If   $I_1=\underset{0}{\overset{\pi /2}{\int }}f\left(\sin 2x\right)\sin xdx$  and  $I_2=\underset{0}{\overset{\pi /4}{\int }}f\left(\cos 2x\right)\cos xdx$   then  $\frac{I_1}{I_2}=$   
 

$$\begin{gathered} I_1=\underset{0}{\overset{\pi /2}{\int }}f\left(\sin 2(\frac{\pi }{2}-x)\right)\sin (\frac{\pi }{2}-x)dx \\ 2I_1=\underset{0}{\overset{\pi /2}{\int }}f\left(\sin 2x\right)(\sin x+\cos x)dx \\ \sqrt{2}\underset{0}{\overset{\pi /2}{\int }}f\left(\sin 2x\right)\cos (x-\frac{\pi }{4})dx \\ x-\frac{\pi }{4}=t =\sqrt{2}\underset{-\pi /4}{\overset{\pi /4}{\int }}f\left(\cos 2t\right)\cos tdt \\ 2I_1=2\sqrt{2}\underset{0}{\overset{\pi /4}{\int }}f\left(\cos 2t\right)\cos tdt \\ 2I_1=2\sqrt{2}{I}_2 \frac{I_1}{I_2}=\sqrt{2} \end{gathered}$$