Given $f\left(x\right)=x^2, g\left(x\right)=\cos x$

i.e. $g\left(f\left(x\right)\right)=\cos x^2$

For $18x^2-9\pi x+{\pi }^2=0$

roots are $\frac{9\mathrm{\pi }\pm \sqrt{81{\mathrm{\pi }}^2-72{\mathrm{\pi }}^2}}{36}=\frac{{\displaystyle 9\mathrm{\pi }\pm 3\mathrm{\pi }}}{{\displaystyle 36}}=\frac{\mathrm{\pi }}{3}, \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 6}}$

i.e. $\alpha =\frac{\mathrm{\pi }}{6}, \beta =\frac{\mathrm{\pi }}{3}$

So, ${\int }_{\alpha }^{\beta }x\left(g\circ f\left(x\right)\right)dx={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}x\cos x^{2}dx$

Let $x^2=t; xdx=\frac{dt}{2}$

$\frac{1}{2}{\int }_{\frac{{\mathrm{\pi }}^2}{36}}^{\frac{{\mathrm{\pi }}^2}{9}}\cos tdt=\frac{1}{2}\left(\sin \frac{{\mathrm{\pi }}^2}{9}-\sin \frac{{\mathrm{\pi }}^2}{36}\right)$