∫1/（sinx+cosx）dx=∫dx/√2sin(x+π/4)=-(√2/2)∫dcos(x+π/4)/sin^2(x+π/4)=-(√2/4）{∫dcos(x+π/4)/[1-cos(x+π/4)]+∫dcos(x+π/4)/[1+cos(x+π/4)]}=-(√2/4)ln{[1+cos(x+π/4)]/[1-cos(x+π/4)]}+c=(√2/4)ln{[1-cos(x+π/4)]/[1+cos(x+π/4)]}+c
