Integraal van $$$\frac{\sin{\left(\pi n y \right)}}{2}$$$ met betrekking tot $$$y$$$

Bepaal $$$\int \frac{\sin{\left(\pi n y \right)}}{2}\, dy$$$.

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

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

Zij $$$u=\pi n y$$$.

Dan $$$du=\left(\pi n y\right)^{\prime }dy = \pi n dy$$$ (de stappen zijn te zien »), en dan geldt dat $$$dy = \frac{du}{\pi n}$$$.

De integraal wordt

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

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

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

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

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

We herinneren eraan dat $$$u=\pi n y$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

$$$\int \frac{\sin{\left(\pi n y \right)}}{2}\, dy = - \frac{\cos{\left(\pi n y \right)}}{2 \pi n} + C$$$A