Answer
$$ \frac{7}{2}$$
Work Step by Step
Given
$$\int_{\pi / 6}^{2 \pi} \cos x d x$$
Since
\begin{aligned}
\cos(x) &\geq 0\ \ \ \ \text{for }\ \ \pi/6\leq x\leq \pi/2 \\
\cos(x) &\leq 0\ \ \ \ \text{for }\ \ \pi/2\leq x\leq 3\pi/2\\
\cos(x) &\geq 0\ \ \ \ \text{for }\ \ 3\pi/2\leq x\leq 2\pi
\end{aligned}
Then area given by
\begin{aligned} \text{Area}&= \int_{\pi/6}^{2\pi}\cos(x)dx\\
&= \int_{\pi/6}^{\pi/2}\cos(x)dx- \int_{\pi/2}^{3\pi/2}\cos(x)dx+ \int_{3\pi/2}^{2\pi}\cos(x)dx\\
&= \sin (x)\bigg|_{\pi/6}^{\pi/2} - \sin (x)\bigg|_{\pi/2}^{3\pi/2}+ \sin (x)\bigg|_{3\pi/2}^{2\pi}\\
&= \sin(\pi/2) -\sin(\pi/6) -[\sin(3\pi/2)-\sin(\pi/2)]+ \sin(2\pi)-\sin(3\pi/2)\\
&= \frac{7}{2}
\end{aligned}