Question

导数的问题

什么的导数是x^2cosx

Answer 1

你好 ，这要两次用到分部积分法：
解：∫x^2cosxdx
=∫x^2dsinx
=x^2sinx-∫sinxdx^2
=x^2sinx-2∫xsinxdx
=x^2sinx+2∫xdcosx
=x^2sinx+2xcosx-2∫cosxdx
=x^2sinx+2xcosx-2sinx+C

Answer 2

∫x^2cosxdx
=∫x^2dsinx
=x^2sinx-∫sinxdx^2
=x^2sinx-2∫xsinxdx
=x^2sinx+2∫xdcosx
=x^2sinx+2xcosx-2∫cosxdx
=x^2sinx+2xcosx-2sinx+C

Answer 3

F(x) = x^2sinx + 2xcosx - 2sinx
F'(x) = x^2cosx + 2xsinx + (-2x)sinx + 2cosx -2cosx
= x^2cosx