Substitute #y = sinx#, #dy = cosx dx#. As the integrand is defined only for #y in [-1,1]#, accordingly #x# varies in #[-pi/2,pi/2]#. In this interval #cosx# is always positive, so:
#int sqrt(1-y^2) dy = int sqrt(1-sin^2x) cosxdx = int cos^2xdx#
Now using the trigonometric identity:
#cos^2 theta = (1+cos2 theta)/2#
we have:
#int sqrt(1-y^2) dy = int (1+cos2x)/2 dx#
#int sqrt(1-y^2) dy = 1/2 int dx +1/2 int cos2x dx#
#int sqrt(1-y^2) dy = x/2 + (sin 2x)/4 +C#
#int sqrt(1-y^2) dy = x/2 + (sinxcosx)/2 +C#
To undo the substitution, note that:
#x = arcsin y#
#sinx = y#
#cosx = sqrt(1-y^2)#
so finally:
#int sqrt(1-y^2) dy = (arcsiny + ysqrt(1-y^2))/2 +C#
