Integral von $$$\cos{\left(5 t \right)} \cos{\left(10 t \right)}$$$
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Ihre Eingabe
Bestimme $$$\int \cos{\left(5 t \right)} \cos{\left(10 t \right)}\, dt$$$.
Lösung
Schreiben Sie den Integranden mithilfe der Formel $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ mit $$$\alpha=5 t$$$ und $$$\beta=10 t$$$ um.:
$${\color{red}{\int{\cos{\left(5 t \right)} \cos{\left(10 t \right)} d t}}} = {\color{red}{\int{\left(\frac{\cos{\left(5 t \right)}}{2} + \frac{\cos{\left(15 t \right)}}{2}\right)d t}}}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(t \right)} = \cos{\left(5 t \right)} + \cos{\left(15 t \right)}$$$ an:
$${\color{red}{\int{\left(\frac{\cos{\left(5 t \right)}}{2} + \frac{\cos{\left(15 t \right)}}{2}\right)d t}}} = {\color{red}{\left(\frac{\int{\left(\cos{\left(5 t \right)} + \cos{\left(15 t \right)}\right)d t}}{2}\right)}}$$
Gliedweise integrieren:
$$\frac{{\color{red}{\int{\left(\cos{\left(5 t \right)} + \cos{\left(15 t \right)}\right)d t}}}}{2} = \frac{{\color{red}{\left(\int{\cos{\left(5 t \right)} d t} + \int{\cos{\left(15 t \right)} d t}\right)}}}{2}$$
Sei $$$u=5 t$$$.
Dann $$$du=\left(5 t\right)^{\prime }dt = 5 dt$$$ (die Schritte sind » zu sehen), und es gilt $$$dt = \frac{du}{5}$$$.
Somit,
$$\frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\int{\cos{\left(5 t \right)} d t}}}}{2} = \frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{5} d u}}}}{2}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{5}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:
$$\frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{5} d u}}}}{2} = \frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{5}\right)}}}{2}$$
Das Integral des Kosinus ist $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{10} = \frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{10}$$
Zur Erinnerung: $$$u=5 t$$$:
$$\frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{10} = \frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{\sin{\left({\color{red}{\left(5 t\right)}} \right)}}{10}$$
Sei $$$u=15 t$$$.
Dann $$$du=\left(15 t\right)^{\prime }dt = 15 dt$$$ (die Schritte sind » zu sehen), und es gilt $$$dt = \frac{du}{15}$$$.
Daher,
$$\frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\int{\cos{\left(15 t \right)} d t}}}}{2} = \frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{15} d u}}}}{2}$$
Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{15}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:
$$\frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{15} d u}}}}{2} = \frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{15}\right)}}}{2}$$
Das Integral des Kosinus ist $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{30} = \frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\sin{\left(u \right)}}}}{30}$$
Zur Erinnerung: $$$u=15 t$$$:
$$\frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left({\color{red}{u}} \right)}}{30} = \frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left({\color{red}{\left(15 t\right)}} \right)}}{30}$$
Daher,
$$\int{\cos{\left(5 t \right)} \cos{\left(10 t \right)} d t} = \frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left(15 t \right)}}{30}$$
Fügen Sie die Integrationskonstante hinzu:
$$\int{\cos{\left(5 t \right)} \cos{\left(10 t \right)} d t} = \frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left(15 t \right)}}{30}+C$$
Antwort
$$$\int \cos{\left(5 t \right)} \cos{\left(10 t \right)}\, dt = \left(\frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left(15 t \right)}}{30}\right) + C$$$A