Integralen av $$$- 3 \sin{\left(\frac{x}{3} \right)}$$$

Bestäm $$$\int \left(- 3 \sin{\left(\frac{x}{3} \right)}\right)\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=-3$$$ och $$$f{\left(x \right)} = \sin{\left(\frac{x}{3} \right)}$$$:

$${\color{red}{\int{\left(- 3 \sin{\left(\frac{x}{3} \right)}\right)d x}}} = {\color{red}{\left(- 3 \int{\sin{\left(\frac{x}{3} \right)} d x}\right)}}$$

Låt $$$u=\frac{x}{3}$$$ vara.

Då $$$du=\left(\frac{x}{3}\right)^{\prime }dx = \frac{dx}{3}$$$ (stegen kan ses »), och vi har att $$$dx = 3 du$$$.

Integralen blir

$$- 3 {\color{red}{\int{\sin{\left(\frac{x}{3} \right)} d x}}} = - 3 {\color{red}{\int{3 \sin{\left(u \right)} d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=3$$$ och $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$- 3 {\color{red}{\int{3 \sin{\left(u \right)} d u}}} = - 3 {\color{red}{\left(3 \int{\sin{\left(u \right)} d u}\right)}}$$

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

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

Kom ihåg att $$$u=\frac{x}{3}$$$:

$$9 \cos{\left({\color{red}{u}} \right)} = 9 \cos{\left({\color{red}{\left(\frac{x}{3}\right)}} \right)}$$

Alltså,

$$\int{\left(- 3 \sin{\left(\frac{x}{3} \right)}\right)d x} = 9 \cos{\left(\frac{x}{3} \right)}$$

Lägg till integrationskonstanten:

$$\int{\left(- 3 \sin{\left(\frac{x}{3} \right)}\right)d x} = 9 \cos{\left(\frac{x}{3} \right)}+C$$

$$$\int \left(- 3 \sin{\left(\frac{x}{3} \right)}\right)\, dx = 9 \cos{\left(\frac{x}{3} \right)} + C$$$A