Answer
\[\frac{1}{2}\sin 2x-\frac{1}{8}\cos 4x+C\]
Work Step by Step
Let \[I=\int (\cos x+\sin x)^2\cos 2x \:dx\]
\[I=\int \left(\cos^2 x+\sin^2 x+2\sin x\cos x\right)\cos 2x\:dx\]
\[\left[2\sin x\cos x=\sin 2x\right]\]
\[I=\int [1+\sin 2x]\cos 2x\: dx\]
\[I=\int \cos 2x \:dx+\frac{1}{2}\int 2\sin 2x \cos 2x\:dx\]
\[\left[2\sin\theta\cos\theta=\sin 2\theta\right]\]
\[I=\int\cos 2x\:dx+\frac{1}{2}\int \sin 4x\:dx\]
\[I=\frac{1}{2}\sin 2x-\frac{1}{8}\cos 4x+C\]
Where $C$ ia constant of integration
Hence \[I=\frac{1}{2}\sin 2x-\frac{1}{8}\cos 4x+C\:\:.\]
