Integraal van $$$2 x \cos{\left(3 x \right)}$$$
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Uw invoer
Bepaal $$$\int 2 x \cos{\left(3 x \right)}\, dx$$$.
Oplossing
Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=2$$$ en $$$f{\left(x \right)} = x \cos{\left(3 x \right)}$$$:
$${\color{red}{\int{2 x \cos{\left(3 x \right)} d x}}} = {\color{red}{\left(2 \int{x \cos{\left(3 x \right)} d x}\right)}}$$
Voor de integraal $$$\int{x \cos{\left(3 x \right)} d x}$$$, gebruik partiële integratie $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.
Zij $$$\operatorname{u}=x$$$ en $$$\operatorname{dv}=\cos{\left(3 x \right)} dx$$$.
Dan $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{\cos{\left(3 x \right)} d x}=\frac{\sin{\left(3 x \right)}}{3}$$$ (de stappen zijn te zien »).
Dus,
$$2 {\color{red}{\int{x \cos{\left(3 x \right)} d x}}}=2 {\color{red}{\left(x \cdot \frac{\sin{\left(3 x \right)}}{3}-\int{\frac{\sin{\left(3 x \right)}}{3} \cdot 1 d x}\right)}}=2 {\color{red}{\left(\frac{x \sin{\left(3 x \right)}}{3} - \int{\frac{\sin{\left(3 x \right)}}{3} d x}\right)}}$$
Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{3}$$$ en $$$f{\left(x \right)} = \sin{\left(3 x \right)}$$$:
$$\frac{2 x \sin{\left(3 x \right)}}{3} - 2 {\color{red}{\int{\frac{\sin{\left(3 x \right)}}{3} d x}}} = \frac{2 x \sin{\left(3 x \right)}}{3} - 2 {\color{red}{\left(\frac{\int{\sin{\left(3 x \right)} d x}}{3}\right)}}$$
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,
$$\frac{2 x \sin{\left(3 x \right)}}{3} - \frac{2 {\color{red}{\int{\sin{\left(3 x \right)} d x}}}}{3} = \frac{2 x \sin{\left(3 x \right)}}{3} - \frac{2 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{3}$$
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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$\frac{2 x \sin{\left(3 x \right)}}{3} - \frac{2 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{3} = \frac{2 x \sin{\left(3 x \right)}}{3} - \frac{2 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}}{3}$$
De integraal van de sinus is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\frac{2 x \sin{\left(3 x \right)}}{3} - \frac{2 {\color{red}{\int{\sin{\left(u \right)} d u}}}}{9} = \frac{2 x \sin{\left(3 x \right)}}{3} - \frac{2 {\color{red}{\left(- \cos{\left(u \right)}\right)}}}{9}$$
We herinneren eraan dat $$$u=3 x$$$:
$$\frac{2 x \sin{\left(3 x \right)}}{3} + \frac{2 \cos{\left({\color{red}{u}} \right)}}{9} = \frac{2 x \sin{\left(3 x \right)}}{3} + \frac{2 \cos{\left({\color{red}{\left(3 x\right)}} \right)}}{9}$$
Dus,
$$\int{2 x \cos{\left(3 x \right)} d x} = \frac{2 x \sin{\left(3 x \right)}}{3} + \frac{2 \cos{\left(3 x \right)}}{9}$$
Vereenvoudig:
$$\int{2 x \cos{\left(3 x \right)} d x} = \frac{2 \left(3 x \sin{\left(3 x \right)} + \cos{\left(3 x \right)}\right)}{9}$$
Voeg de integratieconstante toe:
$$\int{2 x \cos{\left(3 x \right)} d x} = \frac{2 \left(3 x \sin{\left(3 x \right)} + \cos{\left(3 x \right)}\right)}{9}+C$$
Antwoord
$$$\int 2 x \cos{\left(3 x \right)}\, dx = \frac{2 \left(3 x \sin{\left(3 x \right)} + \cos{\left(3 x \right)}\right)}{9} + C$$$A