If $$f: \mathbb{R} \to \mathbb{R}$$ be a continuous function satisfying $$\int_0^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$$, then the value of $$\alpha$$ is _______

We are given the equation $$\int_0^{\pi/2} f(\sin 2x) \sin x \, dx + \alpha \int_0^{\pi/4} f(\cos 2x) \cos x \, dx = 0$$ for all continuous functions $$f:\mathbb{R}\to\mathbb{R}$$, and we wish to determine the constant $$\alpha$$.

Next, we let $$I_1 = \int_0^{\pi/2} f(\sin 2x) \sin x \, dx.$$ By substituting $$x\mapsto \tfrac{\pi}{2}-x$$ (with $$dx=-du$$, and noting that when $$x=0$$, $$u=\tfrac{\pi}{2}$$, and when $$x=\tfrac{\pi}{2}$$, $$u=0$$), one finds

$$I_1 = \int_{\pi/2}^{0} f(\sin(\pi-2x))\cos x\,(-dx) = \int_0^{\pi/2} f(\sin 2x)\cos x \, dx.$$

Therefore, $$I_1$$ can also be written as $$\int_0^{\pi/2} f(\sin 2x)\cos x \, dx$$.

Next, we split this expression at $$\tfrac{\pi}{4}$$:

$$I_1 = \int_0^{\pi/4} f(\sin 2x)\cos x \, dx + \int_{\pi/4}^{\pi/2} f(\sin 2x)\cos x \, dx.$$

In the second integral, we use the substitution $$u=\tfrac{\pi}{2}-x$$ (so that when $$x=\tfrac{\pi}{4}$$, $$u=\tfrac{\pi}{4}$$, and when $$x=\tfrac{\pi}{2}$$, $$u=0$$, with $$dx=-du$$); this gives

$$\int_{\pi/4}^{\pi/2} f(\sin 2x)\cos x \, dx = \int_{\pi/4}^{0} f(\sin(\pi-2u))\sin u\,(-du) = \int_0^{\pi/4} f(\sin 2u)\sin u \, du.$$

Hence

$$I_1 = \int_0^{\pi/4} f(\sin 2x)\bigl(\cos x+\sin x\bigr)\,dx\quad\cdots(1).$$

Now let $$I_2 = \int_0^{\pi/4} f(\cos 2x)\cos x \, dx.$$ By setting $$x\mapsto \tfrac{\pi}{4}-x$$ (so that when $$x=0$$, $$u=\tfrac{\pi}{4}$$, and when $$x=\tfrac{\pi}{4}$$, $$u=0$$, with $$dx=-du$$), one obtains

$$I_2 = \int_{\pi/4}^{0} f\!\Bigl(\cos\bigl(\tfrac{\pi}{2}-2x\bigr)\Bigr) \cos\!\Bigl(\tfrac{\pi}{4}-x\Bigr)\,(-dx) = \int_0^{\pi/4} f(\sin 2x)\cos\!\Bigl(\tfrac{\pi}{4}-x\Bigr)\,dx.$$

Since $$\cos\bigl(\tfrac{\pi}{4}-x\bigr)=\frac{\cos x+\sin x}{\sqrt2}$$, it follows that

$$I_2 = \frac{1}{\sqrt2}\int_0^{\pi/4} f(\sin 2x)\bigl(\cos x+\sin x\bigr)\,dx = \frac{I_1}{\sqrt2}\quad\text{[by (1)].}$$

Substituting these into the original equation $$I_1+\alpha I_2=0$$ gives

$$I_1 + \alpha\cdot\frac{I_1}{\sqrt2}=0 \quad\Longrightarrow\quad I_1\Bigl(1+\tfrac{\alpha}{\sqrt2}\Bigr)=0.$$

Because this must hold for all continuous $$f$$, and $$I_1$$ can be nonzero, we conclude

$$1+\frac{\alpha}{\sqrt2}=0\quad\implies\quad\alpha=-\sqrt2.$$