Notice the symmetry $g\left(\frac{\pi }{2}-x\right)=\sqrt{\frac{\pi }{2}\left(\frac{\pi }{2}-x\right)-{\left(\frac{\pi }{2}-x\right)}^2}=\sqrt{\frac{\pi x}{2}-x^2}=g(x).$

Also $f\left(\frac{\pi }{2}-x\right)={\sin }^2\left(\frac{\pi }{2}-x\right)={\cos }^{2}x.$

Let $I={\int }_0^{\pi /2}f(x)g(x)dx={\int }_0^{\pi /2}{\sin }^{2}x\cdot g(x)dx.$ Apply the king-property ${\int }_0^{a}F(x)dx={\int }_0^{a}F(a-x)dx$:

$I={\int }_0^{\pi /2}{\cos }^{2}x\cdot g(x)dx.$

Adding the two expressions for $I$: $2I={\int }_0^{\pi /2}g(x)dx,$ which is exactly the statement that $2\int fg-\int g=0$.

$\therefore$ The required value is $0$.