If f:ℝ→ℝ be a continuous function satisfying ∫0π/2f(sin2x)sinx dx+α∫0π/4f(cos2x)cosx dx=0, then the value of α is [2023]
(1)
I=∫0π/4f(sin2x)sinx dx+∫π/4π/2f(sin2x)sinx dx+α∫0π/4f(cos2x)cosx dx=0
Apply ∫0af(x) dx=∫0af(a-x) dx in the first part and put x-π4=t in the second part, we get
I=∫0π/4f(cos2x)sin(π4-x)dx+∫0π/4f(cos2t)sin(π4+t)dt+α∫0π/4f(cos2x)cosx dx=0
=∫0π/4f(cos2x)[2sinπ4·cosx+αcosx]dx=0
(α+2)∫0π/4f(cos2x)cosx dx=0 ∴ α=-2
[∵ In (0,π4), f(cos(2x)) and cosx are not zero.]