Integraal van $$$\sin{\left(6 c \right)}$$$

Bepaal $$$\int \sin{\left(6 c \right)}\, dc$$$.

Zij $$$u=6 c$$$.

Dan $$$du=\left(6 c\right)^{\prime }dc = 6 dc$$$ (de stappen zijn te zien »), en dan geldt dat $$$dc = \frac{du}{6}$$$.

Dus,

$${\color{red}{\int{\sin{\left(6 c \right)} d c}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{6}$$$ en $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}} = {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{6}\right)}}$$

De integraal van de sinus is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{6} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{6}$$

We herinneren eraan dat $$$u=6 c$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{6} = - \frac{\cos{\left({\color{red}{\left(6 c\right)}} \right)}}{6}$$

Dus,

$$\int{\sin{\left(6 c \right)} d c} = - \frac{\cos{\left(6 c \right)}}{6}$$

Voeg de integratieconstante toe:

$$\int{\sin{\left(6 c \right)} d c} = - \frac{\cos{\left(6 c \right)}}{6}+C$$

$$$\int \sin{\left(6 c \right)}\, dc = - \frac{\cos{\left(6 c \right)}}{6} + C$$$A