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

Bepaal $$$\int \sin^{4}{\left(x \right)}\, dx$$$.

Pas de machtsreductieformule $$$\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ toe met $$$\alpha=x$$$:

$${\color{red}{\int{\sin^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- \frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{8} + \frac{3}{8}\right)d x}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{8}$$$ en $$$f{\left(x \right)} = - 4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3$$$:

$${\color{red}{\int{\left(- \frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{8} + \frac{3}{8}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(- 4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3\right)d x}}{8}\right)}}$$

Integreer termgewijs:

$$\frac{{\color{red}{\int{\left(- 4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3\right)d x}}}}{8} = \frac{{\color{red}{\left(\int{3 d x} - \int{4 \cos{\left(2 x \right)} d x} + \int{\cos{\left(4 x \right)} d x}\right)}}}{8}$$

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=3$$$:

$$- \frac{\int{4 \cos{\left(2 x \right)} d x}}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{3 d x}}}}{8} = - \frac{\int{4 \cos{\left(2 x \right)} d x}}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\left(3 x\right)}}}{8}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=4$$$ en $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{4 \cos{\left(2 x \right)} d x}}}}{8} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\left(4 \int{\cos{\left(2 x \right)} d x}\right)}}}{8}$$

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,

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{2} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$

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

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$

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

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

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

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

Zij $$$u=4 x$$$.

Dan $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{4}$$$.

Dus,

$$\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{8} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}$$

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

$$\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$

We herinneren eraan dat $$$u=4 x$$$:

$$\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{32}$$

Dus,

$$\int{\sin^{4}{\left(x \right)} d x} = \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}$$

Vereenvoudig:

$$\int{\sin^{4}{\left(x \right)} d x} = \frac{12 x - 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}$$

Voeg de integratieconstante toe:

$$\int{\sin^{4}{\left(x \right)} d x} = \frac{12 x - 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}+C$$

$$$\int \sin^{4}{\left(x \right)}\, dx = \frac{12 x - 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32} + C$$$A