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

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

Herschrijf $$$\sin\left(2 x \right)\cos\left(x \right)$$$ 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 $$$\alpha=2 x$$$ en $$$\beta=x$$$:

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

Werk de uitdrukking uit:

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

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 $$$f{\left(x \right)} = \sin{\left(x \right)} \cos{\left(2 x \right)} + \sin{\left(3 x \right)} \cos{\left(2 x \right)}$$$:

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

Integreer termgewijs:

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

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 $$$\alpha=x$$$ en $$$\beta=2 x$$$:

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

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

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

Integreer termgewijs:

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

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

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

Zij $$$u=3 x$$$.

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

Dus,

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

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

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

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

$$\frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{12} = \frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{12}$$

We herinneren eraan dat $$$u=3 x$$$:

$$\frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{12} = \frac{\cos{\left(x \right)}}{4} + \frac{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(3 x\right)}} \right)}}{12}$$

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 $$$\beta=2 x$$$:

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}}}{2} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{2}$$

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 $$$f{\left(x \right)} = \sin{\left(x \right)} + \sin{\left(5 x \right)}$$$:

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{2} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}{2}\right)}}}{2}$$

Integreer termgewijs:

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}}}{4} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(\int{\sin{\left(x \right)} d x} + \int{\sin{\left(5 x \right)} d x}\right)}}}{4}$$

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

$$\frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{\int{\sin{\left(5 x \right)} d x}}{4} + \frac{{\color{red}{\int{\sin{\left(x \right)} d x}}}}{4} = \frac{\cos{\left(x \right)}}{4} - \frac{\cos{\left(3 x \right)}}{12} + \frac{\int{\sin{\left(5 x \right)} d x}}{4} + \frac{{\color{red}{\left(- \cos{\left(x \right)}\right)}}}{4}$$

Zij $$$u=5 x$$$.

Dan $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{5}$$$.

Dus,

$$- \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(5 x \right)} d x}}}}{4} = - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{4}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{5}$$$ en $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$- \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{4} = - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{4}$$

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

$$- \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{20} = - \frac{\cos{\left(3 x \right)}}{12} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{20}$$

We herinneren eraan dat $$$u=5 x$$$:

$$- \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left({\color{red}{u}} \right)}}{20} = - \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left({\color{red}{\left(5 x\right)}} \right)}}{20}$$

Dus,

$$\int{\sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)} d x} = - \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left(5 x \right)}}{20}$$

Voeg de integratieconstante toe:

$$\int{\sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)} d x} = - \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left(5 x \right)}}{20}+C$$

$$$\int \sin{\left(2 x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)}\, dx = \left(- \frac{\cos{\left(3 x \right)}}{12} - \frac{\cos{\left(5 x \right)}}{20}\right) + C$$$A