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

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

Voor de integraal $$$\int{\cos{\left(\frac{2}{x} \right)} d x}$$$, gebruik partiële integratie $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Zij $$$\operatorname{u}=\cos{\left(\frac{2}{x} \right)}$$$ en $$$\operatorname{dv}=dx$$$.

Dan $$$\operatorname{du}=\left(\cos{\left(\frac{2}{x} \right)}\right)^{\prime }dx=\frac{2 \sin{\left(\frac{2}{x} \right)}}{x^{2}} dx$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{1 d x}=x$$$ (de stappen zijn te zien »).

Dus,

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

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

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

Zij $$$u=\frac{2}{x}$$$.

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

Dus,

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

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

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

Deze integraal (Sinusintegraal) heeft geen gesloten vorm:

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

We herinneren eraan dat $$$u=\frac{2}{x}$$$:

$$x \cos{\left(\frac{2}{x} \right)} + 2 \operatorname{Si}{\left({\color{red}{u}} \right)} = x \cos{\left(\frac{2}{x} \right)} + 2 \operatorname{Si}{\left({\color{red}{\left(\frac{2}{x}\right)}} \right)}$$

Dus,

$$\int{\cos{\left(\frac{2}{x} \right)} d x} = x \cos{\left(\frac{2}{x} \right)} + 2 \operatorname{Si}{\left(\frac{2}{x} \right)}$$

Voeg de integratieconstante toe:

$$\int{\cos{\left(\frac{2}{x} \right)} d x} = x \cos{\left(\frac{2}{x} \right)} + 2 \operatorname{Si}{\left(\frac{2}{x} \right)}+C$$

$$$\int \cos{\left(\frac{2}{x} \right)}\, dx = \left(x \cos{\left(\frac{2}{x} \right)} + 2 \operatorname{Si}{\left(\frac{2}{x} \right)}\right) + C$$$A