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

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=- \frac{1}{3}$$$ och $$$f{\left(x \right)} = \cos{\left(3 x \right)}$$$:

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

Låt $$$u=3 x$$$ vara.

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

Alltså,

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

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

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

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

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

Kom ihåg att $$$u=3 x$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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