Answer
$\displaystyle \frac{\pi}{2}$
(converges)
Work Step by Step
Discontinuity at x=1. Type 1.
$\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}=\lim_{t\rightarrow 1^{-}}\int_{0}^{t}\frac{dx}{\sqrt{1-x^{2}}}$
$=\displaystyle \lim_{t\rightarrow 1-}\left[\arcsin x\right]_{0}^{t}$
$=\displaystyle \lim_{t\rightarrow 1^{-}}\arcsin t-\lim_{t\rightarrow 1^{-}}\arcsin 0$
$=\displaystyle \frac{\pi}{2}-0$
$=\displaystyle \frac{\pi}{2}$
(converges)