计算定积分∫x^2/√(1-x^2)上限1/2,下限0
问题描述：
计算定积分∫x^2/√(1-x^2)上限1/2,下限0
答
∫(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)
答
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