Question

计算定积分∫x^2/√(1-x^2)上限1/2,下限0 拜托啊

Answer 1

令x=sint
∫x^2/√(1-x^2)dx
=∫sin²t/cost*costdt(上限π/6,下限0，下同)
=∫sin²tdt
=1/2∫(1-cos2t)dt
=1/2*t-sin2t/4

=1/2*(π/6-0)-(sinπ/3-sin0)/4
=π/12-根号3/8
=(2π-3根号3)/24

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Answer 2

∫(0->1/2) x^2/√(1-x^2) dx

let
x= sina
dx= cosa da
x=0, a=0
x=1/2, a=π/6
∫(0->1/2) x^2/√(1-x^2) dx
=∫(0->π/6) (sina)^2 da
= (1/2) ∫(0->π/6)( 1-cos2a )da
=(1/2)[ a - sin(2a)/2 ] (0->π/6)
=(1/2)( π/6 - √3/4)