计算定积分∫(上限1下限-1) (2x^2+x^3*cosx)/(1+√(1-x^2))dx

∵x^3*cosx/(1+√(1-x^2))是奇函数
∴∫[x^3*cosx/(1+√(1-x^2))]dx=0
故∴∫[(2x^2+x^3*cosx)/(1+√(1-x^2))]dx
=2∫[x^2/(1+√(1-x^2))]dx+∫[x^3*cosx/(1+√(1-x^2))]dx
=2∫[x^2/(1+√(1-x^2))]dx
=2∫[x^2(1-√(1-x^2))/(1-(1-x^2))]dx (有理化分母)
=2∫[1-√(1-x^2)]dx
=2∫dx-2∫√(1-x^2)dx
=4-2∫√(1-x^2)dx
=4-4∫√(1-x^2)dx
=4-4∫(cost)^2dt (令x=sint)
=4-2∫[1+cos(2t)]dt (应用倍角公式)
=4-2(π/2)
=4-π.