求不定积分∫(2-sinx）/（2+cosx )dx

∫(2-sinx）/（2+cosx )dx
=∫2dx/(2+cosx)-∫sinxdx/(2+cosx)
=∫2dx/[1+2cos²(x/2)]+∫d(2+cosx)/(2+cosx)
=4∫sec²(x/2)d(x/2)/[sec²(x/2)+2]+ln(2+cosx)
=4∫dtan(x/2)/[3+tan²(x/2)]+ln(2+cosx)
=(4/√3)*arctan[tan(x/2)/√3]+ln(2+cosx)+C