Answer
$$\ln \left| {\sin 2x + 2} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sec x\cos 2x}}{{\sin x + \sec x}}} dx \cr
& {\text{Use reciprocal identity }}\sec \theta = \frac{1}{{\cos \theta }} \cr
& = \int {\frac{{\left( {\frac{1}{{\cos x}}} \right)\cos 2x}}{{\sin x + \sec x}}} dx \cr
& = \int {\frac{{\cos 2x}}{{\cos x\left( {\sin x + \sec x} \right)}}} dx \cr
& {\text{Multiply and sinplify}} \cr
& = \int {\frac{{\cos 2x}}{{\cos x\sin x + 1}}} dx \cr
& {\text{Multiply the numerator and denominator by 2}} \cr
& = \int {\frac{{2\cos 2x}}{{2\cos x\sin x + 2}}} dx \cr
& {\text{Recall that }}\sin 2x = 2\sin x\cos x \cr
& = \int {\frac{{2\cos 2x}}{{\sin 2x + 2}}} dx \cr
& {\text{Let }}u = \sin 2x + 2,{\text{ }}du = 2\cos 2xdx \cr
& = \int {\frac{{du}}{u}} \cr
& = \ln \left| u \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \ln \left| {\sin 2x + 2} \right| + C \cr} $$
