Question

How do you integrate `(1-x^2)^.5`?

Answer 1

`int (1-x^2)^0.5 dx = 1/2 x sqrt(1-x^2) + 1/2 sin^(-1)x + C`

Explanation:

Let `x = sin theta`

Then:

`int (1-x^2)^0.5 dx = int (1-sin^2 theta)^0.5 ((d sin theta)/(d theta)) d theta`

`color(white)(int (1-x^2)^0.5 dx) = int (cos^2 theta)^0.5 (cos theta) d theta`

`color(white)(int (1-x^2)^0.5 dx) = int cos^2 theta d theta`

`color(white)(int (1-x^2)^0.5 dx) = int 1/2(cos 2 theta + 1) d theta`

`color(white)(int (1-x^2)^0.5 dx) = 1/4 sin 2 theta + theta/2 + C`

`color(white)(int (1-x^2)^0.5 dx) = 1/2 sin theta cos theta + theta/2 + C`

`color(white)(int (1-x^2)^0.5 dx) = 1/2 x sqrt(1-x^2) + 1/2 sin^(-1)x + C`