Question

How do you integrate by substitution `int x/(sqrt(1-x^2) dx`?

Answer 1

You don't. By inspection, the answer is `(-1/2)sqrt(1-x^2)+c`

Explanation:

It's of the form `(f'(x))/sqrt(f(x))`. In other words, the thing on the top is (nearly) the derivative of the thing under the square root. When you differentiate `sqrt(f(x)` you get `(1/2)(f'(x))/sqrt(f(x))` (chain rule).

Answer 2

`intx/sqrt(1-x^2)dx=-sqrt(1-x^2)+C`

Explanation:

By substitution:

Let `u = 1-x^2`. Then `du = -2xdx`. Substituting, we get

`intx/sqrt(1-x^2)dx = -1/2int1/sqrt(1-x^2)(-2x)dx`

`=-1/2int1/sqrt(u)du`

`=-1/2intu^(-1/2)du`

`=-1/2((u^(-1/2+1))/(-1/2+1))+C`

`=-1/2((u^(1/2))/(1/2))+C`

`=-sqrt(u)+C`

`=-sqrt(1-x^2)+C`