Integral de $$$9 \sin{\left(3 x \right)}$$$

Halla $$$\int 9 \sin{\left(3 x \right)}\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=9$$$ y $$$f{\left(x \right)} = \sin{\left(3 x \right)}$$$:

$${\color{red}{\int{9 \sin{\left(3 x \right)} d x}}} = {\color{red}{\left(9 \int{\sin{\left(3 x \right)} d x}\right)}}$$

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

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

Entonces,

$$9 {\color{red}{\int{\sin{\left(3 x \right)} d x}}} = 9 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}$$

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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$9 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}} = 9 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}$$

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

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

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

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

Por lo tanto,

$$\int{9 \sin{\left(3 x \right)} d x} = - 3 \cos{\left(3 x \right)}$$

Añade la constante de integración:

$$\int{9 \sin{\left(3 x \right)} d x} = - 3 \cos{\left(3 x \right)}+C$$

$$$\int 9 \sin{\left(3 x \right)}\, dx = - 3 \cos{\left(3 x \right)} + C$$$A