Answer
$0$
Work Step by Step
We have
$$
\int_{0}^{2\pi}f(x)dx=\int_{0}^{\pi}f(x)dx+\int_{\pi}^{2\pi}f(x)dx\\
=\int_{0}^{\pi}\cos x dx+\int_{\pi}^{2\pi}\cos x -\sin 2x dx\\
=\sin x|_{0}^{\pi}+(\sin x+\frac{1}{2}\cos 2x)|_{\pi}^{2\pi} \\
=0+\frac{1}{2}(1-1)=0
$$
