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

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

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

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

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,

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

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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

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

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

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

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

Dus,

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

Voeg de integratieconstante toe:

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

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