# Given dy/dx=square root(25-x^2) what is y ?

### 1 Answer | Add Yours

To determine the primitive of the given function dy, we'll calculate the indefinite integral of dy.

Int dy = Int sqrt(25 - x^2)dx

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 = 25 Int sqrt(1 - t^2)dt

We'll substitute t = sin v.

We'll differentiate both sides:

dt = cos v dv

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

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

25 Int sqrt(1 - (sin v)^2)cos v dv = 25 Int sqrt[(cos v)^2]cos v dv

25 Int sqrt[(cos v)^2]cos v dv = 25 Int [(cos v)^2] dv

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

25 Int [(cos v)^2] dv = 25 Int (1 + cos 2v)/2 dv

25 Int (1 + cos 2v)/2 dv = (25/2) Int dv + (25/2) Int cos 2v dv

25 Int (1 + cos 2v)/2 dv = 25*v/2 + 25*sin 2v/4 + C

Int f(x)dx = 25*v/2 + 25*sin 2v/4 + C

**The primitive function is y = (25/2)*{[arcsin (x/5)] + sin [2arcsin (x/5)]/2} + C.**