$\displaystyle

\int \frac{dz}{\sqrt{z^2+\frac{1}{4}}}=2\int \frac{\frac{1}{2}\frac{dx}{\cos ^2 x}}{\sqrt{\tan^2 x+1}}=\int \frac{dx}{\cos x}\\

$

$\displaystyle

z=\frac{1}{2}\tan x\\

$

$\displaystyle

dz=\frac{1}{2}\frac{dx}{\cos ^2 x}

$

what to do next??