Integral of $$$2 \sin{\left(3 x^{2} \right)}$$$

Find $$$\int 2 \sin{\left(3 x^{2} \right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = \sin{\left(3 x^{2} \right)}$$$:

$${\color{red}{\int{2 \sin{\left(3 x^{2} \right)} d x}}} = {\color{red}{\left(2 \int{\sin{\left(3 x^{2} \right)} d x}\right)}}$$

Let $$$u=\sqrt{3} x$$$.

Then $$$du=\left(\sqrt{3} x\right)^{\prime }dx = \sqrt{3} dx$$$ (steps can be seen »), and we have that $$$dx = \frac{\sqrt{3} du}{3}$$$.

So,

$$2 {\color{red}{\int{\sin{\left(3 x^{2} \right)} d x}}} = 2 {\color{red}{\int{\frac{\sqrt{3} \sin{\left(u^{2} \right)}}{3} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{\sqrt{3}}{3}$$$ and $$$f{\left(u \right)} = \sin{\left(u^{2} \right)}$$$:

$$2 {\color{red}{\int{\frac{\sqrt{3} \sin{\left(u^{2} \right)}}{3} d u}}} = 2 {\color{red}{\left(\frac{\sqrt{3} \int{\sin{\left(u^{2} \right)} d u}}{3}\right)}}$$

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

$$\frac{2 \sqrt{3} {\color{red}{\int{\sin{\left(u^{2} \right)} d u}}}}{3} = \frac{2 \sqrt{3} {\color{red}{\left(\frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} u}{\sqrt{\pi}}\right)}{2}\right)}}}{3}$$

Recall that $$$u=\sqrt{3} x$$$:

$$\frac{\sqrt{6} \sqrt{\pi} S\left(\frac{\sqrt{2} {\color{red}{u}}}{\sqrt{\pi}}\right)}{3} = \frac{\sqrt{6} \sqrt{\pi} S\left(\frac{\sqrt{2} {\color{red}{\sqrt{3} x}}}{\sqrt{\pi}}\right)}{3}$$

Therefore,

$$\int{2 \sin{\left(3 x^{2} \right)} d x} = \frac{\sqrt{6} \sqrt{\pi} S\left(\frac{\sqrt{6} x}{\sqrt{\pi}}\right)}{3}$$

Add the constant of integration:

$$\int{2 \sin{\left(3 x^{2} \right)} d x} = \frac{\sqrt{6} \sqrt{\pi} S\left(\frac{\sqrt{6} x}{\sqrt{\pi}}\right)}{3}+C$$

$$$\int 2 \sin{\left(3 x^{2} \right)}\, dx = \frac{\sqrt{6} \sqrt{\pi} S\left(\frac{\sqrt{6} x}{\sqrt{\pi}}\right)}{3} + C$$$A