※ 引述《PttFund (批踢踢基金只進不出)》之銘言：
: If g(u) is continuous on [0,1], then show that
: π/2 π/2
: ∫ g(sin(2u)) cosu du = ∫ g((cosv)^2) cosv dv.
: 0 0
π/2
∫ g((cosv)^2) cosv dv (1)
0
1
令 (cosv)^2 = x => -2cos(v)sin(v)dv = dx => (1) = ∫ [g(x)/(2 √(1-x))] dx
0
π/2 π/4 π/2
∫ g(sin(2u)) cos u du = ∫ g(sin(2u)) cos u du + ∫ g(sin(2u)) cos u du (2)
0 0 π/4
令 sin(2u) = x => 2cos(2u)du = dx
1
(2) = ∫ [(g(x) √(1 + √( 1 - x^2))) / (2 √(2(1 - x^2)))] dx
0
1
+ ∫ [(g(x) √(1 - √(1 - x^2))) / (2 √(2(1 - x^2)))] dx
0
1
故 (2) = ∫ g(x)(A(x)/B(x)) dx, 其中
0
A(x) = √(1 + √(1 - x^2)) + √(1 - √(1 - x^2))
B(x) = 2 √(2(1 - x^2))
令 A(x)/B(x) = t => A^2 = (B^2)t^2
2 + 2x = 8(1-x^2)t^2 => t^2 = 1/4(1-x) => t = ±1/(2 √(1-x)) (負不合)
1
故 (2) = ∫ [g(x)/(2 √(1-x))] dx = (1) ... Q.E.D.
0
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