If $f:\mathrm{ℝ}\to \mathrm{ℝ}$ be a continuous function satisfying ${\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(\sin 2x\right)\sin x dx+\alpha {\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(\cos 2x\right)\cos x dx=0$, then the value of $\alpha$ is

Given,

${\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(\sin 2x\right)\sin x dx+\alpha {\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(\cos 2x\right)\cos x dx=0$

Now, let $I={\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(\sin 2x\right).\sin x dx$

Now using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx+{\int }_c^{b}f\left(x\right)dx$ we get,
$\Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(\sin 2x\right) \sin x dx+{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}f\left(\sin 2x\right).\sin x dx$

Now using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ we get,

$\Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(\cos 2x\right) \sin \left(\frac{\mathrm{\pi }}{4}-x\right)dx+{\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(\sin 2\left(\frac{\mathrm{\pi }}{4}+x\right)\right) \sin \left(\frac{\mathrm{\pi }}{4}+x\right)dx$

$\Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(\cos 2x\right) \left(\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}} \sin x\right)dx+{\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(\cos 2x\right) \left(\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}} \sin x\right)dx$
$\Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(\cos 2x\right)\left(\sqrt{2} \cos x\right)dx$

So, the putting the value of $I$ in ${\int }_0^{\frac{\mathrm{\pi }}{2}}f\left(\sin 2x\right)\sin x dx+\alpha {\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(\cos 2x\right)\cos x dx=0$ we get, $\alpha =-\sqrt{2}$