Prove that limt→∞∫1tsin⁡(x)sin⁡(x2)dx converges

You can write
$\sin (x)\sin (x^2)=\frac{1}{2}[\cos (x^2-x)-\cos (x^2+x)]$and let $u=x^2-x,v=x^2+x$ to obtain
$$\begin{gathered} {\int }_1^t\sin (x)\sin (x^2)dx=\frac{1}{2}{\int }_1^t[\cos (x^2-x)-\cos (x^2+x)] \\ =\frac{1}{4}[{\int }_0^{t^2-t}\frac{\cos (u)}{\sqrt{u+\frac{1}{4}}}du-{\int }_2^{t^2+t}\frac{\cos (v)}{\sqrt{v+\frac{1}{4}}}dv] \end{gathered}$$
The convergence of this expression as $t\to \infty$ is ensured by Dirichlet's test for integrals or integration by parts.