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

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

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 $$$\alpha=x$$$ an:

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

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

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

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=10$$$ an:

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

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

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

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

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

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:

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

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

$$\frac{5 x}{16} + \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{3 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{5 x}{16} + \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{3 {\color{red}{\sin{\left(u \right)}}}}{64}$$

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

$$\frac{5 x}{16} + \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{3 \sin{\left({\color{red}{u}} \right)}}{64} = \frac{5 x}{16} + \frac{\int{15 \cos{\left(2 x \right)} d x}}{32} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{3 \sin{\left({\color{red}{\left(4 x\right)}} \right)}}{64}$$

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

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

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 lässt sich umschreiben als

$$\frac{5 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{15 {\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{32} = \frac{5 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{32}$$

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:

$$\frac{5 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{32} = \frac{5 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{15 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{32}$$

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

$$\frac{5 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{15 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{5 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{15 {\color{red}{\sin{\left(u \right)}}}}{64}$$

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

$$\frac{5 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{15 \sin{\left({\color{red}{u}} \right)}}{64} = \frac{5 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{32} + \frac{15 \sin{\left({\color{red}{\left(2 x\right)}} \right)}}{64}$$

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

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

Also,

$$\frac{5 x}{16} + \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\cos{\left(6 x \right)} d x}}}}{32} = \frac{5 x}{16} + \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{32}$$

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:

$$\frac{5 x}{16} + \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{32} = \frac{5 x}{16} + \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{32}$$

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

$$\frac{5 x}{16} + \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{192} = \frac{5 x}{16} + \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\sin{\left(u \right)}}}}{192}$$

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

$$\frac{5 x}{16} + \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left({\color{red}{u}} \right)}}{192} = \frac{5 x}{16} + \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left({\color{red}{\left(6 x\right)}} \right)}}{192}$$

Daher,

$$\int{\cos^{6}{\left(x \right)} d x} = \frac{5 x}{16} + \frac{15 \sin{\left(2 x \right)}}{64} + \frac{3 \sin{\left(4 x \right)}}{64} + \frac{\sin{\left(6 x \right)}}{192}$$

Vereinfachen:

$$\int{\cos^{6}{\left(x \right)} d x} = \frac{60 x + 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} + \sin{\left(6 x \right)}}{192}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\cos^{6}{\left(x \right)} d x} = \frac{60 x + 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} + \sin{\left(6 x \right)}}{192}+C$$

$$$\int \cos^{6}{\left(x \right)}\, dx = \frac{60 x + 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} + \sin{\left(6 x \right)}}{192} + C$$$A