第2个回答 2013-07-09
u=tan(x/2), -u=tan(-x/2)
sinx=2sin(x/2)cos(x/2)=2tan(x/2)(cos(x/2))^2
=2tan(x/2)/[1+(tan(x/2))^2]=2u/(1+u^2)
cosx=2(cos(x/2))^2-1=2/(1+u^2) -1=(1-u^2)/(1+u^2)
du/dx=1/[2(cos(x/2))^2]=(1+u^2)/2
dx=2du/(1+u^2)
∫R(sinx,cosx)dx=∫R(2u/(1+u^2), (1-u^2)/(1+u^2)) 2du/(1+u^2)
=∫R(2(-u)/(1+u^2), (1-u^2)/(1+u^2)) 2d(-u)/(1+u^2)
=∫R(-sinx,-cosx)dx本回答被提问者采纳