Integralen av $$$\sin^{4}{\left(2 x \right)}$$$

Bestäm $$$\int \sin^{4}{\left(2 x \right)}\, dx$$$.

Låt $$$u=2 x$$$ vara.

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

Integralen blir

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(u \right)} = \sin^{4}{\left(u \right)}$$$:

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

Använd potensreduceringsformeln $$$\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ med $$$\alpha= u $$$:

$$\frac{{\color{red}{\int{\sin^{4}{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\int{\left(- \frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}}{2}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{8}$$$ och $$$f{\left(u \right)} = - 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3$$$:

$$\frac{{\color{red}{\int{\left(- \frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}}{2} = \frac{{\color{red}{\left(\frac{\int{\left(- 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}{8}\right)}}}{2}$$

Integrera termvis:

$$\frac{{\color{red}{\int{\left(- 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}}}{16} = \frac{{\color{red}{\left(\int{3 d u} - \int{4 \cos{\left(2 u \right)} d u} + \int{\cos{\left(4 u \right)} d u}\right)}}}{16}$$

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=3$$$:

$$- \frac{\int{4 \cos{\left(2 u \right)} d u}}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} + \frac{{\color{red}{\int{3 d u}}}}{16} = - \frac{\int{4 \cos{\left(2 u \right)} d u}}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} + \frac{{\color{red}{\left(3 u\right)}}}{16}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=4$$$ och $$$f{\left(u \right)} = \cos{\left(2 u \right)}$$$:

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{4 \cos{\left(2 u \right)} d u}}}}{16} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\left(4 \int{\cos{\left(2 u \right)} d u}\right)}}}{16}$$

Låt $$$v=2 u$$$ vara.

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

Alltså,

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{4} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{4}$$

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

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{4} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{4}$$

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

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{8} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\sin{\left(v \right)}}}}{8}$$

Kom ihåg att $$$v=2 u$$$:

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{\sin{\left({\color{red}{v}} \right)}}{8} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{8}$$

Låt $$$v=4 u$$$ vara.

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

Integralen blir

$$\frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\int{\cos{\left(4 u \right)} d u}}}}{16} = \frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{16}$$

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

$$\frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{16} = \frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{16}$$

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

$$\frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{64} = \frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\sin{\left(v \right)}}}}{64}$$

Kom ihåg att $$$v=4 u$$$:

$$\frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{\sin{\left({\color{red}{v}} \right)}}{64} = \frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{\sin{\left({\color{red}{\left(4 u\right)}} \right)}}{64}$$

Kom ihåg att $$$u=2 x$$$:

$$- \frac{\sin{\left(2 {\color{red}{u}} \right)}}{8} + \frac{\sin{\left(4 {\color{red}{u}} \right)}}{64} + \frac{3 {\color{red}{u}}}{16} = - \frac{\sin{\left(2 {\color{red}{\left(2 x\right)}} \right)}}{8} + \frac{\sin{\left(4 {\color{red}{\left(2 x\right)}} \right)}}{64} + \frac{3 {\color{red}{\left(2 x\right)}}}{16}$$

Alltså,

$$\int{\sin^{4}{\left(2 x \right)} d x} = \frac{3 x}{8} - \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}$$

Förenkla:

$$\int{\sin^{4}{\left(2 x \right)} d x} = \frac{24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}}{64}$$

Lägg till integrationskonstanten:

$$\int{\sin^{4}{\left(2 x \right)} d x} = \frac{24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}}{64}+C$$

$$$\int \sin^{4}{\left(2 x \right)}\, dx = \frac{24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}}{64} + C$$$A