>>236
∫(sinx)/{1-(sinx)^2}dx
=(1/2)∫( 1/(1-sinx) - 1/(1+sinx) ) dx
∫1/(1±sinx)dx
=∫1/(1±2t/(1+t^2)) 2dt/(1+t^2)：tan(x/2)=t
=∫2/(1±t)^2 dt
=-2/(1±t)
=-2/(1±tan(x/2))