Integral von $$$5 \cos{\left(x \right)} \cos{\left(5 x \right)}$$$

Bestimme $$$\int 5 \cos{\left(x \right)} \cos{\left(5 x \right)}\, dx$$$.

Schreibe $$$\cos\left(x \right)\cos\left(5 x \right)$$$ 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 $$$\alpha=x$$$ und $$$\beta=5 x$$$ um:

$${\color{red}{\int{5 \cos{\left(x \right)} \cos{\left(5 x \right)} d x}}} = {\color{red}{\int{\left(\frac{5 \cos{\left(4 x \right)}}{2} + \frac{5 \cos{\left(6 x \right)}}{2}\right)d x}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = 5 \cos{\left(4 x \right)} + 5 \cos{\left(6 x \right)}$$$ an:

$${\color{red}{\int{\left(\frac{5 \cos{\left(4 x \right)}}{2} + \frac{5 \cos{\left(6 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(5 \cos{\left(4 x \right)} + 5 \cos{\left(6 x \right)}\right)d x}}{2}\right)}}$$

Gliedweise integrieren:

$$\frac{{\color{red}{\int{\left(5 \cos{\left(4 x \right)} + 5 \cos{\left(6 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{5 \cos{\left(4 x \right)} d x} + \int{5 \cos{\left(6 x \right)} d x}\right)}}}{2}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=5$$$ und $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$ an:

$$\frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\int{5 \cos{\left(4 x \right)} d x}}}}{2} = \frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{{\color{red}{\left(5 \int{\cos{\left(4 x \right)} d x}\right)}}}{2}$$

Sei $$$u=4 x$$$.

Dann $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{4}$$$.

Das Integral lässt sich umschreiben als

$$\frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{5 {\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{2} = \frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{5 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:

$$\frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{5 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2} = \frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{5 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{2}$$

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

$$\frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{5 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = \frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{5 {\color{red}{\sin{\left(u \right)}}}}{8}$$

Zur Erinnerung: $$$u=4 x$$$:

$$\frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{5 \sin{\left({\color{red}{u}} \right)}}{8} = \frac{\int{5 \cos{\left(6 x \right)} d x}}{2} + \frac{5 \sin{\left({\color{red}{\left(4 x\right)}} \right)}}{8}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=5$$$ und $$$f{\left(x \right)} = \cos{\left(6 x \right)}$$$ an:

$$\frac{5 \sin{\left(4 x \right)}}{8} + \frac{{\color{red}{\int{5 \cos{\left(6 x \right)} d x}}}}{2} = \frac{5 \sin{\left(4 x \right)}}{8} + \frac{{\color{red}{\left(5 \int{\cos{\left(6 x \right)} d x}\right)}}}{2}$$

Sei $$$u=6 x$$$.

Dann $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{6}$$$.

Somit,

$$\frac{5 \sin{\left(4 x \right)}}{8} + \frac{5 {\color{red}{\int{\cos{\left(6 x \right)} d x}}}}{2} = \frac{5 \sin{\left(4 x \right)}}{8} + \frac{5 {\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{2}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{6}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:

$$\frac{5 \sin{\left(4 x \right)}}{8} + \frac{5 {\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{2} = \frac{5 \sin{\left(4 x \right)}}{8} + \frac{5 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{2}$$

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

$$\frac{5 \sin{\left(4 x \right)}}{8} + \frac{5 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{12} = \frac{5 \sin{\left(4 x \right)}}{8} + \frac{5 {\color{red}{\sin{\left(u \right)}}}}{12}$$

Zur Erinnerung: $$$u=6 x$$$:

$$\frac{5 \sin{\left(4 x \right)}}{8} + \frac{5 \sin{\left({\color{red}{u}} \right)}}{12} = \frac{5 \sin{\left(4 x \right)}}{8} + \frac{5 \sin{\left({\color{red}{\left(6 x\right)}} \right)}}{12}$$

Daher,

$$\int{5 \cos{\left(x \right)} \cos{\left(5 x \right)} d x} = \frac{5 \sin{\left(4 x \right)}}{8} + \frac{5 \sin{\left(6 x \right)}}{12}$$

Vereinfachen:

$$\int{5 \cos{\left(x \right)} \cos{\left(5 x \right)} d x} = \frac{5 \left(3 \sin{\left(4 x \right)} + 2 \sin{\left(6 x \right)}\right)}{24}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{5 \cos{\left(x \right)} \cos{\left(5 x \right)} d x} = \frac{5 \left(3 \sin{\left(4 x \right)} + 2 \sin{\left(6 x \right)}\right)}{24}+C$$

$$$\int 5 \cos{\left(x \right)} \cos{\left(5 x \right)}\, dx = \frac{5 \left(3 \sin{\left(4 x \right)} + 2 \sin{\left(6 x \right)}\right)}{24} + C$$$A