Integraal van $$$s^{2} \sin{\left(x^{2} \right)}$$$ met betrekking tot $$$x$$$

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

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

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

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

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

Dus,

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

Voeg de integratieconstante toe:

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

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