Integral de $$$48 \sin{\left(3 t \right)}$$$

Halla $$$\int 48 \sin{\left(3 t \right)}\, dt$$$.

Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=48$$$ y $$$f{\left(t \right)} = \sin{\left(3 t \right)}$$$:

$${\color{red}{\int{48 \sin{\left(3 t \right)} d t}}} = {\color{red}{\left(48 \int{\sin{\left(3 t \right)} d t}\right)}}$$

Sea $$$u=3 t$$$.

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

La integral se convierte en

$$48 {\color{red}{\int{\sin{\left(3 t \right)} d t}}} = 48 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}$$

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

$$48 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}} = 48 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}$$

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

$$16 {\color{red}{\int{\sin{\left(u \right)} d u}}} = 16 {\color{red}{\left(- \cos{\left(u \right)}\right)}}$$

Recordemos que $$$u=3 t$$$:

$$- 16 \cos{\left({\color{red}{u}} \right)} = - 16 \cos{\left({\color{red}{\left(3 t\right)}} \right)}$$

Por lo tanto,

$$\int{48 \sin{\left(3 t \right)} d t} = - 16 \cos{\left(3 t \right)}$$

Añade la constante de integración:

$$\int{48 \sin{\left(3 t \right)} d t} = - 16 \cos{\left(3 t \right)}+C$$

$$$\int 48 \sin{\left(3 t \right)}\, dt = - 16 \cos{\left(3 t \right)} + C$$$A