Answer
$\dfrac{\pi}{4}$
Work Step by Step
Plug $x =\frac{1}{2} \sin \theta \implies dx=\frac{1}{2} \cos \theta d \theta $
Then
$\int \dfrac{ \cos \theta d \theta }{\sqrt{1-(1/4) \sin^2 \theta}} =\int \dfrac{ \cos \theta d \theta }{\sqrt {\cos^2 \theta}} = \int d \theta$
Now, integrate with limits to have
$\int_{0}^{1/2 \sqrt 2} d \theta=[sin^{-1} (2x)]_{0}^{1/2 \sqrt 2}$
Thus, $\int \dfrac{ \cos \theta d \theta }{\sqrt{1-(1/4) \sin^2 \theta}} =\dfrac{\pi}{4}$
