 | t=tan(x/2) とおくと
sinx=2t/(1+t^2)
cosx=(1-t^2)/(1+t^2)
dx=2/(1+t^2)dt
なので
∫(2-cosx)/(1+sinx)dx
=∫{2-(1-t^2)/(1+t^2)}/{1+2t/(1+t^2)}・2/(1+t^2)dt
=∫2(3t^2+1)/{(1+t^2)(t+1)^2}dt
=∫-2/(t+1)+4/(t+1)^2+2t/(1+t^2) dt
=-2log|t+1|-4/(t+1)+log(1+t^2)+C
=-2log|tan(x/2)+1|-4/{tan(x/2)+1}+log{1+tan(x/2)^2}+C
=-2log|tan(x/2)+1|-4/{tan(x/2)+1}-2log|cos(x/2)|+C
=-2log|sin(x/2)+cos(x/2)|-4/{tan(x/2)+1}+C
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