Answer
$$\frac{9\pi -16}{8}$$
Work Step by Step
Given $$r=2+\cos \theta$$
From the given figure, the region bounded by $ \pi /2\leq \theta \leq \pi$
Area in polar form is given by
\begin{align*}
\text{Area}&=\dfrac{1}{2}\int_{\pi/2}^{\pi}r^2d\theta \\
&=\dfrac{1}{2}\int_{\pi/2}^{\pi}(2+\cos \theta )^2d\theta \\
&=\dfrac{1}{2}\int_{\pi/2}^{\pi}(4+4\cos \theta+\cos^2 \theta ) d\theta \\
&=\dfrac{1}{2}\int_{\pi/2}^{\pi}(4+4\cos \theta+\frac{1}{2}+\frac{1}{2}\cos2 \theta ) d\theta \\
&=\frac{9}{2}\theta+4\sin \theta +\frac{1}{4}\sin2 \theta \bigg|_{\pi/2}^{\pi}\\
&=\frac{9\pi -16}{8}
\end{align*}
