Integraal van $$$4 \sin^{2}{\left(\theta \right)}$$$

Bepaal $$$\int 4 \sin^{2}{\left(\theta \right)}\, d\theta$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ toe met $$$c=4$$$ en $$$f{\left(\theta \right)} = \sin^{2}{\left(\theta \right)}$$$:

$${\color{red}{\int{4 \sin^{2}{\left(\theta \right)} d \theta}}} = {\color{red}{\left(4 \int{\sin^{2}{\left(\theta \right)} d \theta}\right)}}$$

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

$$4 {\color{red}{\int{\sin^{2}{\left(\theta \right)} d \theta}}} = 4 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 \theta \right)}}{2}\right)d \theta}}}$$

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

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

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, d\theta = c \theta$$$ toe met $$$c=1$$$:

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

Zij $$$u=2 \theta$$$.

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

Dus,

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

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

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

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

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

We herinneren eraan dat $$$u=2 \theta$$$:

$$2 \theta - \sin{\left({\color{red}{u}} \right)} = 2 \theta - \sin{\left({\color{red}{\left(2 \theta\right)}} \right)}$$

Dus,

$$\int{4 \sin^{2}{\left(\theta \right)} d \theta} = 2 \theta - \sin{\left(2 \theta \right)}$$

Voeg de integratieconstante toe:

$$\int{4 \sin^{2}{\left(\theta \right)} d \theta} = 2 \theta - \sin{\left(2 \theta \right)}+C$$

$$$\int 4 \sin^{2}{\left(\theta \right)}\, d\theta = \left(2 \theta - \sin{\left(2 \theta \right)}\right) + C$$$A