求第2小题定积分
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∫(-1/2->1/2) (arcsinx)^2/√(1-x^2) dx
=2∫(0->1/2) (arcsinx)^2/√(1-x^2) dx
=(2/3)∫(0->1/2) d(arcsinx)^3
=(2/3)[(arcsinx)^3]|(0->1/2)
=(2/3)(π/6)^3
=(1/324)π^3
(2)
∫(π/6->π/2) (cosu)^2 du
=(1/2)∫(π/6->π/2) (1+cos2u) du
=(1/2) [u+(1/2)sin2u] |(π/6->π/2)
=(1/2) [π/3 -(1/2)(√3/2) ]
=π/6 - √3/8
=2∫(0->1/2) (arcsinx)^2/√(1-x^2) dx
=(2/3)∫(0->1/2) d(arcsinx)^3
=(2/3)[(arcsinx)^3]|(0->1/2)
=(2/3)(π/6)^3
=(1/324)π^3
(2)
∫(π/6->π/2) (cosu)^2 du
=(1/2)∫(π/6->π/2) (1+cos2u) du
=(1/2) [u+(1/2)sin2u] |(π/6->π/2)
=(1/2) [π/3 -(1/2)(√3/2) ]
=π/6 - √3/8
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