题目

∫(x上限1,下限0)x^2/√(2-x^2)dx

提问时间：2021-03-01

答案

令 x=√2sint,则
I =∫(x上限1,下限0) x^2/√(2-x^2)dx
=∫(t上限π/4,下限0) 4(sint)^2(cost)^2dt
=∫(t上限π/4,下限0) (sin2t)^2dt
= (1/2)∫(t上限π/4,下限0) (1-cos4t)dt
= (1/2)[t-sin4t/4](t上限π/4,下限0) = π/8.