Integraal van $$$\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)}$$$

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

Herschrijf $$$\sin\left(x \right)\sin\left(2 x \right)$$$ met behulp van de formule $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ met $$$\alpha=x$$$ en $$$\beta=2 x$$$:

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

Werk de uitdrukking uit:

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

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 $$$f{\left(x \right)} = \sin{\left(3 x \right)} \cos{\left(x \right)} - \sin{\left(3 x \right)} \cos{\left(3 x \right)}$$$:

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

Integreer termgewijs:

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

Herschrijf de integraand met behulp van de formule $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$, met $$$\alpha=3 x$$$ en $$$\beta=x$$$:

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

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 $$$f{\left(x \right)} = \sin{\left(2 x \right)} + \sin{\left(4 x \right)}$$$:

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

Integreer termgewijs:

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

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,

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

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

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

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

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

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

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

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,

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

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

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

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{16} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{16}$$

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{16} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(4 x\right)}} \right)}}{16}$$

Zij $$$u=\sin{\left(3 x \right)}$$$.

Dan $$$du=\left(\sin{\left(3 x \right)}\right)^{\prime }dx = 3 \cos{\left(3 x \right)} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$\cos{\left(3 x \right)} dx = \frac{du}{3}$$$.

Dus,

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}}}{2} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\int{\frac{u}{3} d u}}}}{2}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{3}$$$ en $$$f{\left(u \right)} = u$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\int{\frac{u}{3} d u}}}}{2} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\left(\frac{\int{u d u}}{3}\right)}}}{2}$$

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=1$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\int{u d u}}}}{6}=- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{6}=- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{6}$$

We herinneren eraan dat $$$u=\sin{\left(3 x \right)}$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{u}}^{2}}{12} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\sin{\left(3 x \right)}}}^{2}}{12}$$

Dus,

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x} = - \frac{\sin^{2}{\left(3 x \right)}}{12} - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16}$$

Voeg de integratieconstante toe:

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x} = - \frac{\sin^{2}{\left(3 x \right)}}{12} - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16}+C$$

$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)}\, dx = \left(- \frac{\sin^{2}{\left(3 x \right)}}{12} - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16}\right) + C$$$A