简单计算一下即可，答案如图所示
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第1个回答 2017-07-20
=∫(1-2sin²x-2sinxcosx)/(cosx+sinx)dx
=∫1/√2sin(x+π/4)-2sinxdx
=-1/√2∫1/(1-cos²(x+π/4))dcos(x+π/4)+2cosx
=-(1/2√2)ln(1+cos(x+π/4))/(1-cos(x+π/4))+2cosx+C本回答被网友采纳
=∫1/√2sin(x+π/4)-2sinxdx
=-1/√2∫1/(1-cos²(x+π/4))dcos(x+π/4)+2cosx
=-(1/2√2)ln(1+cos(x+π/4))/(1-cos(x+π/4))+2cosx+C本回答被网友采纳
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