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

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

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

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

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

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

Expand the expression:

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

Gliedweise integrieren:

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

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

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

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

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

Daher,

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

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

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

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

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

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

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

Schreibe $$$\sin\left(2 x \right)\cos\left(4 x \right)$$$ mithilfe der Formel $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ mit $$$\alpha=2 x$$$ und $$$\beta=4 x$$$ um:

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

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

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

Gliedweise integrieren:

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

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

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

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,

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

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

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

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{48} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{48}$$

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} - \frac{\cos{\left({\color{red}{u}} \right)}}{48} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} - \frac{\cos{\left({\color{red}{\left(6 x\right)}} \right)}}{48}$$

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}}}{4} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{2}\right)}}}{4}$$

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

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

Daher,

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{8} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\left(- \frac{\cos{\left(2 x \right)}}{2}\right)}}}{8}$$

Daher,

$$\int{\frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2} d x} = - \frac{\cos{\left(2 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{48}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2} d x} = - \frac{\cos{\left(2 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{48}+C$$

$$$\int \frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2}\, dx = \left(- \frac{\cos{\left(2 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{48}\right) + C$$$A