Integral de $$$- \sin{\left(3 x - 2 \right)}$$$

Halla $$$\int \left(- \sin{\left(3 x - 2 \right)}\right)\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=-1$$$ y $$$f{\left(x \right)} = \sin{\left(3 x - 2 \right)}$$$:

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

Sea $$$u=3 x - 2$$$.

Entonces $$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{3}$$$.

Por lo tanto,

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

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

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

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

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

Recordemos que $$$u=3 x - 2$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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