Integral von $$$6 \cos{\left(3 t \right)}$$$

Bestimme $$$\int 6 \cos{\left(3 t \right)}\, dt$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ mit $$$c=6$$$ und $$$f{\left(t \right)} = \cos{\left(3 t \right)}$$$ an:

$${\color{red}{\int{6 \cos{\left(3 t \right)} d t}}} = {\color{red}{\left(6 \int{\cos{\left(3 t \right)} d t}\right)}}$$

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,

$$6 {\color{red}{\int{\cos{\left(3 t \right)} d t}}} = 6 {\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}$$

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

$$6 {\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}} = 6 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{3}\right)}}$$

Das Integral des Kosinus ist $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

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

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

$$2 \sin{\left({\color{red}{u}} \right)} = 2 \sin{\left({\color{red}{\left(3 t\right)}} \right)}$$

Daher,

$$\int{6 \cos{\left(3 t \right)} d t} = 2 \sin{\left(3 t \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{6 \cos{\left(3 t \right)} d t} = 2 \sin{\left(3 t \right)}+C$$

$$$\int 6 \cos{\left(3 t \right)}\, dt = 2 \sin{\left(3 t \right)} + C$$$A