∫1/（1+sinx）dx
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用万能代换
∫1/（1+sinx）dx
=∫[sin^2(x/2)+cos^2(x/2)]/[sin^2(x/2)+cos^2(x/2)+2sin(x/2)cos(x/2)]dx
=∫[1+tan^2(x/2)]/[tan^2(x/2)+1+2tan(x/2)]dx
=∫sec^2(x/2)/[tan(x/2)+1]^2dx
=2∫1/[tan(x/2)+1]^2dtan(x/2)
=-2/[tan(x/2)+1]+C
∫1/（1+sinx）dx
=∫[sin^2(x/2)+cos^2(x/2)]/[sin^2(x/2)+cos^2(x/2)+2sin(x/2)cos(x/2)]dx
=∫[1+tan^2(x/2)]/[tan^2(x/2)+1+2tan(x/2)]dx
=∫sec^2(x/2)/[tan(x/2)+1]^2dx
=2∫1/[tan(x/2)+1]^2dtan(x/2)
=-2/[tan(x/2)+1]+C
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