Answer
$ f_{(x)} = \dfrac{\tan{(2x)}}{2} + 2$
Work Step by Step
$f_{(x)}' = \sec^2{(2x)} \longrightarrow f_{(x)} = \int \sec^2{(2x)}\ dx $
Let $u = 2x \longrightarrow du = 2\ dx \longrightarrow dx = \dfrac{du}{2}$
$\int \sec^2{(2x)}\ dx = \int \sec^2{(u)}\ \dfrac{du}{2} = \dfrac{1}{2}\int \sec^2{(u)}\ du = \dfrac{1}{2} \times \tan{(u)} + C$
$\dfrac{\tan{(u)}}{2} + C = \dfrac{\tan{(2x)}}{2} + C \longrightarrow f_{(x)} = \dfrac{\tan{(2x)}}{2} + C$
Since the graph passes through the point $(\frac{\pi}{2}, 2)$, then
$f_{(\frac{\pi}{2})} = \dfrac{\tan{(2\times\frac{\pi}{2})}}{2} + C = 2 \longrightarrow \dfrac{\tan{\pi}}{2} + C = 2 \longrightarrow \dfrac{0}{2} + C = 2 \longrightarrow C = 2$
Hence $ f_{(x)} = \dfrac{\tan{(2x)}}{2} + 2$
