Answer
Convergent $\;,\;$ $\large\frac{π}{2}$
Work Step by Step
Let \[I=\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}\]
Since 1 is point of infinite discontinuity of integrand $\large\frac{1}{\sqrt{1-x^2}}$
\[\Rightarrow I=\lim_{t\rightarrow 1^-}\int_{0}^{t}\frac{dx}{\sqrt{1-x^2}}\;\;\;\ldots (1)\]
\[I=\lim_{t\rightarrow 1^-}\left[\sin^{-1}x\right]_{0}^{t}\]
\[I=\lim_{t\rightarrow 1^-}\left[\sin^{-1}t\right]=\frac{π}{2}\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=\frac{π}{2}$.
