1=sin^2(x/2)+cos^2(x/2)
1/sinx=(sin^2(x/2)/(2sin(x/2)*cos(x/2))+(cos^2(x/2)/(2sin(x/2)*cos(x/2))=
=(sin(x/2)/2cos(x/2)) + (cos(x/2)/2sin(x/2))
∫ dx/sinx= ∫ (sin(x/2)dx/2cos(x/2)) + ∫ (cos(x/2)dx/2sin(x/2))=
оба интеграла методом замены приводятся к интегралу ∫du/u)=-ln|cos(x/2)|+ln|sin(x/2)+C=
=ln|tg(x/2))+C