\input amstex \int_0^1 \frac{\sqrt{1-y^2}}{1+y^2} dy \overset{y=\sin z}\to= \int_0^{\pi/2}  \frac{\cos^2 z \,dz}{1+\sin^2 z}= \int_0^{\pi/2} \left(\frac{2}{1+\sin^2 z }-1\right)dz 
= \int_0^{\pi/2}\frac{2dz}{1+\sin^2z}-\frac{\pi}{2} \overset{t=\tan z}\to=
\int_0^{+\infty}\frac{2dt}{1+2t^2}-\frac{\pi}{2}=\sqrt{2}\Bigl[\arctan(t\sqrt{2})\Bigr]_{t=0}^{t \to +\infty}
=\frac{\pi\sqrt{2}}{2}-\frac{\pi}{2}