# Evaluate the indefinite integral of y=square root of 16-x^2.

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### 1 Answer

Int f(x)dx = Int sqrt(16 - x^2)dx (y = f(x))

We'll factorize by 16:

Int sqrt[16(1 - x^2/16)]dx = 4Int sqrt[1 - (x/4)^2]dx

We'll substitute x/4 = t.

We'll differentiate both sides:

dx/4 = dt

dx = 4dt

4Int sqrt[1 - (x/4)^2]dx = 16 Int sqrt(1 - t^2)dt

We'll substitute t = sin v.

We'll differentiate both sides:

dt = cos v dv

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

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

16 Int sqrt(1 - (sin v)^2)cos v dv = 16 Int sqrt[(cos v)^2]cos v dv

16 Int sqrt[(cos v)^2]cos v dv = 16 Int [(cos v)^2] dv

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

16 Int [(cos v)^2] dv = 16 Int (1 + cos 2v)/2 dv

16 Int (1 + cos 2v)/2 dv = (16/2) Int dv + 8 Int cos 2v dv

16 Int (1 + cos 2v)/2 dv = 8v + 4 sin 2v + C

Int f(x)dx = 8v + 4 sin 2v + C

**Int f(x)dx = 8 arcsin (x/4) + 4 sin 2arcsin (x/4)+ C**