Integral von $$$\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)}$$$

Bestimme $$$\int \sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)}\, d\theta$$$.

Schreibe den Integranden mithilfe der Doppelwinkel-Formel $$$\sin\left(\theta \right)\cos\left(\theta \right)=\frac{1}{2}\sin\left( 2 \theta \right)$$$ um:

$${\color{red}{\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\frac{\sin^{6}{\left(2 \theta \right)}}{64} d \theta}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ mit $$$c=\frac{1}{64}$$$ und $$$f{\left(\theta \right)} = \sin^{6}{\left(2 \theta \right)}$$$ an:

$${\color{red}{\int{\frac{\sin^{6}{\left(2 \theta \right)}}{64} d \theta}}} = {\color{red}{\left(\frac{\int{\sin^{6}{\left(2 \theta \right)} d \theta}}{64}\right)}}$$

Wende die Potenzreduktionsformel $$$\sin^{6}{\left(\alpha \right)} = - \frac{15 \cos{\left(2 \alpha \right)}}{32} + \frac{3 \cos{\left(4 \alpha \right)}}{16} - \frac{\cos{\left(6 \alpha \right)}}{32} + \frac{5}{16}$$$ mit $$$\alpha=2 \theta$$$ an:

$$\frac{{\color{red}{\int{\sin^{6}{\left(2 \theta \right)} d \theta}}}}{64} = \frac{{\color{red}{\int{\left(- \frac{15 \cos{\left(4 \theta \right)}}{32} + \frac{3 \cos{\left(8 \theta \right)}}{16} - \frac{\cos{\left(12 \theta \right)}}{32} + \frac{5}{16}\right)d \theta}}}}{64}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ mit $$$c=\frac{1}{32}$$$ und $$$f{\left(\theta \right)} = - 15 \cos{\left(4 \theta \right)} + 6 \cos{\left(8 \theta \right)} - \cos{\left(12 \theta \right)} + 10$$$ an:

$$\frac{{\color{red}{\int{\left(- \frac{15 \cos{\left(4 \theta \right)}}{32} + \frac{3 \cos{\left(8 \theta \right)}}{16} - \frac{\cos{\left(12 \theta \right)}}{32} + \frac{5}{16}\right)d \theta}}}}{64} = \frac{{\color{red}{\left(\frac{\int{\left(- 15 \cos{\left(4 \theta \right)} + 6 \cos{\left(8 \theta \right)} - \cos{\left(12 \theta \right)} + 10\right)d \theta}}{32}\right)}}}{64}$$

Gliedweise integrieren:

$$\frac{{\color{red}{\int{\left(- 15 \cos{\left(4 \theta \right)} + 6 \cos{\left(8 \theta \right)} - \cos{\left(12 \theta \right)} + 10\right)d \theta}}}}{2048} = \frac{{\color{red}{\left(\int{10 d \theta} - \int{15 \cos{\left(4 \theta \right)} d \theta} + \int{6 \cos{\left(8 \theta \right)} d \theta} - \int{\cos{\left(12 \theta \right)} d \theta}\right)}}}{2048}$$

Wenden Sie die Konstantenregel $$$\int c\, d\theta = c \theta$$$ mit $$$c=10$$$ an:

$$- \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\int{\cos{\left(12 \theta \right)} d \theta}}{2048} + \frac{{\color{red}{\int{10 d \theta}}}}{2048} = - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\int{\cos{\left(12 \theta \right)} d \theta}}{2048} + \frac{{\color{red}{\left(10 \theta\right)}}}{2048}$$

Sei $$$u=12 \theta$$$.

Dann $$$du=\left(12 \theta\right)^{\prime }d\theta = 12 d\theta$$$ (die Schritte sind » zu sehen), und es gilt $$$d\theta = \frac{du}{12}$$$.

Somit,

$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\cos{\left(12 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{12} d u}}}}{2048}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{12}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:

$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{12} d u}}}}{2048} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{12}\right)}}}{2048}$$

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

$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{24576} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\sin{\left(u \right)}}}}{24576}$$

Zur Erinnerung: $$$u=12 \theta$$$:

$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\sin{\left({\color{red}{u}} \right)}}{24576} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\sin{\left({\color{red}{\left(12 \theta\right)}} \right)}}{24576}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ mit $$$c=15$$$ und $$$f{\left(\theta \right)} = \cos{\left(4 \theta \right)}$$$ an:

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{15 \cos{\left(4 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\left(15 \int{\cos{\left(4 \theta \right)} d \theta}\right)}}}{2048}$$

Sei $$$u=4 \theta$$$.

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

Somit,

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\cos{\left(4 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2048}$$

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:

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2048} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{2048}$$

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

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{8192} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\sin{\left(u \right)}}}}{8192}$$

Zur Erinnerung: $$$u=4 \theta$$$:

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 \sin{\left({\color{red}{u}} \right)}}{8192} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 \sin{\left({\color{red}{\left(4 \theta\right)}} \right)}}{8192}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ mit $$$c=6$$$ und $$$f{\left(\theta \right)} = \cos{\left(8 \theta \right)}$$$ an:

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{{\color{red}{\int{6 \cos{\left(8 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{{\color{red}{\left(6 \int{\cos{\left(8 \theta \right)} d \theta}\right)}}}{2048}$$

Sei $$$u=8 \theta$$$.

Dann $$$du=\left(8 \theta\right)^{\prime }d\theta = 8 d\theta$$$ (die Schritte sind » zu sehen), und es gilt $$$d\theta = \frac{du}{8}$$$.

Somit,

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\cos{\left(8 \theta \right)} d \theta}}}}{1024} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{1024}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{8}$$$ und $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ an:

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{1024} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{8}\right)}}}{1024}$$

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

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{8192} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\sin{\left(u \right)}}}}{8192}$$

Zur Erinnerung: $$$u=8 \theta$$$:

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 \sin{\left({\color{red}{u}} \right)}}{8192} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 \sin{\left({\color{red}{\left(8 \theta\right)}} \right)}}{8192}$$

Daher,

$$\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} + \frac{3 \sin{\left(8 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576}$$

Vereinfachen:

$$\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta} = - \frac{- 120 \theta + 45 \sin{\left(4 \theta \right)} - 9 \sin{\left(8 \theta \right)} + \sin{\left(12 \theta \right)}}{24576}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta} = - \frac{- 120 \theta + 45 \sin{\left(4 \theta \right)} - 9 \sin{\left(8 \theta \right)} + \sin{\left(12 \theta \right)}}{24576}+C$$

$$$\int \sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)}\, d\theta = - \frac{- 120 \theta + 45 \sin{\left(4 \theta \right)} - 9 \sin{\left(8 \theta \right)} + \sin{\left(12 \theta \right)}}{24576} + C$$$A