Integral von $$$\sin{\left(\alpha \left(\beta + x\right) \right)}$$$ nach $$$x$$$

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

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

Dann $$$du=\left(\alpha \left(\beta + x\right)\right)^{\prime }dx = \alpha dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{\alpha}$$$.

Also,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{\alpha}$$$ und $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ an:

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

Das Integral des Sinus lautet $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

Zur Erinnerung: $$$u=\alpha \left(\beta + x\right)$$$:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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