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

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

Skriv om integranden med hjälp av formeln $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ tillsammans med $$$\alpha=2 x$$$ och $$$\beta=4 x$$$:

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

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

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

Integrera termvis:

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

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 kan omskrivas som

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

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

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

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

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

Kom ihåg att $$$u=2 x$$$:

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

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

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

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

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

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

$$\frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{12} = \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{12}$$

Kom ihåg att $$$u=6 x$$$:

$$\frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{12} = \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{\left(6 x\right)}} \right)}}{12}$$

Alltså,

$$\int{\cos{\left(2 x \right)} \cos{\left(4 x \right)} d x} = \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(6 x \right)}}{12}$$

Lägg till integrationskonstanten:

$$\int{\cos{\left(2 x \right)} \cos{\left(4 x \right)} d x} = \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(6 x \right)}}{12}+C$$

$$$\int \cos{\left(2 x \right)} \cos{\left(4 x \right)}\, dx = \left(\frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(6 x \right)}}{12}\right) + C$$$A