Integral of $$$\sin{\left(2 x^{2} \right)}$$$

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

Let $$$u=\sqrt{2} x$$$.

Then $$$du=\left(\sqrt{2} x\right)^{\prime }dx = \sqrt{2} dx$$$ (steps can be seen »), and we have that $$$dx = \frac{\sqrt{2} du}{2}$$$.

So,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{\sqrt{2}}{2}$$$ and $$$f{\left(u \right)} = \sin{\left(u^{2} \right)}$$$:

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

This integral (Fresnel Sine Integral) does not have a closed form:

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

Recall that $$$u=\sqrt{2} x$$$:

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

Therefore,

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

Add the constant of integration:

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

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