∫dx/(sinx+cosx)
=∫dx/[√2sin(x+π/4)]
=(-1/√2)∫dcos(x+π/4)/[(1-cos(x+π/4))(1+cos(x+π/4))]
=(-1/√2)ln|1+cos(x+π/4)|/|sin(x+π/4| +C可以寫得詳細(xì)點嗎？看不懂啊謝謝！sinx+cosx=√2*[(√2/2)sinx+(√2/2)cosx]=√2sin(x+π/4)∫dx/sinx=-∫sinxdx/sinx^2=-∫dcosx/[(1-cosx)(1+cosx)]=(-1/2)∫dcosx/(1+cosx)+(-1/2)∫dcosx/(1-cosx)=(-1/2)ln(1+cosx)-(-1/2)ln(1-cosx) =(-1/2)ln|1+cosx|/|1-cosx| =(-1/2)ln(1+cosx)^2/sinx^2 =(-1)ln|1+cosx|/|sinx|+C