Integralen av $$$\sin{\left(9 x \right)}$$$

Bestäm $$$\int \sin{\left(9 x \right)}\, dx$$$.

Låt $$$u=9 x$$$ vara.

Då $$$du=\left(9 x\right)^{\prime }dx = 9 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{9}$$$.

Alltså,

$${\color{red}{\int{\sin{\left(9 x \right)} d x}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{9} d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{9}$$$ och $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\sin{\left(u \right)}}{9} d u}}} = {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{9}\right)}}$$

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

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{9} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{9}$$

Kom ihåg att $$$u=9 x$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{9} = - \frac{\cos{\left({\color{red}{\left(9 x\right)}} \right)}}{9}$$

Alltså,

$$\int{\sin{\left(9 x \right)} d x} = - \frac{\cos{\left(9 x \right)}}{9}$$

Lägg till integrationskonstanten:

$$\int{\sin{\left(9 x \right)} d x} = - \frac{\cos{\left(9 x \right)}}{9}+C$$

$$$\int \sin{\left(9 x \right)}\, dx = - \frac{\cos{\left(9 x \right)}}{9} + C$$$A