find limit

Integrate{[(sin(y/2))^2-(sin(x/2))^2]^(-0.5), 0, y}dx

Consider y=+0
In the integral interval, 0<x<y
As y approaches 0, x also approaches 0 ... (1)
sin(x/2) / sin(y/2) <1

Take the transformation sin(y/2)sin(z)=sin(x/2)
sin(y/2)cos(z)dz=(1/2)cos(x/2)dx
dx={sin(y/2)cos(z)/[(1/2)cos(x/2)]}dz

when x=0, z=0
when x=y, z=兀/2

[(sin(y/2))^2-(sin(x/2))^2]^(0.5)
=sin(y/2)[1-(sin(z))^2]^(0.5)
=sin(y/2)cos(z)

Integrate{[(sin(y/2))^2-(sin(x/2))^2]^(-0.5), 0, y}dx
=Integrate{sin(y/2)cos(z)/[(1/2)cos(x/2)]/sin(y/2)cos(z), 0, 兀/2}dz
=Integrate{2/cos(x/2), 0, 兀/2}dz ...(2)

From (1) as x->0 cos(x/2)->1
(2) becomes Integrate{2, 0, 兀/2}dz = 兀

I believe the case for y=-0 should be similar

To: 螞蟻雄兵
No, the answer is not F(0) because discontinuity occurs at y=0.
Actually, the answer is Pi, but I do not know how to prove it rigorously.
Can any body help?
Maybe you can try to find F(0.1), F(0.01), F(0.001), ... by numerical
method, verifying the answer(Pi)
But try to prove it.