Integraal van $$$8 \sin{\left(2 x \right)}$$$

Bepaal $$$\int 8 \sin{\left(2 x \right)}\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=8$$$ en $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$:

$${\color{red}{\int{8 \sin{\left(2 x \right)} d x}}} = {\color{red}{\left(8 \int{\sin{\left(2 x \right)} d x}\right)}}$$

Zij $$$u=2 x$$$.

Dan $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{2}$$$.

Dus,

$$8 {\color{red}{\int{\sin{\left(2 x \right)} d x}}} = 8 {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$8 {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}} = 8 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}$$

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

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

We herinneren eraan dat $$$u=2 x$$$:

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

Dus,

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

Voeg de integratieconstante toe:

$$\int{8 \sin{\left(2 x \right)} d x} = - 4 \cos{\left(2 x \right)}+C$$

$$$\int 8 \sin{\left(2 x \right)}\, dx = - 4 \cos{\left(2 x \right)} + C$$$A