We can determine the following equations of $\int\,\cos^2\,2x\,dx\,=\,\frac{1}{2}\,x\,+\,\frac{1}{8}\,\sin\,4x\,+\,C_1$ and $\int\,\cos^3\,2x\,dx\,=\,\frac{1}{2}\,\sin\,2x\,-\,\frac{1}{6}\,\sin^3\,2x\,+\,C_2,$ where $C_1$ and $C_2$ are constants. Then, solve the next obvious step, and we arrive with the solution of $\frac{1}{8}\int\,dx\,+\,\frac{3}{8}\int\,\cos\,2x\,dx\,+\,\frac{3}{8}\int\,\cos^2\,2x\,dx\,+\,\frac{1}{8}\int\,\cos^3\,2x\,dx\,=\,\frac{5}{16}\,x\,+\,\frac{1}{4}\,\sin\,2x\,+\,\frac{3}{64}\,\sin\,4x\,-\,\frac{1}{48}\,\sin^3\,2x\,+\,C,$ where $C$ is an constant.

$\frac{3}{8}\int\,\cos\,2x\,dx\,=\,\frac{3}{16}\,\sin\,2x+K,$ where $K$ is an constant. So yes, this part is added to $\frac{1}{16}\,\sin\,2x$ during the process of getting the final solution.