What is the antiderivative of the function given by y=(36-x^2)^1/2?

To determine the antiderivative, or the primitive of the
given function f(x) = y, we'll calculate the indefinite integral of
f(x).

Int f(x)dx = Int sqrt(36 -
x^2)dx

We'll factorize by
36:

Int sqrt[36(1 - x^2/36)]dx = 6Int sqrt[1 -
(x/6)^2]dx

We'll substitute x/6 =
t.

We'll differentiate both
sides:

dx/6 = dt

dx =
6dt

6Int sqrt[1 - (x/6)^2]dx = 36 Int sqrt(1 -
t^2)dt

We'll substitute t = sin
v.

We'll differentiate both
sides:

dt = cos v dv

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

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

36 Int sqrt(1 - (sin v)^2)cos v dv = 36 Int
sqrt[(cos v)^2]cos v dv

36 Int sqrt[(cos v)^2]cos v dv = 36
Int [(cos v)^2] dv

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

36 Int [(cos v)^2] dv = 36 Int (1 + cos 2v)/2
dv

36 Int (1 + cos 2v)/2 dv = (36/2) Int dv + 18 Int cos 2v
dv

36 Int (1 + cos 2v)/2 dv = 18v + 9 sin 2v +
C

Int f(x)dx = 18v + 9 sin 2v +
C

The antiderivative of the function is F(x)
= Int f(x)dx = 18*arcsin (x/6) + 9*sin [2arcsin (x/6)]+
C.