Proof only by transformation that : $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $

<Here is what I found

Employing the change of variables $2u =x^2$ We get $$I=\int_0^\infty \cos(x^2) dx =\frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx$$ $$ J=\int_0^\infty \sin(x^2) dx=\frac{1}{\sqrt{2}}\int^\infty_0\frac{\sin(2x)}{\sqrt{x}}\,dx $$

Summary: We will prove that $J\ge 0$ and $I\ge 0$ so that, proving that $I=J$ is equivalent to $$ \color{blue}{0= (I+J)(I-J)=I^2 -J^2 =\lim_{t \to 0}I_t^2-J^2_t}$$ Where, $$I_t = \int_0^\infty e^{-tx^2}\cos(x^2) dx~~~~\text{and}~~~ J_t = \int_0^\infty e^{-tx^2}\sin(x^2) dx$$ $t\mapsto I_t$ and $t\mapsto J_t$ are clearly continuous due to the present of the integrand factor $e^{-tx^2}$.

However, By Fubini we have,

\begin{split} I_t^2-J^2_t&=& \left(\int_0^\infty e^{-tx^2}\cos(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\cos(y^2) dy\right) \\&-& \left(\int_0^\infty e^{-tx^2}\sin(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\sin(y^2) dy\right) \\ &=& \int_0^\infty \int_0^\infty e^{-t(x^2+y^2)}\cos(x^2+y^2)dxdy\\ &=&\int_0^{\frac\pi2}\int_0^\infty re^{-tr^2}\cos r^2 drd\theta\\&=&\frac\pi4 Re\left( \int_0^\infty \left[\frac{1}{i-t}e^{(i-t)r^2}\right]' dr\right)\\ &=&\color{blue}{\frac\pi4\frac{t}{1+t^2}\to 0~~as ~~~t\to 0} \end{split}

To end the proof: Let us show that $I> 0$ and $J> 0$. Performing an integration by part we obtain $$J = \frac{1}{\sqrt{2}} \int^\infty_0\frac{\sin(2x)}{x^{1/2}}\,dx=\frac{1}{\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx\color{red}{>0}$$ Given that $\color{red}{\sin 2x= 2\sin x\cos x =(\sin^2x)'}$. Similarly we have, $$I = \frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx=\frac{1}{2\sqrt{2}}\underbrace{\left[\frac{\sin 2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{4\sqrt{2}} \int^\infty_0\frac{\sin 2 x}{x^{3/2}}\,dx\\= \frac{1}{4\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{3}{8\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx\color{red}{>0}$$

Conclusion: $~~~I^2-J^2 =0$, $I>0$ and $J>0$ impliy $I=J$. Note that we did not attempt to compute neither the value of $~~I$ nor $J$.

Extra-to-the answer However using similar technique in above prove one can easily arrives at the following $$\color{blue}{I_tJ_t = \frac\pi8\frac{1}{t^2+1}}$$ from which one get the following explicit value of $$\color{red}{I^2=J^2= IJ = \lim_{t\to 0}I_tJ_t =\frac{\pi}{8}}$$

See also here for more on (The Fresnel Integrals Revisited)

Note by change of variable it suffices to show

$$\int^\infty_0\frac{\cos(x)}{\sqrt{x}}\,dx =\int^\infty_0\frac{\sin(x)}{\sqrt{x}}\,dx $$

Consider the following function

$$f(z)=z^{-1/2}\,e^{iz}$$

Where we choose the principal root for $ z^{-1/2}=e^{-1/2\log(z)}$. By integrating around the following contour

$$\int_{C_r}f(z)\,dz+\int_{r}^R f(x)\,dx+\int_{\gamma}f(z)\,dz+\int^{iR}_{ir}f(x)\,dx = 0$$

Taking the integral around the small quarter circle with $r\to 0$ $$\left| \int_{C_r}f(z)\,dz\right|\leq \left|\sqrt{r}\int^{\pi/2}_{0}e^{it/2} e^{rie^{it}}\,dt\right| \leq \sqrt{r}\int^{\pi/2}_{0}\left|e^{-r\sin(t)}\right|\,dt\sim 0$$

On $\gamma(t)=(1-t)R+iRt$ where $0\leq t \leq 1$

$$\left|\int_{\gamma}f(z)\,dz\right| = \left| R(i-1)\int^1_0e^{-1/2\log(R(1-t)+iRt)}e^{i(1-t)R-Rt}\,dt\right| \\ \leq \frac{\sqrt{2}}{\sqrt{R}} \int^1_0 \frac{e^{-Rt}}{\sqrt[4]{(1-t)^2+t^2}}\,dt$$

Hence we have

$$\left|\int_{\gamma}f(z)\,dz\right| \leq \frac{\sqrt{2}}{\sqrt{R}} \int^1_0 e^{-Rt}\,dt=\frac{\sqrt{2}}{R\sqrt{R}}\left(1-e^{-R}\right)\sim_{\infty}0$$

Finally what is remaining when $r\to 0$ and $R \to \infty$

$$\int^\infty_0 \frac{e^{ix}}{\sqrt{x}}\,dx =i \int^{\infty}_{0}(ix)^{-1/2}e^{-x}\,dx$$

Note that $i^{-1/2}=e^{-i\pi/4}$

$$\int^\infty_0\frac{e^{ix}}{\sqrt{x}}\,dx = ie^{-i\pi/4}I = \frac{I}{\sqrt{2}}+i\frac{I}{\sqrt{2}}$$

By equating the real part with the real part and the imaginary part with the imaginary part we reach $$\int^\infty_0\frac{\cos(x)}{\sqrt{x}}\,dx =\int^\infty_0\frac{\sin(x)}{\sqrt{x}}\,dx = \frac{I}{\sqrt{2}} $$

Although $I$ is easy to evaluate using the gamma function, we didn't have to evaluate it to show equivalence.

Since $e^{iz^2}$ is entire, by Cauchy's Integral Theorem, we have $$ \int_0^R e^{iz^2}\,\mathrm{d}z =\int_0^{(1+i)R} e^{iz^2}\,\mathrm{d}z+\int_{(1+i)R}^R e^{iz^2}\,\mathrm{d}z\tag1 $$ where, using the parameterization $z=R(1+it)$, we have the estimate $$ \begin{align} \left|\,\int_{(1+i)R}^R e^{iz^2}\,\mathrm{d}z\,\right| &\le R\int_0^1e^{-2R^2t}\,\mathrm{d}t\\ &\le\frac1{2R}\tag2 \end{align} $$ and using the reparameterization $z\mapsto(1+i)z$, $$ \begin{align} \int_0^{(1+i)R}e^{iz^2}\,\mathrm{d}z &=(1+i)\int_0^Re^{-2z^2}\,\mathrm{d}z\tag3 \end{align} $$ Combining $(1)$, $(2)$, and $(3)$, while letting $R\to\infty$, validates the following change of variables: $\boldsymbol{\color{#C00}{z\mapsto(1+i)z}}$. $$ \begin{align} \int_0^\infty\left(\cos\left(z^2\right)+i\sin\left(z^2\right)\right)\mathrm{d}z &=\boldsymbol{\color{#C00}{\int_0^\infty e^{iz^2}\,\mathrm{d}z}}\\ &\boldsymbol{\color{#C00}{=(1+i)\int_0^\infty e^{-2z^2}\,\mathrm{d}z}}\tag4 \end{align} $$ Since the real and imaginary parts of $(4)$ are the same, we get that $$ \int_0^\infty\cos\left(z^2\right)\,\mathrm{d}z =\int_0^\infty\sin\left(z^2\right)\,\mathrm{d}z\tag5 $$