Answer
$$\frac{1}{2}\sin 2\theta + \frac{1}{2}{\sin ^2}2\theta + C$$
Work Step by Step
$$\eqalign{
& \int {\cos 2\theta {{\left( {\sin \theta + \cos \theta } \right)}^2}} d\theta \cr
& {\text{Expand the binomial}} \cr
& = \int {\cos 2\theta \left( {{{\sin }^2}\theta + 2\sin \theta \cos \theta + {{\cos }^2}\theta } \right)} d\theta \cr
& {\text{Simplify}} \cr
& = \int {\cos 2\theta \left( {1 + 2\sin \theta \cos \theta } \right)} d\theta \cr
& {\text{Where sin2}}\theta = {\text{2sin}}\theta {\text{cos}}\theta {\text{, then}} \cr
& = \int {\cos 2\theta \left( {1 + {\text{sin2}}\theta } \right)} d\theta \cr
& = \int {\cos 2\theta } d\theta + \int {\cos 2\theta \sin 2\theta } d\theta \cr
& {\text{Multiply and divide by constants}} \cr
& = \frac{1}{2}\int {\cos 2\theta } \left( 2 \right)d\theta + \frac{1}{2}\int {\sin 2\theta \left( {2\cos 2\theta } \right)} d\theta \cr
& {\text{Integrate}} \cr
& {\text{ = }}\frac{1}{2}\sin 2\theta + \frac{1}{2}{\sin ^2}2\theta + C \cr} $$
