=∫ 1/sinx dx

= ∫1/[2sin (x/2) cos (x/2)] dx, double angle formula.

=∫ 1/[sin(x/2)cos(x/2)]d(x/2)

=∫ 1/tan(x/2)* seconds? (x/2) d(x/2)

=∫ 1/tan(x/2) d[tan(x/2)], note ∫sec? (x/2)d(x/2)=tan(x/2)+C

=ln|tan(x/2)|+C, this is the first answer.

Further simplify:

=ln|sin(x/2)/cos(x/2)|+C

=ln|2sin(x/2)cos(x/2)/[2cos？ (x/2)]|+C, and solve the double angle formula.

=ln|sinx/( 1+cosx)|+C

=ln|sinx( 1-cosx)/sin？ x|+C

=ln|( 1-cosx)/sinx|+C

=ln|cscx-cotx|+C, this is the second answer.

For reference only.