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计算定积分∫x^2/√(1-x^2)上限1/2,下限0
人气:375 ℃ 时间:2020-04-01 02:41:43
解答
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
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