Integral of $$$\cos{\left(\omega t^{2} \right)}$$$ with respect to $$$t$$$

Find $$$\int \cos{\left(\omega t^{2} \right)}\, dt$$$.

Let $$$u=\sqrt{\omega} t$$$.

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

The integral becomes

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{\sqrt{\omega}}$$$ and $$$f{\left(u \right)} = \cos{\left(u^{2} \right)}$$$:

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

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

$$\frac{{\color{red}{\int{\cos{\left(u^{2} \right)} d u}}}}{\sqrt{\omega}} = \frac{{\color{red}{\left(\frac{\sqrt{2} \sqrt{\pi} C\left(\frac{\sqrt{2} u}{\sqrt{\pi}}\right)}{2}\right)}}}{\sqrt{\omega}}$$

Recall that $$$u=\sqrt{\omega} t$$$:

$$\frac{\sqrt{2} \sqrt{\pi} C\left(\frac{\sqrt{2} {\color{red}{u}}}{\sqrt{\pi}}\right)}{2 \sqrt{\omega}} = \frac{\sqrt{2} \sqrt{\pi} C\left(\frac{\sqrt{2} {\color{red}{\sqrt{\omega} t}}}{\sqrt{\pi}}\right)}{2 \sqrt{\omega}}$$

Therefore,

$$\int{\cos{\left(\omega t^{2} \right)} d t} = \frac{\sqrt{2} \sqrt{\pi} C\left(\frac{\sqrt{2} \sqrt{\omega} t}{\sqrt{\pi}}\right)}{2 \sqrt{\omega}}$$

Add the constant of integration:

$$\int{\cos{\left(\omega t^{2} \right)} d t} = \frac{\sqrt{2} \sqrt{\pi} C\left(\frac{\sqrt{2} \sqrt{\omega} t}{\sqrt{\pi}}\right)}{2 \sqrt{\omega}}+C$$

$$$\int \cos{\left(\omega t^{2} \right)}\, dt = \frac{\sqrt{2} \sqrt{\pi} C\left(\frac{\sqrt{2} \sqrt{\omega} t}{\sqrt{\pi}}\right)}{2 \sqrt{\omega}} + C$$$A