Find f.
f'(x) = 9/(square root of (1-x^2)) ,
f(1/2) = 1
f(x) = ??????
dy/dx = 9 integral dx/(1-x^2)^.5
y = 9 sin^-1 x + c
note sin pi/6 = sin 30 deg = 1/2
so
y(1/2) = 9 (pi/6) + c = 1
so
c = 1 - 3 pi/2
so
y = 9 sin^-1 x + (2-3 ) pi/2