Use the prescribed cosine substitution in a difference of squares

Since $du=-2\sin2\theta\,d\theta$, the integral becomes $-\frac12\int\dfrac1{2^2-u^2}du=-\frac18\ln\left|\dfrac{2+u}{2-u}\right|+C$. Since $-1\leq\cos2\theta\leq1$, both $2+u$ and $2-u$ are positive. Thus the source removes the bars and obtains $-\frac18\ln\left(\dfrac{2+\cos2\theta}{2-\cos2\theta}\right)+C$.