Integraal van $$$\cos{\left(4 x - 2 \right)} - 1$$$

Bepaal $$$\int \left(\cos{\left(4 x - 2 \right)} - 1\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(\cos{\left(4 x - 2 \right)} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{\cos{\left(4 x - 2 \right)} d x}\right)}}$$

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:

$$\int{\cos{\left(4 x - 2 \right)} d x} - {\color{red}{\int{1 d x}}} = \int{\cos{\left(4 x - 2 \right)} d x} - {\color{red}{x}}$$

Zij $$$u=4 x - 2$$$.

Dan $$$du=\left(4 x - 2\right)^{\prime }dx = 4 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{4}$$$.

De integraal kan worden herschreven als

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

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

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

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

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

We herinneren eraan dat $$$u=4 x - 2$$$:

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

Dus,

$$\int{\left(\cos{\left(4 x - 2 \right)} - 1\right)d x} = - x + \frac{\sin{\left(4 x - 2 \right)}}{4}$$

Voeg de integratieconstante toe:

$$\int{\left(\cos{\left(4 x - 2 \right)} - 1\right)d x} = - x + \frac{\sin{\left(4 x - 2 \right)}}{4}+C$$

$$$\int \left(\cos{\left(4 x - 2 \right)} - 1\right)\, dx = \left(- x + \frac{\sin{\left(4 x - 2 \right)}}{4}\right) + C$$$A