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(D) Let $I = \int \limits_0^{\pi / 2} f(\sin 2x) \sin x \, dx + \alpha \int \limits_0^{\pi / 4} f(\cos 2x) \cos x \, dx = 0$. Split the first integral

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$-\sqrt{2}$

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(D) Let $I = \int \limits_0^{\pi / 2} f(\sin 2x) \sin x \, dx + \alpha \int \limits_0^{\pi / 4} f(\cos 2x) \cos x \, dx = 0$. Split the first integral at $\frac{\pi}{4}$: $I = \int \limits_0^{\pi / 4} f(\sin 2x) \sin x \, dx + \int \limits_{\pi / 4}^{\pi / 2} f(\sin 2x) \sin x \, dx + \alpha \int \limits_0^{\pi / 4} f(\cos 2x) \cos x \, dx = 0$. In the first integral,use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$: $\int_0^{\pi / 4} f(\sin 2x) \sin x \, dx = \int_0^{\pi / 4} f(\sin 2(\frac{\pi}{4}-x)) \sin(\frac{\pi}{4}-x) \, dx = \int_0^{\pi / 4} f(\cos 2x) \sin(\frac{\pi}{4}-x) \, dx$. In the second integral,substitute $x = \frac{\pi}{2} - t$,so $dx = -dt$: $\int_{\pi / 4}^{\pi / 2} f(\sin 2x) \sin x \, dx = \int_{\pi / 4}^0 f(\sin 2(\frac{\pi}{2}-t)) \sin(\frac{\pi}{2}-t) (-dt) = \int_0^{\pi / 4} f(\sin(\pi-2t)) \cos t \, dt = \int_0^{\pi / 4} f(\sin 2t) \cos t \, dt$. Wait,using $x = \frac{\pi}{4} + t$ for the second integral: $\int_{\pi / 4}^{\pi / 2} f(\sin 2x) \sin x \, dx = \int_0^{\pi / 4} f(\sin 2(\frac{\pi}{4}+t)) \sin(\frac{\pi}{4}+t) \, dt = \int_0^{\pi / 4} f(\cos 2t) \sin(\frac{\pi}{4}+t) \, dt$. Combining these: $\int_0^{\pi / 4} f(\cos 2x) [\sin(\frac{\pi}{4}-x) + \sin(\frac{\pi}{4}+x) + \alpha \cos x] \, dx = 0$. Using $\sin(A-B) + \sin(A+B) = 2 \sin A \cos B$: $\sin(\frac{\pi}{4}-x) + \sin(\frac{\pi}{4}+x) = 2 \sin(\frac{\pi}{4}) \cos x = 2(\frac{1}{\sqrt{2}}) \cos x = \sqrt{2} \cos x$. So,$\int_0^{\pi / 4} f(\cos 2x) [\sqrt{2} \cos x + \alpha \cos x] \, dx = 0$. $(\sqrt{2} + \alpha) \int_0^{\pi / 4} f(\cos 2x) \cos x \, dx = 0$. Thus,$\alpha = -\sqrt{2}$.

If $f : R \rightarrow R$ is a continuous function satisfying $\int \limits_0^{\pi / 2} f(\sin 2x) \cdot \sin x \, dx + \alpha \int \limits_0^{\pi / 4} f(\cos 2x) \cdot \cos x \, dx = 0$,then $\alpha$ is equal to

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