Answer
$= \frac{\pi}{3}$
Work Step by Step
$\int^{\frac{1}{\sqrt 2}}_{\frac{1}{2}} \frac{4}{\sqrt {1-x^{2}}} dx$
$= 4\int^{\frac{1}{\sqrt 2}}_{\frac{1}{2}} \frac{1}{\sqrt {1-x^{2}}} dx$
$= 4 [\arcsin x |^{\frac{1}{\sqrt 2}}_{\frac{1}{2}}]$
$= 4[\arcsin (\frac{1}{\sqrt 2}) - \arcsin (\frac{1}{2})]$
$= 4[(\frac{\pi}{4}) - (\frac{\pi}{6})]$
$= 4[(\frac{6\pi}{24}) - (\frac{4\pi}{24})]$
$= 4[(\frac{2\pi}{24}) ]$
$= 4(\frac{\pi}{12})$
$= \frac{4\pi}{12}$
$= \frac{\pi}{3}$
