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]...