 | tan(x/2)=tとおくと
sinx=2t/(1+t^2)
dx=2/(1+t^2)dt
なので
∫1/(1-sinx)^2 dx
=∫1/(1-2t/(1+t^2))^2・2/(1+t^2) dt
=2∫(1+t^2)/(t-1)^4 dt
=2∫(1+(u+1)^2)/u^4 du （t-1=uとおいた）
=2∫(u^2+2u+2)/u^4 du
=2∫1/u^2+2/u^3+2/u^4 du
=-2/u-2/u^2-4/(3u^3)+C
=-2(3u^2+3u+2)/(3u^3)+C
=-2(3(t-1)^2+3(t-1)+2)/(3(t-1)^3)+C
=-2(3t^2-3t+2)/(3(t-1)^3)+C
=-2(3(tan(x/2))^2-3tan(x/2)+2)/(3(tan(x/2)-1)^3)+C
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