Integralen av $$$\frac{\cos^{6}{\left(x \right)}}{2}$$$

Bestäm $$$\int \frac{\cos^{6}{\left(x \right)}}{2}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(x \right)} = \cos^{6}{\left(x \right)}$$$:

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

Använd potensreduceringsformeln $$$\cos^{6}{\left(\alpha \right)} = \frac{15 \cos{\left(2 \alpha \right)}}{32} + \frac{3 \cos{\left(4 \alpha \right)}}{16} + \frac{\cos{\left(6 \alpha \right)}}{32} + \frac{5}{16}$$$ med $$$\alpha=x$$$:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{32}$$$ och $$$f{\left(x \right)} = 15 \cos{\left(2 x \right)} + 6 \cos{\left(4 x \right)} + \cos{\left(6 x \right)} + 10$$$:

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

Integrera termvis:

$$\frac{{\color{red}{\int{\left(15 \cos{\left(2 x \right)} + 6 \cos{\left(4 x \right)} + \cos{\left(6 x \right)} + 10\right)d x}}}}{64} = \frac{{\color{red}{\left(\int{10 d x} + \int{15 \cos{\left(2 x \right)} d x} + \int{6 \cos{\left(4 x \right)} d x} + \int{\cos{\left(6 x \right)} d x}\right)}}}{64}$$

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=10$$$:

$$\frac{\int{15 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{{\color{red}{\int{10 d x}}}}{64} = \frac{\int{15 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{6 \cos{\left(4 x \right)} d x}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{{\color{red}{\left(10 x\right)}}}{64}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=6$$$ och $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$:

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

Låt $$$u=4 x$$$ vara.

Då $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{4}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{4}$$$ och $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

Integralen av cosinus är $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{5 x}{32} + \frac{\int{15 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{3 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{128} = \frac{5 x}{32} + \frac{\int{15 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{3 {\color{red}{\sin{\left(u \right)}}}}{128}$$

Kom ihåg att $$$u=4 x$$$:

$$\frac{5 x}{32} + \frac{\int{15 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{3 \sin{\left({\color{red}{u}} \right)}}{128} = \frac{5 x}{32} + \frac{\int{15 \cos{\left(2 x \right)} d x}}{64} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{3 \sin{\left({\color{red}{\left(4 x\right)}} \right)}}{128}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=15$$$ och $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:

$$\frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{{\color{red}{\int{15 \cos{\left(2 x \right)} d x}}}}{64} = \frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{{\color{red}{\left(15 \int{\cos{\left(2 x \right)} d x}\right)}}}{64}$$

Låt $$$u=2 x$$$ vara.

Då $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{2}$$$.

Alltså,

$$\frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{15 {\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{64} = \frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{64}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{64} = \frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{15 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{64}$$

Integralen av cosinus är $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{15 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{128} = \frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{15 {\color{red}{\sin{\left(u \right)}}}}{128}$$

Kom ihåg att $$$u=2 x$$$:

$$\frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{15 \sin{\left({\color{red}{u}} \right)}}{128} = \frac{5 x}{32} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\int{\cos{\left(6 x \right)} d x}}{64} + \frac{15 \sin{\left({\color{red}{\left(2 x\right)}} \right)}}{128}$$

Låt $$$u=6 x$$$ vara.

Då $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{6}$$$.

Integralen blir

$$\frac{5 x}{32} + \frac{15 \sin{\left(2 x \right)}}{128} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\int{\cos{\left(6 x \right)} d x}}}}{64} = \frac{5 x}{32} + \frac{15 \sin{\left(2 x \right)}}{128} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{64}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{6}$$$ och $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{5 x}{32} + \frac{15 \sin{\left(2 x \right)}}{128} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{6} d u}}}}{64} = \frac{5 x}{32} + \frac{15 \sin{\left(2 x \right)}}{128} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{6}\right)}}}{64}$$

Integralen av cosinus är $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{5 x}{32} + \frac{15 \sin{\left(2 x \right)}}{128} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{384} = \frac{5 x}{32} + \frac{15 \sin{\left(2 x \right)}}{128} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{{\color{red}{\sin{\left(u \right)}}}}{384}$$

Kom ihåg att $$$u=6 x$$$:

$$\frac{5 x}{32} + \frac{15 \sin{\left(2 x \right)}}{128} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\sin{\left({\color{red}{u}} \right)}}{384} = \frac{5 x}{32} + \frac{15 \sin{\left(2 x \right)}}{128} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\sin{\left({\color{red}{\left(6 x\right)}} \right)}}{384}$$

Alltså,

$$\int{\frac{\cos^{6}{\left(x \right)}}{2} d x} = \frac{5 x}{32} + \frac{15 \sin{\left(2 x \right)}}{128} + \frac{3 \sin{\left(4 x \right)}}{128} + \frac{\sin{\left(6 x \right)}}{384}$$

Förenkla:

$$\int{\frac{\cos^{6}{\left(x \right)}}{2} d x} = \frac{60 x + 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} + \sin{\left(6 x \right)}}{384}$$

Lägg till integrationskonstanten:

$$\int{\frac{\cos^{6}{\left(x \right)}}{2} d x} = \frac{60 x + 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} + \sin{\left(6 x \right)}}{384}+C$$

$$$\int \frac{\cos^{6}{\left(x \right)}}{2}\, dx = \frac{60 x + 45 \sin{\left(2 x \right)} + 9 \sin{\left(4 x \right)} + \sin{\left(6 x \right)}}{384} + C$$$A