TITLE: Integral of $\sqrt{1-x^2}$ using integration by parts
QUESTION [14 upvotes]: I was asked to solve this indefinite integral using Integration by parts.
$$\int \sqrt{1-x^2} dx$$
I know how to solve if use the substitution $x=\sin(t)$ but I'm looking for the Integration by parts way. 
any help would be very appreciated.

REPLY [4 votes]: A different approach, building up from first principles, withot using cos or sin to get the identity,
$$\arcsin(z) = \int\frac1{\sqrt{1-x^2}}dx$$
where the integrals is from 0 to z.
With the integration by parts given in previous answers, this gives the result.
The distance around a unit circle traveled from the y axis for a distance on the x axis = $\arcsin(x)$. 
$$\arcsin(z) = \int\frac{ds}{dx}dx$$
Pythagoras gives the distance in terms of change in y and change in x.
$$ds^2 = dx^2 + dy^2$$ 
$$\frac{ds}{dx} = \sqrt{1 + {\frac{dy}{dx}}^2}$$
x and y are on a unit circle.
$$1 = x^2 + y^2$$
Rearranging to solve for y and differentiating.
$$\frac{dy}{dx} = \frac{x}{\sqrt{1 - x^2}}$$
Substituting in the above,
$$\frac{ds}{dx} = \frac1{\sqrt{1 - x^2}}$$
And subsituting in the integral gives, 
$$\arcsin(z) = \int\frac1{\sqrt{1-x^2}}dx$$