If $\mathrm{f}:\mathrm{ℝ}\to \mathrm{ℝ}$ be a continuous function satisfying $\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{f}\left(\sin 2\mathrm{x}\right)\mathrm{sinxdx}+\mathrm{\alpha }\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\mathrm{f}\left(\cos 2\mathrm{x}\right)\mathrm{cosx}\mathrm{dx}=0$ then the value of $\mathrm{\alpha }$ is

Let ${\mathrm{I}}_1+{\mathrm{\alpha I}}_2=0$, where ${\mathrm{I}}_1=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\mathrm{f}\left(\sin 2\mathrm{x}\right)\mathrm{sinxdx}=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\mathrm{f}\left(\sin 2\mathrm{x}\right)\mathrm{cosxdx}$
${\mathrm{I}}_2=\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\mathrm{f}\left(\cos 2\mathrm{x}\right)\mathrm{cosxdx}=\frac{1}{\sqrt{2}}\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\mathrm{f}\left(\sin 2\mathrm{x}\right)(\mathrm{cosx}+\mathrm{sinx})\mathrm{dx}$
$2{\mathrm{I}}_1=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\mathrm{f}\left(\sin 2\mathrm{x}\right)(\mathrm{sinx}+\mathrm{cosx})\mathrm{dx}=2\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\mathrm{f}\left(\sin 2\mathrm{x}\right)(\mathrm{sinx}+\mathrm{cosx})\mathrm{dx}$
$=2\sqrt{2}{\mathrm{I}}_2 \therefore {\mathrm{I}}_1-\sqrt{2}{\mathrm{I}}_2=0 \Rightarrow \mathrm{\alpha }=-\sqrt{2}$