Question

高数积分求详解

Answer 1

原式=∫(-1,1) 2x^2/[1+√(1-x^2)]dx+∫(-1,1) x/[1+√(1-x^2)]dx
因为2x^2/[1+√(1-x^2)]是偶函数，x/[1+√(1-x^2)]是奇函数
所以原式=2*∫(0,1) 2x^2/[1+√(1-x^2)]dx+0
=4*∫(0,1) x^2/[1+√(1-x^2)]dx
令x=sint，则dx=costdt
原式=4*∫(0,π/2) [sin^2t/(1+cost)]*costdt
=4*∫(0,π/2) [(1-cos^2t)/(1+cost)]*costdt
=4*∫(0,π/2) (1-cost)*costdt
=∫(0,π/2) (4cost-4cos^2t)dt
=∫(0,π/2) (4cost-2-2cos2t)dt
=(4sint-2t-sin2t)|(0,π/2)
=4-π

Answer 2

Answer 3

Answer 4

=4-π

方法如下图所示，请认真查看，祝学习愉快：