Question

∫x√(1-x^2)dx=？ 求详细过程！

Answer 1

∫x√(1-x^2)dx

let
x=siny
dx=cosydy
∫x√(1-x^2)dx

=∫siny(cosy)^2dy
=-(cosy)^3/3 + C
=-(1/3)(1-x^2)^(3/2) + C

Answer 2

凑微分法
∫x/√(1-x^2)dx
=-1/2∫d(1-x^2)/√(1-x^2)
=-1/2∫[(1-x^2)^(-1/2)]d(1-x^2)
=-1/2*2*(1-x^2)^(1/2)+C
=
-√(1-x^2)+C

追问

不是x除以，是乘以

追答

∫x√(1-x^2)dx=1/2∫2x√(1-x^2)dx=-1/2∫√(1-x^2)d(1-x^2)
=-1/4√(1-x^2)+C

追问

答案是：-1/3(1-x^2)^3/2+c

追答

∫x√(1-x^2)dx=1/2∫2x√(1-x^2)dx=-1/2∫√(1-x^2)d(1-x^2)=-1/3(1-x^2)^3/2+c