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2020-11-15 · 知道合伙人教育行家
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∫ sinx/(1+sinx) dx
=∫ [1 - 1/(1+sinx)] dx
=x - ∫ dx/(1+sinx)
=x - ∫ dx/[1+2sin(x/2)cos(x/2) ]
分子分母同时除 ( cos(x/2))^2
=x -∫ [sec(x/2)]^2/[ [sec(x/2)]^2+2tan(x/2) ] dx
=x -2∫ dtan(x/2) /[ [tan(x/2)]^2+2tan(x/2) +1 ]
=x -2∫ du /( u^2+2u+1)
=x -2∫ du /( u+1)^2
=x +2/( u+1) +C
=x +2/[ tan(x/2) +1] +C
=∫ [1 - 1/(1+sinx)] dx
=x - ∫ dx/(1+sinx)
=x - ∫ dx/[1+2sin(x/2)cos(x/2) ]
分子分母同时除 ( cos(x/2))^2
=x -∫ [sec(x/2)]^2/[ [sec(x/2)]^2+2tan(x/2) ] dx
=x -2∫ dtan(x/2) /[ [tan(x/2)]^2+2tan(x/2) +1 ]
=x -2∫ du /( u^2+2u+1)
=x -2∫ du /( u+1)^2
=x +2/( u+1) +C
=x +2/[ tan(x/2) +1] +C
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