} - \frac{{\color{red}{\left(- \cos{\left(x \right)}\right)}}}{4}$$

Sea $$$u=3 x$$$.

Entonces $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{3}$$$.

Por lo tanto,

$$\frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(3 x \right)} d x}}}}{4} = \frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{4}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{3}$$$ y $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{4} = \frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}}{4}$$

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

$$\frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{12} = \frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{12}$$

Recordemos que $$$u=3 x$$$:

$$\frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{12} = \frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(3 x\right)}} \right)}}{12}$$

Reescribe el integrando utilizando la fórmula $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ con $$$\alpha=3 x$$$ y $$$\beta=2 x$$$:

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}}}{2} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{2}$$

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 $$$f{\left(x \right)} = \sin{\left(x \right)} + \sin{\left(5 x \right)}$$$:

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{2} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}{2}\right)}}}{2}$$

Integra término a término:

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}}}{4} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(\int{\sin{\left(x \right)} d x} + \int{\sin{\left(5 x \right)} d x}\right)}}}{4}$$

La integral del seno es $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{\int{\sin{\left(5 x \right)} d x}}{4} + \frac{{\color{red}{\int{\sin{\left(x \right)} d x}}}}{4} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{\int{\sin{\left(5 x \right)} d x}}{4} + \frac{{\color{red}{\left(- \cos{\left(x \right)}\right)}}}{4}$$

Sea $$$u=5 x$$$.

Entonces $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{5}$$$.

Por lo tanto,

$$- \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(5 x \right)} d x}}}}{4} = - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{4}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{5}$$$ y $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$- \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{4} = - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{4}$$

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

$$- \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{20} = - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{20}$$

Recordemos que $$$u=5 x$$$:

$$- \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left({\color{red}{u}} \right)}}{20} = - \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left({\color{red}{\left(5 x\right)}} \right)}}{20}$$

Por lo tanto,

$$\int{\sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)} d x} = - \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left(5 x \right)}}{20}$$

Añade la constante de integración:

$$\int{\sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)} d x} = - \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left(5 x \right)}}{20}+C$$

$$$\int \sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)}\, dx = \left(- \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left(5 x \right)}}{20}\right) + C$$$A