计算定积分∫(上限1下限-1) (2x^2+x^9*cosx)/(1+√(1-x^2))dx

1个回答

  • 因为这一项,x^9*cosx/(1+√(1-x^2)是个奇函数,所以在-1,1上的积分为0

    所以元积分=4∫(0到1) x^2/(1+√(1-x^2))dx

    令x=sint

    元积分=4∫(0到π/2) (sint)^2 cost/(1+cost)dt

    =4∫(0到π/2) [1-(cost)^2] cost/(1+cost)dt

    =4∫(0到π/2) [1-(cost)] costdt

    =4∫(0到π/2) [cost-(cost)^2] dt

    =4sint|(0到π/2)-2∫(0到π/2) (cos2t+1) dt

    =4-(sin2t+2t)|(0到π/2)

    =4-π