Integraal van $$$u \sin^{2}{\left(3 x \right)}$$$ met betrekking tot $$$x$$$

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

Pas de machtsreductieformule $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ toe met $$$\alpha=3 x$$$:

$${\color{red}{\int{u \sin^{2}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{u \left(1 - \cos{\left(6 x \right)}\right)}{2} d x}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(x \right)} = u \left(1 - \cos{\left(6 x \right)}\right)$$$:

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

Expand the expression:

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

Integreer termgewijs:

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

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

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=u$$$ en $$$f{\left(x \right)} = \cos{\left(6 x \right)}$$$:

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

Zij $$$v=6 x$$$.

Dan $$$dv=\left(6 x\right)^{\prime }dx = 6 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{dv}{6}$$$.

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ toe met $$$c=\frac{1}{6}$$$ en $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

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

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

$$\frac{u x}{2} - \frac{u {\color{red}{\int{\cos{\left(v \right)} d v}}}}{12} = \frac{u x}{2} - \frac{u {\color{red}{\sin{\left(v \right)}}}}{12}$$

We herinneren eraan dat $$$v=6 x$$$:

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

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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