用万能代替
∫1／（sinx+cosx）dx
=∫1／{2tan(x/2)/[1+tan^2(x/2)]+[1-tan^2(x/2)]/[1+tan^2(x/2)]}dx
=∫[1+tan^2(x/2)]／[2tan(x/2)+1-tan^2(x/2)]dx
=-∫1／[-2tan(x/2)-1+tan^2(x/2)]dtan(x/2)
=-∫1／{[tan(x/2)-1]^2-2}dtan(x/2)
=-1/(2√2)∫{1／[tan(x/2)-1-√2]-1／[tan(x/2)-1+√2]}dtan(x/2)
=-1/(2√2)ln[tan(x/2)-1-√2]+1/2ln[tan(x/2)-1+√2]+C