Here's the general approach.
For the first integral, assume [tex]t>0[/tex]. I'm also interpreting it to say
[tex]I_1=\displaystyle\int_0^\infty\sin(tx^2)\,\mathrm dx[/tex]
(otherwise the integral simply diverges).
Writing [tex]I_1=I_1(t)[/tex], take the Laplace transform to get
[tex]\mathcal L_s\{I_1(t)\}=\displaystyle\int_0^\infty\left(\int_0^\infty\sin(tx^2)\,\mathrm dx\right)e^{-st}\,\mathrm dt[/tex]
[tex]\mathcal L_s\{I_1(t)\}=\displaystyle\int_0^\infty\left(\underbrace{\int_0^\infty \sin(tx^2)e^{-st}}_{\mathcal L_s\{\sin(tx^2)\}}\,\mathrm dt\right)\,\mathrm dx[/tex]
[tex]\mathcal L_s\{I_1(t)\}=\displaystyle\int_0^\infty\frac{x^2}{s^2+x^4}\,\mathrm dx[/tex]
which is an elementary integral that can be computed by decomposing into partial fractions, or by employing a proper trigonometric substitution. I leave that calculation to you; you should end up with
[tex]\mathcal L_s\{I_1(t)\}=\dfrac\pi{2\sqrt{2s}}[/tex]
and taking the inverse Laplace transform yields the answer,
[tex]I_1(t)=\mathcal L_t^{-1}\left\{\dfrac\pi{2\sqrt{2s}}\right\}=\dfrac{\sqrt\pi}{2\sqrt{2t}}[/tex]
Hopefully that should give you a decent idea of how this method works.
For the remaining integrals, you will need to introduce a parameter before proceeding. Here's what you can try:
[tex]I_2(t)=\displaystyle\int_0^\infty\frac{\sin(tx)}x\,\mathrm dx[/tex]
[tex]I_3(t)=\displaystyle\int_0^\infty\frac{1-\cos(tx)}x\,\mathrm dx[/tex]
[tex]I_4(t)=\displaystyle\int_0^\infty\frac{\sin^2(tx)}x\,\mathrm dx[/tex]
For the fifth integral, you can make the observation that the given integrand already resembles the Laplace transform of [tex]\dfrac{\sin t}t[/tex]:
[tex]\underbrace{\mathcal L_s\left\{\dfrac{\sin t}t\right\}}_{Y(s)}=\displaystyle\int_0^\infty e^{-st}\underbrace{\frac{\sin t}t}_{y(t)}\,\mathrm dt\implies \int_0^\infty e^{-4t}\frac{\sin t}t\,\mathrm dt=Y(4)[/tex]
Finally, for the last integral, you can use
[tex]I_6(t)=\displaystyle\int_0^\infty e^{-tx^2}\,\mathrm dx[/tex]
