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

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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.**