Integral de $$$- \frac{3 \sin{\left(\frac{x}{2} - 1 \right)}}{2}$$$

Halla $$$\int \left(- \frac{3 \sin{\left(\frac{x}{2} - 1 \right)}}{2}\right)\, dx$$$.

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

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

Sea $$$u=\frac{x}{2} - 1$$$.

Entonces $$$du=\left(\frac{x}{2} - 1\right)^{\prime }dx = \frac{dx}{2}$$$ (los pasos pueden verse »), y obtenemos que $$$dx = 2 du$$$.

Por lo tanto,

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

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

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

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 $$$u=\frac{x}{2} - 1$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

$$$\int \left(- \frac{3 \sin{\left(\frac{x}{2} - 1 \right)}}{2}\right)\, dx = 3 \cos{\left(\frac{x}{2} - 1 \right)} + C$$$A