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

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

Zij $$$u=x + 2$$$.

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

De integraal wordt

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

Deze integraal (Fresnel-sinusintegraal) heeft geen gesloten vorm:

$${\color{red}{\int{\sin{\left(u^{2} \right)} d u}}} = {\color{red}{\left(\frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} u}{\sqrt{\pi}}\right)}{2}\right)}}$$

We herinneren eraan dat $$$u=x + 2$$$:

$$\frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} {\color{red}{u}}}{\sqrt{\pi}}\right)}{2} = \frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} {\color{red}{\left(x + 2\right)}}}{\sqrt{\pi}}\right)}{2}$$

Dus,

$$\int{\sin{\left(\left(x + 2\right)^{2} \right)} d x} = \frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} \left(x + 2\right)}{\sqrt{\pi}}\right)}{2}$$

Voeg de integratieconstante toe:

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

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