Integral von $$$\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}$$$

Bestimme $$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}\, dx$$$.

Schreibe $$$\sin\left(x \right)\sin\left(2 x \right)$$$ mithilfe der Formel $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ mit $$$\alpha=x$$$ und $$$\beta=2 x$$$ um:

$${\color{red}{\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \cos{\left(x \right)} d x}}}$$

Expandiere den Ausdruck:

$${\color{red}{\int{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \cos{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{2} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{2}\right)d x}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = \cos^{2}{\left(x \right)} - \cos{\left(x \right)} \cos{\left(3 x \right)}$$$ an:

$${\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{2} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\cos^{2}{\left(x \right)} - \cos{\left(x \right)} \cos{\left(3 x \right)}\right)d x}}{2}\right)}}$$

Gliedweise integrieren:

$$\frac{{\color{red}{\int{\left(\cos^{2}{\left(x \right)} - \cos{\left(x \right)} \cos{\left(3 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(- \int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x} + \int{\cos^{2}{\left(x \right)} d x}\right)}}}{2}$$

Wende die Potenzreduktionsformel $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ mit $$$\alpha=x$$$ an:

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos^{2}{\left(x \right)} d x}}}}{2} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{2}$$

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

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{2} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}}{2}$$

Gliedweise integrieren:

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}}}{4} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}}{4}$$

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=1$$$ an:

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{\int{1 d x}}}}{4} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{x}}}{4}$$

Sei $$$u=2 x$$$.

Dann $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{2}$$$.

Das Integral wird zu

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{4} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{4}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{4} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{4}$$

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

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{8}$$

Zur Erinnerung: $$$u=2 x$$$:

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{8} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

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 $$$\alpha=x$$$ und $$$\beta=3 x$$$ um.:

$$\frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}}}{2} = \frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{2}$$

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

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

Gliedweise integrieren:

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

Das Integral $$$\int{\cos{\left(2 x \right)} d x}$$$ wurde bereits berechnet:

$$\int{\cos{\left(2 x \right)} d x} = \frac{\sin{\left(2 x \right)}}{2}$$

Daher,

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

Sei $$$v=4 x$$$.

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

Daher,

$$\frac{x}{4} - \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{4} = \frac{x}{4} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{4}$$

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

$$\frac{x}{4} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{4} = \frac{x}{4} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{4}$$

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

$$\frac{x}{4} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{16} = \frac{x}{4} - \frac{{\color{red}{\sin{\left(v \right)}}}}{16}$$

Zur Erinnerung: $$$v=4 x$$$:

$$\frac{x}{4} - \frac{\sin{\left({\color{red}{v}} \right)}}{16} = \frac{x}{4} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{16}$$

Daher,

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} d x} = \frac{x}{4} - \frac{\sin{\left(4 x \right)}}{16}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} d x} = \frac{x}{4} - \frac{\sin{\left(4 x \right)}}{16}+C$$

$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}\, dx = \left(\frac{x}{4} - \frac{\sin{\left(4 x \right)}}{16}\right) + C$$$A