Integral von $$$\sin{\left(3 t \right)}$$$

Bestimme $$$\int \sin{\left(3 t \right)}\, dt$$$.

Sei $$$u=3 t$$$.

Dann $$$du=\left(3 t\right)^{\prime }dt = 3 dt$$$ (die Schritte sind » zu sehen), und es gilt $$$dt = \frac{du}{3}$$$.

Also,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ an:

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

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

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

Zur Erinnerung: $$$u=3 t$$$:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

$$$\int \sin{\left(3 t \right)}\, dt = - \frac{\cos{\left(3 t \right)}}{3} + C$$$A