Evaluating $\int_{\mathbb{R}}\sin(x^2)\ dx$ via complex numbers

Question

I found a rather interesting integral in my favourite maths book of all time "Les Mathématiciens de A à Z" under the name of Fresnel's integral. It is the one in the title of this post and I was wondering whether I could get some help as to the final stages of my evaluation: Let $f(x)=e^{-ix^2}=\cos(x^2)-i\sin(x^2)$. Integrating this function on $\mathbb{R}$ will yield: $$I=\int_{\mathbb{R}}f(x)\ dx=\int_{\mathbb{R}}\cos(x^2)\ dx-i\int_{\mathbb{R}}\sin(x^2)\ dx$$ $$\Rightarrow\Re(I)=\int_{\mathbb{R}}\cos(x^2)\ dx \wedge\Im(I)=-\int_{\mathbb{R}}\sin(x^2)\ dx$$ What we'll do here is square $I$ and play around with the new expression: $$I^2=\iint_{\mathbb{R}}e^{-i(x^2+y^2)}\ dx\ dy$$ Converting to polar coordinates yield: $$I^2=\int_0^{2\pi}\int_{\mathbb{R}^+}re^{-ir^2}\ dr\ d\theta$$ Substituting $u=r^2\Rightarrow du=2r\ dr$ yields: $$I^2=\frac{1}{2}\int_0^{2\pi}\int_{\mathbb{R}^+}e^{-iu}\ du\ d\theta$$ $$=-\frac{i}{2}\int_0^{2\pi}d\theta$$ $$=-\pi i$$ So this means that we have: $$I=\sqrt{I^2}=\pm\sqrt{-\pi i}=\pm\sqrt{\pi}e^{\frac{3\pi i}{4}}=\pm\sqrt{\pi}(-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}})$$ $$=\mp\sqrt{\frac{\pi}{2}}\pm i\sqrt{\frac{\pi}{2}}$$ From this we can deduce, by comparing the real and imaginary parts of $I$ and our final expression,we yield: $$\int_{\mathbb{R}}\cos(x^2)\ dx=\mp\sqrt{\frac{\pi}{2}}\wedge\int_{\mathbb{R}}\sin(x^2)\ dx=\mp\sqrt{\frac{\pi}{2}}$$ So we have the absolute values of our Fresnel integrals (and these do check out numerically). However, how can I find the correct corresponding sign of each? Help would be greatly appreciated (and hopefully my evaluation contains no embarassingly obvious mistakes...).