Integral de $$$\sin{\left(3 x \right)} \cos{\left(x \right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \sin{\left(3 x \right)} \cos{\left(x \right)}\, dx$$$.
Solución
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 $$$\beta=x$$$:
$${\color{red}{\int{\sin{\left(3 x \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(4 x \right)}}{2}\right)d x}}}$$
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 $$$f{\left(x \right)} = \sin{\left(2 x \right)} + \sin{\left(4 x \right)}$$$:
$${\color{red}{\int{\left(\frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(4 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(2 x \right)} + \sin{\left(4 x \right)}\right)d x}}{2}\right)}}$$
Integra término a término:
$$\frac{{\color{red}{\int{\left(\sin{\left(2 x \right)} + \sin{\left(4 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(2 x \right)} d x} + \int{\sin{\left(4 x \right)} d x}\right)}}}{2}$$
Sea $$$u=2 x$$$.
Entonces $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{2}$$$.
Por lo tanto,
$$\frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{2} = \frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$\frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2} = \frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{2}$$
La integral del seno es $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4} = \frac{\int{\sin{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}$$
Recordemos que $$$u=2 x$$$:
$$\frac{\int{\sin{\left(4 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{4} = \frac{\int{\sin{\left(4 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$
Sea $$$u=4 x$$$.
Entonces $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{4}$$$.
La integral puede reescribirse como
$$- \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\sin{\left(4 x \right)} d x}}}}{2} = - \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{2}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{4}$$$ y $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$- \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{2} = - \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{4}\right)}}}{2}$$
La integral del seno es $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$- \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = - \frac{\cos{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$
Recordemos que $$$u=4 x$$$:
$$- \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left({\color{red}{u}} \right)}}{8} = - \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left({\color{red}{\left(4 x\right)}} \right)}}{8}$$
Por lo tanto,
$$\int{\sin{\left(3 x \right)} \cos{\left(x \right)} d x} = - \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left(4 x \right)}}{8}$$
Añade la constante de integración:
$$\int{\sin{\left(3 x \right)} \cos{\left(x \right)} d x} = - \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left(4 x \right)}}{8}+C$$
Respuesta
$$$\int \sin{\left(3 x \right)} \cos{\left(x \right)}\, dx = \left(- \frac{\cos{\left(2 x \right)}}{4} - \frac{\cos{\left(4 x \right)}}{8}\right) + C$$$A