Fresnel Integral multiplied with cosine term. $$I=\int_a^b \sin(\alpha-\beta x^2)\cos(x)\, dx.$$
Can anybody tell me, how to solve this integral ?
 I know that this is related to Fresnel Integral if the $\cos(x)$ term is absent. 

If $\beta>0$
$c+d=2\alpha-2\beta x^{2}$
$c-d=2x$
so $c=\alpha+x-\beta x^{2}=(\alpha+\frac{1}{4\beta})-(\frac{1}{2\sqrt{\beta}}-\sqrt{\beta}x)^{2}$ and $d=\alpha-x-\beta x^{2}=(\alpha+\frac{1}{4\beta})-(\frac{1}{2\sqrt{\beta}}+\sqrt{\beta}x)^{2}$
then using that $\sin(c)+\sin(d)=2\sin(\frac{1}{2}(c+d))\cos(\frac{1}{2}(c-d))$ we have:
$\int_{a}^{b}\sin(\alpha-\beta x^{2})\cos(x)dx$
$=\frac{1}{2}\int_{a}^{b}\sin((\alpha+\frac{1}{4\beta})-(\frac{1}{2\sqrt{\beta}}-\sqrt{\beta}x)^{2})dx+\frac{1}{2}\int_{a}^{b}\sin((\alpha+\frac{1}{4\beta})-(\frac{1}{2\sqrt{\beta}}-\sqrt{\beta} x)^{2})dx$
Notice that $\int_{s}^{t}\sin(c-u^{2})du=\int_{s}^{t}\sin(c)\cos(u^{2})du-\int_{s}^{t}\cos(c)\sin(u^{2})du$
$=\sin(c)(C(t)-C(s))-\cos(x)(S(t)-S(s))$
where $S(x)=\int_{0}^{x}\sin(p^{2})dp$ and $C(x)=\int_{0}^{x}\cos(p^{2})dp$
A similar procedure can be done for $\beta<0$ since $\sin(-x)=-\sin(x)$. If $\beta=0$ then this is just $\sin(\alpha)\int_{a}^{b}\cos(x)dx$.
