Integraal van $$$\cos^{2}{\left(3 x \right)}$$$

Bepaal $$$\int \cos^{2}{\left(3 x \right)}\, dx$$$.

Zij $$$u=3 x$$$.

Dan $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{3}$$$.

Dus,

$${\color{red}{\int{\cos^{2}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{3} d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{3}$$$ en $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\cos^{2}{\left(u \right)} d u}}{3}\right)}}$$

Pas de machtsreductieformule $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ toe met $$$\alpha= u $$$:

$$\frac{{\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{3}$$

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 $$$f{\left(u \right)} = \cos{\left(2 u \right)} + 1$$$:

$$\frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{3} = \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}}{3}$$

Integreer termgewijs:

$$\frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{6} = \frac{{\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{6}$$

Pas de constantenregel $$$\int c\, du = c u$$$ toe met $$$c=1$$$:

$$\frac{\int{\cos{\left(2 u \right)} d u}}{6} + \frac{{\color{red}{\int{1 d u}}}}{6} = \frac{\int{\cos{\left(2 u \right)} d u}}{6} + \frac{{\color{red}{u}}}{6}$$

Zij $$$v=2 u$$$.

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

Dus,

$$\frac{u}{6} + \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{6} = \frac{u}{6} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{6}$$

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

$$\frac{u}{6} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{6} = \frac{u}{6} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{6}$$

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

$$\frac{u}{6} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{12} = \frac{u}{6} + \frac{{\color{red}{\sin{\left(v \right)}}}}{12}$$

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

$$\frac{u}{6} + \frac{\sin{\left({\color{red}{v}} \right)}}{12} = \frac{u}{6} + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{12}$$

We herinneren eraan dat $$$u=3 x$$$:

$$\frac{\sin{\left(2 {\color{red}{u}} \right)}}{12} + \frac{{\color{red}{u}}}{6} = \frac{\sin{\left(2 {\color{red}{\left(3 x\right)}} \right)}}{12} + \frac{{\color{red}{\left(3 x\right)}}}{6}$$

Dus,

$$\int{\cos^{2}{\left(3 x \right)} d x} = \frac{x}{2} + \frac{\sin{\left(6 x \right)}}{12}$$

Voeg de integratieconstante toe:

$$\int{\cos^{2}{\left(3 x \right)} d x} = \frac{x}{2} + \frac{\sin{\left(6 x \right)}}{12}+C$$

$$$\int \cos^{2}{\left(3 x \right)}\, dx = \left(\frac{x}{2} + \frac{\sin{\left(6 x \right)}}{12}\right) + C$$$A