To determine the antiderivative, you'll have to evaluate the indefinite integral of the given function y.

Int f(x)dx = Int sqrt(25 - x^2)dx (y = f(x))

We'll factorize by 25:

Int sqrt[25(1 - x^2/25)]dx = 5Int sqrt[1 - (x/5)^2]dx

We'll substitute x/5 = t.

We'll differentiate both sides:

dx/5 = dt

dx = 5dt

5Int sqrt[1 - (x/5)^2]dx = 5 Int sqrt(1 - t^2)dt

We'll substitute t = sin v.

We'll differentiate both sides:

dt = cos v dv

5 Int sqrt(1 - (sin v)^2)cos v dv

But 1 - (sin v)^2 = (cos v)^2 (trigonometry)

5 Int sqrt(1 - (sin v)^2)cos v dv = 5 Int sqrt[(cos v)^2]cos v dv

5 Int sqrt[(cos v)^2]cos v dv = 5 Int [(cos v)^2] dv

But (cos v)^2 = (1 + cos 2v)/2

5 Int [(cos v)^2] dv = 5 Int (1 + cos 2v)/2 dv

5 Int (1 + cos 2v)/2 dv = (5/2) Int dv + (5/2) Int cos 2v dv

5 Int (1 + cos 2v)/2 dv = 5v/2 + 5 sin 2v/2 + C

Int f(x)dx = 5v/2 + 5 sin 2v/2 + C

**Int f(x)dx = 5 [arcsin (x/5)]/2 + 5 [sin 2arcsin (x/5)]/2 + C**