Integraal van $$$\sin^{2}{\left(x \right)} + 1$$$

Bepaal $$$\int \left(\sin^{2}{\left(x \right)} + 1\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(\sin^{2}{\left(x \right)} + 1\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\sin^{2}{\left(x \right)} d x}\right)}}$$

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

$$\int{\sin^{2}{\left(x \right)} d x} + {\color{red}{\int{1 d x}}} = \int{\sin^{2}{\left(x \right)} d x} + {\color{red}{x}}$$

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

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

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 $$$f{\left(x \right)} = 1 - \cos{\left(2 x \right)}$$$:

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

Integreer termgewijs:

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

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

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

Zij $$$u=2 x$$$.

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

Dus,

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

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

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

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

$$\frac{3 x}{2} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{3 x}{2} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

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

$$\frac{3 x}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{3 x}{2} - \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

Dus,

$$\int{\left(\sin^{2}{\left(x \right)} + 1\right)d x} = \frac{3 x}{2} - \frac{\sin{\left(2 x \right)}}{4}$$

Vereenvoudig:

$$\int{\left(\sin^{2}{\left(x \right)} + 1\right)d x} = \frac{6 x - \sin{\left(2 x \right)}}{4}$$

Voeg de integratieconstante toe:

$$\int{\left(\sin^{2}{\left(x \right)} + 1\right)d x} = \frac{6 x - \sin{\left(2 x \right)}}{4}+C$$

$$$\int \left(\sin^{2}{\left(x \right)} + 1\right)\, dx = \frac{6 x - \sin{\left(2 x \right)}}{4} + C$$$A