Answer
$\int\frac{dx}{1-cos x }=-cot(\frac{x}{2})+C$
Work Step by Step
$\int\frac{dx}{1-cos x }$
using the substitution $t=tan(\frac{x}{2})$
then $dx=\frac{2}{1+t^{2}}dt$
and $cosx=\frac{1-t^{2}}{1+t^{2}}$
then the integral is
$\int\frac{dx}{1-cos x }=\int\frac{1}{1-\frac{1-t^{2}}{1+t^{2}}}.\frac{2}{1+t^{2}}dt$=
$2\int\frac{1}{(1+t^{2})-(1-t^{2})}dt=2\int\frac{1}{2t^{2}}dt=\int\frac{1}{t^{2}}dt$
$\int\frac{1}{t^{2}}dt=\frac{-1}{t}+C$
then the integral is
$\int\frac{dx}{1-cos x }=\frac{-1}{tan(\frac{x}{2})}+C=-cot(\frac{x}{2})+C$
