$I_{20}-I_{19}={\int }_0^{\frac{\pi }{2}}\frac{\sin \left(39x\right)-\sin \left(37x\right)}{\sin x}dx$
$\Rightarrow I_{20}-I_{19}={\int }_0^{\frac{\pi }{2}}\frac{2\sin x\cos \left(38x\right)}{\sin x}dx$
$\Rightarrow I_{20}-I_{19}={\int }_0^{\frac{\pi }{2}}2\cos \left(38x\right)dx$
$=\left[\frac{\sin \left(38x\right)}{19}\right]\begin{array}{c}\frac{\pi }{2}\\ 0\end{array}$
$=\frac{\sin \left(\frac{38\pi }{2}\right)}{19}$
$=0$
$\therefore I_{20}=I_{19}$

Hence, ${\left[I_{20}\right]}^3-{\left[I_{19}\right]}^3=0$