求积分 ∫ x arccos xdx 求救啊
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求积分 ∫ x arccos xdx 求救啊
∫x•arccosxdx
= ∫arccosxd(x²/2)
= (1/2)x²•arccosx - (1/2)∫x²d(arccosx)
= (1/2)x²•arccos(x) - (1/2)∫ x²•[-1/√(1 - x²)]dx
= (1/2)x²•arccos(x) + (1/2)∫ x²/√(1 - x²) dx,
令:x = sinθ θ∈[-π/2,π/2],
得:dx = cosθdθ,√(1 - x²)=cosθ.
原积分= (1/2)x²•arccos(x) + (1/2)∫ (sin²θ/cosθ)•cosθdθ
= (1/2)x²•arccos(x) + (1/2)∫ sin²θdθ
= (1/2)x²•arccos(x) + (1/2)∫ (1 - cos2θ)/2dθ
= (1/2)x²•arccos(x) + (1/4)(θ- 1/2•sin2θ) + C
= (1/2)x²•arccos(x) + (1/4)arcsin(x) - (x/4)√(1 - x²)+C
= ∫arccosxd(x²/2)
= (1/2)x²•arccosx - (1/2)∫x²d(arccosx)
= (1/2)x²•arccos(x) - (1/2)∫ x²•[-1/√(1 - x²)]dx
= (1/2)x²•arccos(x) + (1/2)∫ x²/√(1 - x²) dx,
令:x = sinθ θ∈[-π/2,π/2],
得:dx = cosθdθ,√(1 - x²)=cosθ.
原积分= (1/2)x²•arccos(x) + (1/2)∫ (sin²θ/cosθ)•cosθdθ
= (1/2)x²•arccos(x) + (1/2)∫ sin²θdθ
= (1/2)x²•arccos(x) + (1/2)∫ (1 - cos2θ)/2dθ
= (1/2)x²•arccos(x) + (1/4)(θ- 1/2•sin2θ) + C
= (1/2)x²•arccos(x) + (1/4)arcsin(x) - (x/4)√(1 - x²)+C
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