Integraal van $$$4 \sin^{2}{\left(x \right)}$$$
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Uw invoer
Bepaal $$$\int 4 \sin^{2}{\left(x \right)}\, dx$$$.
Oplossing
Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=4$$$ en $$$f{\left(x \right)} = \sin^{2}{\left(x \right)}$$$:
$${\color{red}{\int{4 \sin^{2}{\left(x \right)} d x}}} = {\color{red}{\left(4 \int{\sin^{2}{\left(x \right)} d x}\right)}}$$
Pas de machtsreductieformule $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ toe met $$$\alpha=x$$$:
$$4 {\color{red}{\int{\sin^{2}{\left(x \right)} d x}}} = 4 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}}$$
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 $$$f{\left(x \right)} = 1 - \cos{\left(2 x \right)}$$$:
$$4 {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}} = 4 {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}{2}\right)}}$$
Integreer termgewijs:
$$2 {\color{red}{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}} = 2 {\color{red}{\left(\int{1 d x} - \int{\cos{\left(2 x \right)} d x}\right)}}$$
Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:
$$- 2 \int{\cos{\left(2 x \right)} d x} + 2 {\color{red}{\int{1 d x}}} = - 2 \int{\cos{\left(2 x \right)} d x} + 2 {\color{red}{x}}$$
Zij $$$u=2 x$$$.
Dan $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{2}$$$.
De integraal kan worden herschreven als
$$2 x - 2 {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = 2 x - 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$
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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$2 x - 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = 2 x - 2 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$
De integraal van de cosinus is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$2 x - {\color{red}{\int{\cos{\left(u \right)} d u}}} = 2 x - {\color{red}{\sin{\left(u \right)}}}$$
We herinneren eraan dat $$$u=2 x$$$:
$$2 x - \sin{\left({\color{red}{u}} \right)} = 2 x - \sin{\left({\color{red}{\left(2 x\right)}} \right)}$$
Dus,
$$\int{4 \sin^{2}{\left(x \right)} d x} = 2 x - \sin{\left(2 x \right)}$$
Voeg de integratieconstante toe:
$$\int{4 \sin^{2}{\left(x \right)} d x} = 2 x - \sin{\left(2 x \right)}+C$$
Antwoord
$$$\int 4 \sin^{2}{\left(x \right)}\, dx = \left(2 x - \sin{\left(2 x \right)}\right) + C$$$A