Help, very difficult math on limits

I just know one step for this problem Somebody, help me solve it.

Find lim(a->0) ∫(from -a to a) f(x)/(√a^2-x^2) *dx, where (a>0).
Note that f(x) is not continuous on the region -1
I think it would be easier to change it into polar coordinates and so,
x=asinθ
dx = acosθdθ ...........-pi/2<θpi/2

Have no idea what to do from here. HELP!

What is f(x)?
If the problem is
lim(a->0) ∫(from -a to a) 1/(√a^2-x^2) *dx
I got 0

to compute
∫(1/(√a^2 - x^2) dx
put x = a sin t
dx = a cos t dt

substitute
∫(1/(√a^2 - a^2 sin^2 t) a cos t dt

(1/a)∫(1/√(1 - sin^2 t) a cos t dt

(1/a)∫(1/cos t) a cos t dt

∫dt = t + c

t = asin (x/a)

∫(1/(√a^2 - x^2) dx = asin (x/a) + c

Because of the symmetry the integral between - a and a can be computed as
2 lim (k->a) asin (k/a) = pi

my result to
lim(a->0) ∫(from -a to a)1/(√a^2-x^2) dx
is pi (a>0)

Do you mean not continuous on -1 If you do as you say you get lim(a->0)∫ (from -pi/2 to pi/2) f(acosθ)dθ
=∫from -pi/2 to pi/2 f(0) dθ = pi.f(0)