Integralen av $$$\sin{\left(4 x \right)} \sin{\left(8 x \right)}$$$

Bestäm $$$\int \sin{\left(4 x \right)} \sin{\left(8 x \right)}\, dx$$$.

Skriv om integranden med hjälp av formeln $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ med $$$\alpha=4 x$$$ och $$$\beta=8 x$$$:

$${\color{red}{\int{\sin{\left(4 x \right)} \sin{\left(8 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(4 x \right)}}{2} - \frac{\cos{\left(12 x \right)}}{2}\right)d x}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(x \right)} = \cos{\left(4 x \right)} - \cos{\left(12 x \right)}$$$:

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

Integrera termvis:

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

Låt $$$u=12 x$$$ vara.

Då $$$du=\left(12 x\right)^{\prime }dx = 12 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{12}$$$.

Integralen blir

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{12}$$$ och $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

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

$$\frac{\int{\cos{\left(4 x \right)} d x}}{2} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{24} = \frac{\int{\cos{\left(4 x \right)} d x}}{2} - \frac{{\color{red}{\sin{\left(u \right)}}}}{24}$$

Kom ihåg att $$$u=12 x$$$:

$$\frac{\int{\cos{\left(4 x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{24} = \frac{\int{\cos{\left(4 x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{\left(12 x\right)}} \right)}}{24}$$

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,

$$- \frac{\sin{\left(12 x \right)}}{24} + \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{2} = - \frac{\sin{\left(12 x \right)}}{24} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2}$$

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

$$- \frac{\sin{\left(12 x \right)}}{24} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2} = - \frac{\sin{\left(12 x \right)}}{24} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{2}$$

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

$$- \frac{\sin{\left(12 x \right)}}{24} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = - \frac{\sin{\left(12 x \right)}}{24} + \frac{{\color{red}{\sin{\left(u \right)}}}}{8}$$

Kom ihåg att $$$u=4 x$$$:

$$- \frac{\sin{\left(12 x \right)}}{24} + \frac{\sin{\left({\color{red}{u}} \right)}}{8} = - \frac{\sin{\left(12 x \right)}}{24} + \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{8}$$

Alltså,

$$\int{\sin{\left(4 x \right)} \sin{\left(8 x \right)} d x} = \frac{\sin{\left(4 x \right)}}{8} - \frac{\sin{\left(12 x \right)}}{24}$$

Lägg till integrationskonstanten:

$$\int{\sin{\left(4 x \right)} \sin{\left(8 x \right)} d x} = \frac{\sin{\left(4 x \right)}}{8} - \frac{\sin{\left(12 x \right)}}{24}+C$$

$$$\int \sin{\left(4 x \right)} \sin{\left(8 x \right)}\, dx = \left(\frac{\sin{\left(4 x \right)}}{8} - \frac{\sin{\left(12 x \right)}}{24}\right) + C$$$A