I really wonder how I can prove the following integrals.
$$\int_0^\infty \sin ax^2\cos 2bx\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos \frac{b^2}{a}-\sin\frac{b^2}{a}\right)$$
and
$$\int_0^\infty \cos ax^2\cos 2bx\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos \frac{b^2}{a}+\sin\frac{b^2}{a}\right)$$
I tried $\sin ax^2=\Im(e^{iax^2})$ and $\cos ax^2=\Re(e^{iax^2})$ then I used by parts method but I failed. Obviously tangent half-angle substitution doesn't work. I'm quite sure if we can calculate one of them, the similar technique can be used to calculate the other. Could anyone here please help me to calculate the integrals preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.