Integral von $$$\sin{\left(\frac{x}{k} \right)}$$$ nach $$$x$$$

Bestimme $$$\int \sin{\left(\frac{x}{k} \right)}\, dx$$$.

Sei $$$u=\frac{x}{k}$$$.

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

Daher,

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

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

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

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

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

Zur Erinnerung: $$$u=\frac{x}{k}$$$:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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