Answer
$$ - \frac{1}{4}\cot 2x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 - \cos 4x}}} \cr
& {\text{A matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula }}47 \cr
& \int {\frac{{dx}}{{1 - \cos ax}}} = - \frac{1}{a}\cot \frac{{ax}}{2} + C \cr
& {\text{with }}a = 4 \cr
& \int {\frac{{dx}}{{1 - \cos 4x}}} = - \frac{1}{4}\cot \frac{{4x}}{2} + C \cr
& \int {\frac{{dx}}{{1 - \cos 4x}}} = - \frac{1}{4}\cot 2x + C \cr} $$
