∫[0,1]√(1-X^2)arcsinxdx如何用定积分的分部积分法求，感谢~
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∫(0→1) √(1 - x²)•arcsinx dx
(x = sinz，dx = cosz dz)
∫(0→π/2) (z•cosz)•(cosz dz)
= ∫(0→π/2) z•cos²z dz
= (1/2)∫(0→π/2) (z + z•cos2z) dz
= (1/2)∫(0→π/2) z dz + (1/2)∫(0→π/2) z•coz2z dz
= (1/2)(z²/2)|(0→π/2) + (1/2)(1/2)∫(0→π/2) z d(sin2z)
= (1/4)(π²/4) + (1/4)z•sin2z|(0→π/2) - (1/4)∫(0→π/2) sin2z dz，分部积分法
= (π²/16) - (1/4)(-1/2)cos2z|(0→π/2)
= π²/16 + (1/8)(- 1 - 1)
= π²/16 - 1/4
(x = sinz，dx = cosz dz)
∫(0→π/2) (z•cosz)•(cosz dz)
= ∫(0→π/2) z•cos²z dz
= (1/2)∫(0→π/2) (z + z•cos2z) dz
= (1/2)∫(0→π/2) z dz + (1/2)∫(0→π/2) z•coz2z dz
= (1/2)(z²/2)|(0→π/2) + (1/2)(1/2)∫(0→π/2) z d(sin2z)
= (1/4)(π²/4) + (1/4)z•sin2z|(0→π/2) - (1/4)∫(0→π/2) sin2z dz，分部积分法
= (π²/16) - (1/4)(-1/2)cos2z|(0→π/2)
= π²/16 + (1/8)(- 1 - 1)
= π²/16 - 1/4
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