Sqrt(1-x^2) integral

Sqrt(1-x^2) integral

Indefinite integral
integral sqrt(1 – x^2) dx = 1/2 (sqrt(1 – x^2) x + sin^(-1)(x)) + constant
Expanded form of the integral
1/2 sqrt(1 – x^2) x + 1/2 sin^(-1)(x) + constant
Series expansion of the integral at x=-1
-pi/4 + 2/3 sqrt(2) (x + 1)^(3/2) – (x + 1)^(5/2)/(5 sqrt(2)) + O((x + 1)^(7/2))
(Puiseux series)
Series expansion of the integral at x=0
x – x^3/6 – x^5/40 – x^7/112 + O(x^9)
(Taylor series)
Series expansion of the integral at x=1
(-1)^(floor(-(arg(x – 1))/(2 pi))) (-(i sqrt(x – 1))/sqrt(2) + (i (x – 1)^(3/2))/(12 sqrt(2)) – (3 i (x – 1)^(5/2))/(160 sqrt(2)) + O((x – 1)^(7/2))) + (pi/4 + (i sqrt(x – 1))/sqrt(2) + (5 i (x – 1)^(3/2))/(4 sqrt(2)) + (7 i (x – 1)^(5/2))/(32 sqrt(2)) + O((x – 1)^(7/2)))
Series expansion of the integral at x=∞
(i x^2)/2 + 1/4 (2 i log(1/x) + pi – i (1 + log(4))) + i/(16 x^2) + O((1/x)^4)
(generalized Puiseux series)
Definite integral
integral_(-1)^1 sqrt(1 – x^2) dx = pi/2~~1.5708

Leave a Reply

Your email address will not be published. Required fields are marked *