Integral de $$$\sin{\left(\alpha \left(\beta + x\right) \right)}$$$ con respecto a $$$x$$$

Halla $$$\int \sin{\left(\alpha \left(\beta + x\right) \right)}\, dx$$$.

Sea $$$u=\alpha \left(\beta + x\right)$$$.

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

La integral se convierte en

$${\color{red}{\int{\sin{\left(\alpha \left(\beta + x\right) \right)} d x}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{\alpha} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{\alpha}$$$ y $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

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

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

Recordemos que $$$u=\alpha \left(\beta + x\right)$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{\alpha} = - \frac{\cos{\left({\color{red}{\alpha \left(\beta + x\right)}} \right)}}{\alpha}$$

Por lo tanto,

$$\int{\sin{\left(\alpha \left(\beta + x\right) \right)} d x} = - \frac{\cos{\left(\alpha \left(\beta + x\right) \right)}}{\alpha}$$

Añade la constante de integración:

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

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