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

Bepaal $$$\int \sin{\left(2 q \right)}\, dq$$$.

Zij $$$u=2 q$$$.

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

De integraal wordt

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

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

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

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

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

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

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

Dus,

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

Voeg de integratieconstante toe:

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

$$$\int \sin{\left(2 q \right)}\, dq = - \frac{\cos{\left(2 q \right)}}{2} + C$$$A