∫(0,1)arccos根号tdt答案是∫(0,1)t/√(1-t)dt=4/3

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∫(0,1)arccos根号tdt答案是∫(0,1)t/√(1-t)dt=4/3
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∫(0,1)arccos根号tdt答案是∫(0,1)t/√(1-t)dt=4/3
∫(0,1)arccos根号tdt
答案是∫(0,1)t/√(1-t)dt=4/3

∫(0,1)arccos根号tdt答案是∫(0,1)t/√(1-t)dt=4/3
令x = arccos√t.√t = cosx、t = cos²x、dt = 2cosx(- sinx) dx = - sin2x dx
当t = 0、x = π/2
当t = 1、x = 0
∫(0→1) arccos√t dt
= ∫(π/2→0) x · (- sin2x) dx
= ∫(0→π/2) xsin2x dx
= (- 1/2)∫(0→π/2) x d(cos2x)
= (- 1/2)xcos2x + (1/2)∫(0→π/2) cos2x dx
= (- 1/2)[(π/2)cos(π)] - 0] + (1/4)sin2x
= (- 1/2)(- π/2) + (1/4)[sin(π) - sin(0)]
= π/4

怎样由cosA+cosB=根号2得到有0≤A≤arccos[根号2-1],-65.53≤B≤65.53?