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

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

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

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

Desarrolla la expresión:

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

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 $$$f{\left(x \right)} = \cos^{2}{\left(x \right)} - \cos{\left(x \right)} \cos{\left(3 x \right)}$$$:

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

Integra término a término:

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

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

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos^{2}{\left(x \right)} d x}}}}{2} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{2}$$

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 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$:

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{2} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}}{2}$$

Integra término a término:

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}}}{4} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}}{4}$$

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{\int{1 d x}}}}{4} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{x}}}{4}$$

Sea $$$u=2 x$$$.

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

La integral puede reescribirse como

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

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

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

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

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{8}$$

Recordemos que $$$u=2 x$$$:

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{8} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

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 $$$\alpha=x$$$ y $$$\beta=3 x$$$:

$$\frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}}}{2} = \frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{2}$$

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 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + \cos{\left(4 x \right)}$$$:

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

Integra término a término:

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

La integral $$$\int{\cos{\left(2 x \right)} d x}$$$ ya ha sido calculada:

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

Por lo tanto,

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

Sea $$$v=4 x$$$.

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

La integral puede reescribirse como

$$\frac{x}{4} - \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{4} = \frac{x}{4} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{4}$$

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

$$\frac{x}{4} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{4} = \frac{x}{4} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{4}$$

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

$$\frac{x}{4} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{16} = \frac{x}{4} - \frac{{\color{red}{\sin{\left(v \right)}}}}{16}$$

Recordemos que $$$v=4 x$$$:

$$\frac{x}{4} - \frac{\sin{\left({\color{red}{v}} \right)}}{16} = \frac{x}{4} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{16}$$

Por lo tanto,

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} d x} = \frac{x}{4} - \frac{\sin{\left(4 x \right)}}{16}$$

Añade la constante de integración:

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} d x} = \frac{x}{4} - \frac{\sin{\left(4 x \right)}}{16}+C$$

$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}\, dx = \left(\frac{x}{4} - \frac{\sin{\left(4 x \right)}}{16}\right) + C$$$A