1个回答
展开全部
∫(sinxcosx)/(sinx+cosx)dx
=∫(2sinxcosx)/[2√2sin(x+π/4)]dx
=∫sin2x/[2√2sin(x+π/4)]dx
=∫-cos(2x+π/2)/[2√2sin(x+π/4)]dx
=∫[2sin^2(x+π/4)-1]/[2√2sin(x+π/4)]dx
=∫{√2/2*sin(x+π/4)+√2/4[-1/sin(x+π/4)]}dx
=∫√2/2*sin(x+π/4)dx+∫√2/4[-1/sin(x+π/4)]dx
=-√2/2*cos(x+π/4)+√2/4*cos(x+π/4)/sin(x+π/4)+C
=∫(2sinxcosx)/[2√2sin(x+π/4)]dx
=∫sin2x/[2√2sin(x+π/4)]dx
=∫-cos(2x+π/2)/[2√2sin(x+π/4)]dx
=∫[2sin^2(x+π/4)-1]/[2√2sin(x+π/4)]dx
=∫{√2/2*sin(x+π/4)+√2/4[-1/sin(x+π/4)]}dx
=∫√2/2*sin(x+π/4)dx+∫√2/4[-1/sin(x+π/4)]dx
=-√2/2*cos(x+π/4)+√2/4*cos(x+π/4)/sin(x+π/4)+C
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询