Hint: note that $$I=\int_{0}^{\infty}\cos\left(x^{2}\right)dx\stackrel{x^{2}=u}{=}\frac{1}{2}\int_{0}^{\infty}u^{-1/2}\cos\left(u\right)du $$ $$=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\cos\left(u\right)\int_{0}^{\infty}v^{-1/2}e^{-uv}dvdu $$ and now using the Fubini theorem we can exchange the integrals and get $$I=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}\cos\left(u\right)e^{-uv}dudv=\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}e^{u\left(i-v\right)}dudv\right) $$ $$\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{-1/2}}{v-i}dv\right)=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{1/2}}{v^{2}+1}dv\stackrel{z=\sqrt{v}}{=}\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{z^{2}}{z^{4}+1}dz $$ and the last integral is simple to evaluate using partial fractions.

Hint: note that $$I=\int_{0}^{\infty}\cos\left(x^{2}\right)dx\stackrel{x^{2}=u}{=}\frac{1}{2}\int_{0}^{\infty}u^{-1/2}\cos\left(u\right)du $$ $$=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\cos\left(u\right)\int_{0}^{\infty}v^{-1/2}e^{-uv}dvdu $$ and now using the Fubini theorem we can exchange the integrals and get $$I=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}\cos\left(u\right)e^{-uv}dudv=\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}e^{u\left(i-v\right)}dudv\right) $$ $$\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{-1/2}}{v-i}dv\right)=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{1/2}}{v^{2}+1}dv\stackrel{z=\sqrt{v}}{=}\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{z^{2}}{z^{4}+1}dz $$ and the last integral is simple to evaluate using partial fractions. Note that $$\int_{0}^{\infty}\frac{z^{2}}{z^{4}+1}dz=\frac{1}{2\sqrt{2}}\left(\int_{0}^{\infty}\frac{z}{z^{2}-\sqrt{2}z+1}-\frac{z}{z^{2}+\sqrt{2}z+1}dz\right) $$ $$=\frac{1}{2\sqrt{2}}\left(\lim_{a\rightarrow\infty}\frac{1}{2}\int_{0}^{a}\frac{2z-\sqrt{2}+\sqrt{2}}{z^{2}-\sqrt{2}z+1}-\frac{2z+\sqrt{2}-\sqrt{2}}{z^{2}+\sqrt{2}z+1}dz\right) $$ $$=\frac{1}{2\sqrt{2}}\left(\lim_{a\rightarrow\infty}\frac{1}{2}\int_{0}^{a}\frac{2z-\sqrt{2}}{z^{2}-\sqrt{2}z+1}-\frac{2z+\sqrt{2}-\sqrt{2}}{z^{2}+\sqrt{2}z+1}dz+\int_{0}^{a}\frac{\sqrt{2}}{z^{2}-\sqrt{2}z+1}dz\right)$$ and I think you can conclude by yourself from here.

Hint: note that $$I=\int_{0}^{\infty}\cos\left(x^{2}\right)dx\stackrel{x^{2}=u}{=}\frac{1}{2}\int_{0}^{\infty}u^{-1/2}\cos\left(u\right)du $$ $$=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\cos\left(u\right)\int_{0}^{\infty}v^{-1/2}e^{-uv}dvdu $$ and now using the Fubini theorem we can exchange the integrals and get $$I=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}\cos\left(u\right)e^{-uv}dudv=\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}e^{u\left(i-v\right)}dudv\right) $$ $$\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{-1/2}}{v-i}dv\right)=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{1/2}}{v^{2}+1}dv\stackrel{z=\sqrt{v}}{=}\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{z^{2}}{z^{4}+1}dv $$ and the last integral is simple to evaluate using partial fractions.

Hint: note that $$I=\int_{0}^{\infty}\cos\left(x^{2}\right)dx\stackrel{x^{2}=u}{=}\frac{1}{2}\int_{0}^{\infty}u^{-1/2}\cos\left(u\right)du $$ $$=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\cos\left(u\right)\int_{0}^{\infty}v^{-1/2}e^{-uv}dvdu $$ and now using the Fubini theorem we can exchange the integrals and get $$I=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}\cos\left(u\right)e^{-uv}dudv=\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}e^{u\left(i-v\right)}dudv\right) $$ $$\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{-1/2}}{v-i}dv\right)=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{1/2}}{v^{2}+1}dv\stackrel{z=\sqrt{v}}{=}\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{z^{2}}{z^{4}+1}dz $$ and the last integral is simple to evaluate using partial fractions.