Answer
$$ - \cot \left( {\frac{\theta }{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{d\theta }}{{1 - \cos \theta }}} \cr
& {\text{Use substitution }}\left( 5 \right){\text{ from the page 528}}{\text{, let }}x = \theta \cr
& \,\,u = \tan \left( {\frac{\theta }{2}} \right){\text{,}}\,\,\,\,\,\,d\theta = \frac{2}{{1 + {u^2}}}du,\,\,\,\,\,\,\sin \theta = \frac{{2u}}{{1 + {u^2}}}\,\,\,\,\,{\text{ and}}\,{\text{ }}\,\,\cos \theta = \frac{{1 - {u^2}}}{{1 + {u^2}}} \cr
& \cr
& \int {\frac{{d\theta }}{{1 - \cos \theta }}} = \int {\frac{{\frac{2}{{1 + {u^2}}}du}}{{1 - \frac{{1 - {u^2}}}{{1 + {u^2}}}}}} \cr
& {\text{simplify}} \cr
& = \int {\frac{2}{{1 + {u^2} - 1 + {u^2}}}du} \cr
& = \int {\frac{2}{{2{u^2}}}du} \cr
& = \int {\frac{1}{{{u^2}}}du} \cr
& \cr
& {\text{Integrate}} \cr
& = - \frac{1}{u} + C \cr
& {\text{write in terms of }}x{\text{; substitute }}u = \tan \left( {\frac{\theta }{2}} \right) \cr
& = - \frac{1}{{\tan \left( {\theta /2} \right)}} + C \cr
& = - \cot \left( {\frac{\theta }{2}} \right) + C \cr} $$
