Using the identity:
#sec theta = 1/cos theta#
so that using the parametric formulas we have:
#sec theta = (1+tan^2(theta/2))/(1-tan^2(theta/2))#
and substitute:
#t = tan(theta/2)#
#theta = 2 arctant#
#d theta = 2/(1+t^2)#
Then:
#int 1/(1-sec theta)d theta = 2int 1/(1-(1+t^2)/(1-t^2)) dt/(1+t^2)#
#int 1/(1-sec theta)d theta = 2int (1-t^2)/(1 -t^2 -1-t^2)dt/(1+t^2)#
#int 1/(1-sec theta)d theta = int (t^2-1)/(t^2(t^2+1))dt#
Decompose in partial fractions:
#(t^2-1)/(t^2(t^2+1)) = A/t^2+B/(t^2+1)#
#(t^2-1)/(t^2(t^2+1)) = (A(t^2+1)+Bt^2)/(t^2(t^2+1))#
#t^2-1 = (A+B)t^2 +A#
#{(A=-1), (B=2):}#
#(t^2-1)/(t^2(t^2+1)) = -1/t^2+2/(t^2+1)#
Then:
#int 1/(1-sec theta)d theta = -int dt/t^2 +2 int dt/(1+t^2)#
#int 1/(1-sec theta)d theta = 1/t +2 arctan t +C#
and undoing the substitution:
#int 1/(1-sec theta)d theta = cot(theta/2)+ theta +C#
