All causes and sin are from the same root

Nice problem!!

$\sqrt{x}=t$

$\large{2 \displaystyle \int _{0}^{\infty} \sin (t^2)+\cos (t^2) dt}$

We know that

$\large{\displaystyle \int _{0}^{\infty} e^{-ax}x^{n-1} dx=\frac{\Gamma (n)}{a^{n}}}$ where $\Gamma (n)$ is Gamma function

Replace $a \rightarrow a+ib$ where $i =\sqrt{-1}$

$\large{\displaystyle \int _{0}^{\infty} e^{-ax}e^{-ibx}x^{n-1} dx=\frac{\Gamma (n)}{(a+ib)^{n}}}$

Put $a=r \cos y ~ and ~ b=r \sin y$ .

So that $r^2=a^2+b^2$ and $y=\arctan \left(\frac{b}{a}\right)$

Using de moviers theorem

$\large{\displaystyle \int _{0}^{\infty} e^{-ax}(\cos bx-i \sin bx)x^{n-1} dx=\frac{\Gamma (n)}{r^{n}}(\cos ny+\sin ny)^{-1}}$

Comparing the real and imaginary parts we finally get

$\large{\displaystyle \int _{0}^{\infty} e^{-ax}(\cos bx)x^{n-1} dx=\frac{\Gamma (n)}{r^{n}}(\cos ny)}$

$\large{\displaystyle \int _{0}^{\infty} e^{-ax}(\sin bx)x^{n-1} dx=\frac{\Gamma (n)}{r^{n}}(\sin ny)}$

I will give the answer for $cos$ part only as the $\sin$ part is exactly analogous to it.

Put $a=0$

$\large{\displaystyle \int _{0}^{\infty} e^{-0x}(\cos bx)x^{n-1} dx=\frac{\Gamma (n)}{(0^2+b^2)^{\frac{n}{2}}}(\cos \left(n \arctan \left(\frac{b}{0} \right)\right))}$

$\large{\displaystyle \int _{0}^{\infty} x^{n-1} \cos bxdx=\frac{\Gamma (n)}{b^n} \cos \left(\frac{n \pi}{2}\right)}$

Put $x^n=z$ so that $x^{n-1}dx=\frac{dz}{n}$ and $\large{x=z^{\frac{1}{n}}}$

Then

$\large{\displaystyle \int_{0}^{\infty} \cos(b z^{\frac{1}{n}}) dz=\frac{n \Gamma (n)}{b^n} \cos \left(\frac{n \pi}{2}\right)}$

Here $b=1$ and $n=\frac{1}{2}$

$\large{\displaystyle \int_{0}^{\infty} \cos (x^2)=\frac{\sqrt{\pi}}{2 \sqrt{2}}}$

The final answer is

$\huge{\boxed{\sqrt{2 \pi}}}$