The area above the x-axis and under the curve $y=\sqrt{\frac{1}{x}-1}\text{ for }\frac{1}{2}\le x\le 1$ is

$$\begin{gathered} A={\int }_{\frac{1}{2}}^1\sqrt{\frac{1}{x}-1}dx \\ x={\cos }^2 \theta \Rightarrow dx=-2\sin \theta \cos \theta d\theta \\ x=1\Rightarrow \theta =0, x=\frac{1}{2}\Rightarrow \theta =\frac{\mathrm{\pi }}{4} \\ A={\int }_{\frac{\mathrm{\pi }}{4}}^0\sqrt{se{c}^2\theta -1}(-2\sin \theta \cos \theta )d\theta \\ A=2{\int }_0^{\frac{\mathrm{\pi }}{4}}\tan \theta \sin \theta \cos \theta d\theta \\ A={\int }_0^{\frac{\mathrm{\pi }}{4}}2 {\sin }^2 \theta d\theta \\ A={\int }_0^{\frac{\mathrm{\pi }}{4}}1- \cos 2 \theta d\theta \\ A={\left[\theta +\frac{\sin 2\theta }{2}\right]}_0^{\frac{\mathrm{\pi }}{4}}=\frac{\mathrm{\pi }}{4}+\frac{1}{2} \end{gathered}$$