怎么求∫xsinx/1＋ex dx 在-π/2到 π/2的定积分.ex是e的x次幂.

题目

怎么求∫xsinx/1＋ex dx 在-π/2到 π/2的定积分.ex是e的x次幂.

答案

∫[xsinx/(1+e^x)]dx=∫[xsinx/(1+e^x)]dx+∫[xsinx/(1+e^x)]dx (分成两个积分)
=-∫[xsinx/(1+1/e^x)]dx+∫[xsinx/(1+e^x)]dx (第一个积分用-x代换x)
=∫[xsinx/(1+1/e^x)]dx+∫[xsinx/(1+e^x)]dx
=∫[1/(1+1/e^x)+1/(1+e^x)]xsinxdx
=∫[e^x/(e^x+1)+1/(1+e^x)]xsinxdx
=∫[(e^x+1)/(1+e^x)]xsinxdx
=∫xsinxdx
=∫xd(-cosx)
=(-xcosx)│+∫cosxdx (应用分部积分法)
=-(π/2)cos(π/2)+0*cos0+(sinx)│
=sin(π/2)-sin0
=1.