Note that 1/1-x2→∞1/\sqrt{1-x^{2}}\rightarrow\infty as x→1-x\rightarrow 1-; hence the integral is improper.
Solution. We let x=sintx=\sin t, and observe that sint→1-\sin t\rightarrow 1- as t→(π/2)-t\rightarrow(\pi/2)-; so the limits 0<x<10<x<1 convert to: 0<t<π20<t<\frac{\pi}{2}.
Also, dx/dt=costdx/dt={\cos t} and 1-x2=1-sin2t=cost,\sqrt{1-x^{2}}={\sqrt{1-\sin^{2}t}}\,{=\cos t,} so