请问∫x[(1-x^4)^3/2]dx如何变换成∫(cost)^4dt先换元ξ(x)=x^2,然后设x^2=sint,但是后面如何继续变换啊，百思不得其解∫x[(1-x^4)^3/2]dx=1/2∫(1-x^4)^3/2dx^2，设x^2=sint，1/2∫(1-x^4)^3/2dx^2=1/2∫(1-sin^2t)^3/2·1/co

Question

请问∫x[(1-x^4)^3/2]dx如何变换成∫(cost)^4dt
先换元ξ(x)=x^2,然后设x^2=sint,但是后面如何继续变换啊，百思不得其解
∫x[(1-x^4)^3/2]dx=1/2∫(1-x^4)^3/2dx^2，设x^2=sint，1/2∫(1-x^4)^3/2dx^2=1/2∫(1-sin^2t)^3/2·1/cost dsint这样有没有错误呢

Answer 1

let
x^2= sint
2xdx = cost dt
∫x[(1-x^4)^3/2]dx
=∫ 1/2(1-(sint)^2)^3/2 cost dt
=(1/2)∫ (cost)^4 dt