Integral von $$$\cos^{6}{\left(3 x \right)}$$$

Bestimme $$$\int \cos^{6}{\left(3 x \right)}\, dx$$$.

Sei $$$u=3 x$$$.

Dann $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{3}$$$.

Daher,

$${\color{red}{\int{\cos^{6}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{\cos^{6}{\left(u \right)}}{3} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{3}$$$ und $$$f{\left(u \right)} = \cos^{6}{\left(u \right)}$$$ an:

$${\color{red}{\int{\frac{\cos^{6}{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\cos^{6}{\left(u \right)} d u}}{3}\right)}}$$

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

$$\frac{{\color{red}{\int{\cos^{6}{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\int{\left(\frac{15 \cos{\left(2 u \right)}}{32} + \frac{3 \cos{\left(4 u \right)}}{16} + \frac{\cos{\left(6 u \right)}}{32} + \frac{5}{16}\right)d u}}}}{3}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{32}$$$ und $$$f{\left(u \right)} = 15 \cos{\left(2 u \right)} + 6 \cos{\left(4 u \right)} + \cos{\left(6 u \right)} + 10$$$ an:

$$\frac{{\color{red}{\int{\left(\frac{15 \cos{\left(2 u \right)}}{32} + \frac{3 \cos{\left(4 u \right)}}{16} + \frac{\cos{\left(6 u \right)}}{32} + \frac{5}{16}\right)d u}}}}{3} = \frac{{\color{red}{\left(\frac{\int{\left(15 \cos{\left(2 u \right)} + 6 \cos{\left(4 u \right)} + \cos{\left(6 u \right)} + 10\right)d u}}{32}\right)}}}{3}$$

Gliedweise integrieren:

$$\frac{{\color{red}{\int{\left(15 \cos{\left(2 u \right)} + 6 \cos{\left(4 u \right)} + \cos{\left(6 u \right)} + 10\right)d u}}}}{96} = \frac{{\color{red}{\left(\int{10 d u} + \int{15 \cos{\left(2 u \right)} d u} + \int{6 \cos{\left(4 u \right)} d u} + \int{\cos{\left(6 u \right)} d u}\right)}}}{96}$$

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=10$$$ an:

$$\frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{6 \cos{\left(4 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{10 d u}}}}{96} = \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{6 \cos{\left(4 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\left(10 u\right)}}}{96}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=6$$$ und $$$f{\left(u \right)} = \cos{\left(4 u \right)}$$$ an:

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{6 \cos{\left(4 u \right)} d u}}}}{96} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\left(6 \int{\cos{\left(4 u \right)} d u}\right)}}}{96}$$

Sei $$$v=4 u$$$.

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

Also,

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{\cos{\left(4 u \right)} d u}}}}{16} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{16}$$

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:

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{16} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{16}$$

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

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{64} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\sin{\left(v \right)}}}}{64}$$

Zur Erinnerung: $$$v=4 u$$$:

$$\frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{\sin{\left({\color{red}{v}} \right)}}{64} = \frac{5 u}{48} + \frac{\int{15 \cos{\left(2 u \right)} d u}}{96} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{\sin{\left({\color{red}{\left(4 u\right)}} \right)}}{64}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=15$$$ und $$$f{\left(u \right)} = \cos{\left(2 u \right)}$$$ an:

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\int{15 \cos{\left(2 u \right)} d u}}}}{96} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{{\color{red}{\left(15 \int{\cos{\left(2 u \right)} d u}\right)}}}{96}$$

Sei $$$v=2 u$$$.

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

Somit,

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{32} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32}$$

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

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{32}$$

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

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\int{\cos{\left(v \right)} d v}}}}{64} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 {\color{red}{\sin{\left(v \right)}}}}{64}$$

Zur Erinnerung: $$$v=2 u$$$:

$$\frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 \sin{\left({\color{red}{v}} \right)}}{64} = \frac{5 u}{48} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\int{\cos{\left(6 u \right)} d u}}{96} + \frac{5 \sin{\left({\color{red}{\left(2 u\right)}} \right)}}{64}$$

Sei $$$v=6 u$$$.

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

Somit,

$$\frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\int{\cos{\left(6 u \right)} d u}}}}{96} = \frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{96}$$

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

$$\frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{6} d v}}}}{96} = \frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{6}\right)}}}{96}$$

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

$$\frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{576} = \frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{{\color{red}{\sin{\left(v \right)}}}}{576}$$

Zur Erinnerung: $$$v=6 u$$$:

$$\frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\sin{\left({\color{red}{v}} \right)}}{576} = \frac{5 u}{48} + \frac{5 \sin{\left(2 u \right)}}{64} + \frac{\sin{\left(4 u \right)}}{64} + \frac{\sin{\left({\color{red}{\left(6 u\right)}} \right)}}{576}$$

Zur Erinnerung: $$$u=3 x$$$:

$$\frac{5 \sin{\left(2 {\color{red}{u}} \right)}}{64} + \frac{\sin{\left(4 {\color{red}{u}} \right)}}{64} + \frac{\sin{\left(6 {\color{red}{u}} \right)}}{576} + \frac{5 {\color{red}{u}}}{48} = \frac{5 \sin{\left(2 {\color{red}{\left(3 x\right)}} \right)}}{64} + \frac{\sin{\left(4 {\color{red}{\left(3 x\right)}} \right)}}{64} + \frac{\sin{\left(6 {\color{red}{\left(3 x\right)}} \right)}}{576} + \frac{5 {\color{red}{\left(3 x\right)}}}{48}$$

Daher,

$$\int{\cos^{6}{\left(3 x \right)} d x} = \frac{5 x}{16} + \frac{5 \sin{\left(6 x \right)}}{64} + \frac{\sin{\left(12 x \right)}}{64} + \frac{\sin{\left(18 x \right)}}{576}$$

Vereinfachen:

$$\int{\cos^{6}{\left(3 x \right)} d x} = \frac{180 x + 45 \sin{\left(6 x \right)} + 9 \sin{\left(12 x \right)} + \sin{\left(18 x \right)}}{576}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\cos^{6}{\left(3 x \right)} d x} = \frac{180 x + 45 \sin{\left(6 x \right)} + 9 \sin{\left(12 x \right)} + \sin{\left(18 x \right)}}{576}+C$$

$$$\int \cos^{6}{\left(3 x \right)}\, dx = \frac{180 x + 45 \sin{\left(6 x \right)} + 9 \sin{\left(12 x \right)} + \sin{\left(18 x \right)}}{576} + C$$$A