Integralen av $$$\cos{\left(\frac{2}{x} \right)}$$$

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

För integralen $$$\int{\cos{\left(\frac{2}{x} \right)} d x}$$$, använd partiell integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

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

Då gäller $$$\operatorname{du}=\left(\cos{\left(\frac{2}{x} \right)}\right)^{\prime }dx=\frac{2 \sin{\left(\frac{2}{x} \right)}}{x^{2}} dx$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{1 d x}=x$$$ (stegen kan ses »).

Integralen blir

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=2$$$ och $$$f{\left(x \right)} = \frac{\sin{\left(\frac{2}{x} \right)}}{x}$$$:

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

Låt $$$u=\frac{2}{x}$$$ vara.

Då $$$du=\left(\frac{2}{x}\right)^{\prime }dx = - \frac{2}{x^{2}} dx$$$ (stegen kan ses »), och vi har att $$$\frac{dx}{x^{2}} = - \frac{du}{2}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=-1$$$ och $$$f{\left(u \right)} = \frac{\sin{\left(u \right)}}{u}$$$:

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

Denna integral (Sinusintegralen) har ingen sluten form:

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

Kom ihåg att $$$u=\frac{2}{x}$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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