- #1

Nano-Passion

- 1,291

- 0

∫ (1-x^2)^1/2 dx, -1<= x <= 1

x = sin Θ

dx = cos Θ dΘ

-pi/2 <= Θ <= pi/2

∫ (1-x^2)^1/2 dx = ∫ 1 - sin^2Θ dΘ

u = 2Θ

du = 2 dΘ

-pi <= u <= pi

Then the integral is equal to:

Θ - 1/2Θ - 1/2 sin u(Θ(x)) + C

where Θ(x) = sin^-1x.

Would u(Θ(x)) = 2arcsinx be true?

Which would give us:

arcsin x - 1/2arcsin x - 1/2 sin (2arcsin x) + C

However ugly it is, would this be correct with the adjusted interval? Does it make sense?