赞 共回答了17个问题采纳率：94.1% 举报
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
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
1年前 追问
2
令u = √t，t = u²，dt = 2u du 当t = 0、u = 0 当t = 1、u = 1 ∫(0→1) arccos√t dt = ∫(0→1) arccos(u) · 2u du = ∫(0→1) arccos(u) d(u²) = u²arccos(u) - ∫(0→1) u² d(arccos(u)) <--分部积分法 = [(1)arccos(1) - 0] - ∫(0→1) u² · - 1/√(1 - u²) du = ∫(0→1) u²/√(1 - u²) du --> 令u = sinz，du = cosz dz --> 当u = 0，z = 0 当u = 1，z = π/2 = ∫(0→π/2) sin²z/cosz · cosz dz = ∫(0→π/2) sin²z dz = (1/2)∫(0→π/2) (1 - cos2z) dz = (1/2)[z - (1/2)sin2z] = (1/2)[π/2 - (1/2)sin(π)] = (1/2)(π/2 - 0) = π/4
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