1/(1+cosx)=1/[1+2cos²(x/2)-1]
=sec²(x/2)/2
所以原式=∫e^xsec²(x/2)/2dx+∫e^xsinx*sec²(x/2)/2dx
=∫e^xsec²(x/2)d(x/2)+∫e^x2sin(x/2)cos(x/2)*sec²(x/2)/2dx
=∫e^xdtan(x/2)+∫e^x*sin(x/2)/cos(x/2)dx
=∫e^xdtan(x/2)+∫tan(x/2)de^x
=e^x*tan(x/2)-∫tan(x/2)de^x+∫tan(x/2)de^x
=e^x*tan(x/2)+C