Let $f\left(x\right)$  be an even function, $I_1={\displaystyle {\int }_0^{\frac{\pi }{2}}f\left(\cos 2x\right)\cos xdx,I_2=}{\displaystyle {\int }_0^{\frac{\pi }{4}}f\left(\sin 2x\right)\cos xdx}$ . Then $\frac{I_1}{I_2}=$

$I_1={\displaystyle {\int }_0^{\frac{\pi }{4}}f\left(\cos 2x\right)\cos xdx+}{\displaystyle {\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}f\left(\cos 2x\right)\cos xdx}$

Setting $x=\frac{\pi }{4}-t$  in the first integral and $x=\frac{\pi }{4}+t$  in the second integral,

$I_1={\displaystyle {\int }_0^{\frac{\pi }{4}}f\left(\sin 2t\right)\cos \left(\frac{\pi }{4}-t\right)dt+}{\displaystyle {\int }_0^{\frac{\pi }{4}}f\left(\sin 2t\right)\cos \left(\frac{\pi }{4}+t\right)dt}$

$={\displaystyle {\int }_0^{\frac{\pi }{4}}f\left(\sin 2t\right)\left[\cos \left(\frac{\pi }{4}-t\right)+\cos \left(\frac{\pi }{4}+t\right)\right]dt}$

$=\sqrt{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}f\left(\sin 2t\right)\cos tdt=\sqrt{2}{I}_2}$