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

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

Zij $$$u=\sqrt{x}$$$.

Dan $$$du=\left(\sqrt{x}\right)^{\prime }dx = \frac{1}{2 \sqrt{x}} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$\frac{dx}{\sqrt{x}} = 2 du$$$.

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=2$$$ en $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

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

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

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

$$2 \sin{\left({\color{red}{u}} \right)} = 2 \sin{\left({\color{red}{\sqrt{x}}} \right)}$$

Dus,

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

Voeg de integratieconstante toe:

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

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