Integraal van $$$\frac{\cos{\left(4 t \right)}}{2}$$$

Bepaal $$$\int \frac{\cos{\left(4 t \right)}}{2}\, dt$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(t \right)} = \cos{\left(4 t \right)}$$$:

$${\color{red}{\int{\frac{\cos{\left(4 t \right)}}{2} d t}}} = {\color{red}{\left(\frac{\int{\cos{\left(4 t \right)} d t}}{2}\right)}}$$

Zij $$$u=4 t$$$.

Dan $$$du=\left(4 t\right)^{\prime }dt = 4 dt$$$ (de stappen zijn te zien »), en dan geldt dat $$$dt = \frac{du}{4}$$$.

Dus,

$$\frac{{\color{red}{\int{\cos{\left(4 t \right)} d t}}}}{2} = \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

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

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

We herinneren eraan dat $$$u=4 t$$$:

$$\frac{\sin{\left({\color{red}{u}} \right)}}{8} = \frac{\sin{\left({\color{red}{\left(4 t\right)}} \right)}}{8}$$

Dus,

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

Voeg de integratieconstante toe:

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

$$$\int \frac{\cos{\left(4 t \right)}}{2}\, dt = \frac{\sin{\left(4 t \right)}}{8} + C$$$A