Definite Integral

∫_[-π/2~0] x sinx/(1+e^x) dx +∫_[0~π/2] x sinx/(1+e^x)dx
=∫_[π/2~0] -xsin(-x)/[1+e^(-x)] (-dx)+ ∫_[0~π/2] xsinx/(1+e^x) dx
=∫_[0~π/2] xsinx e^x/(e^x+1) dx +∫_[0~π/2] xsinx/(1+e^x)dx
=∫_[0~π/2] xsinx dx
= -xcosx+ sinx for x=0~π/2
= 1

媽弗勒時﹗媽弗勒時﹗

2012-05-24 08:10:17 補充：
媽弗勒時﹗媽弗勒時﹗