Answer
$$
\int(\cos x+\sin x)^{2} \cos 2 x d x = \frac{1}{2} \sin 2 x-\frac{1}{8} \cos 4 x+C
$$
where $C$ is a constant
Work Step by Step
$$
\begin{aligned} \int(\cos x+\sin x)^{2} \cos 2 x d x &=\int\left(\cos ^{2} x+2 \sin x \cos x+\sin ^{2} x\right) \cos 2 x d x \\
&=\int(1+\sin 2 x) \cos 2 x d x \\
& =\int \cos 2 x d x+\frac{1}{2} \int \sin 4 x d x \\
&=\frac{1}{2} \sin 2 x-\frac{1}{8} \cos 4 x+C
\end{aligned}
$$
where $C$ is a constant