If f: ℝ → ℝ be a continuous function satisfying ∫ _ 0 ^ (π)/(2) f(...

Given,

${∫}_{0}^{\dfrac{\pi}{2}}f\left(\sin 2x\right)sinxdx + \alpha{∫}_{0}^{\dfrac{\pi}{4}}f\left(\cos 2x\right)cosxdx = 0$

Now, let $I = {∫}_{0}^{\dfrac{\pi}{2}}f\left(\sin 2x\right) . sinxdx$

Now using property ${∫}_{a}^{b}f\left(x\right) d x = {∫}_{a}^{c}f\left(x\right) d x + {∫}_{c}^{b}f\left(x\right) d x$ we get,
$⇒ I = {∫}_{0}^{\dfrac{\pi}{4}}f\left(\sin 2x\right)sinxdx + {∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}f\left(\sin 2x\right) . sinxdx$

Now using property ${∫}_{a}^{b}f\left(x\right) d x = {∫}_{a}^{b}f\left(a + b - x\right) d x$ we get,

$⇒ I = {∫}_{0}^{\dfrac{\pi}{4}}f\left(\cos 2x\right)\sin \left(\dfrac{\pi}{4} - x\right)dx + {∫}_{0}^{\dfrac{\pi}{4}}f\left(\sin 2\left(\dfrac{\pi}{4} + x\right)\right)\sin \left(\dfrac{\pi}{4} + x\right)dx$

$⇒ I = {∫}_{0}^{\dfrac{\pi}{4}}f\left(\cos 2x\right)\left(\dfrac{1}{\sqrt{2}}\cos x - \dfrac{1}{\sqrt{2}}\sin x\right)dx + {∫}_{0}^{\dfrac{\pi}{4}}f\left(\cos 2x\right)\left(\dfrac{1}{\sqrt{2}}\cos x + \dfrac{1}{\sqrt{2}}\sin x\right)dx$
$⇒ I = {∫}_{0}^{\dfrac{\pi}{4}}f\left(\cos 2x\right)\left(\sqrt{2}\cos x\right)dx$

So, the putting the value of $I$ in ${∫}_{0}^{\dfrac{\pi}{2}}f\left(\sin 2x\right)sinxdx + \alpha{∫}_{0}^{\dfrac{\pi}{4}}f\left(\cos 2x\right)cosxdx = 0$ we get, $\alpha = - \sqrt{2}$