Integral of $\sqrt{1-x^2}$ using integration by parts 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.
 A: 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, 
$$\arcsin(z) = \int\frac1{\sqrt{1-x^2}}dx$$
A: I'll restate the accepted answer in different notation, which is easier for me to parse: let 
$$u=\sqrt{1-x^2}, \quad dv=dx$$
so that 
$$du=\frac{-x}{\sqrt{1-x^2}}dx,\quad v=x$$
For brevity, write $I=\int \sqrt{1-x^2}\, dx$. Using $\int u\,dv = uv-\int v\,du$, obtain 
$$
I =  x\sqrt{1-x^2} - \int \frac{-x^2}{\sqrt{1-x^2}} \,dx
$$
The last integral does not look  simpler than $I$ itself, but it can be related back to it:
$$
\int \frac{-x^2}{\sqrt{1-x^2}} \,dx = \int \frac{1-x^2}{\sqrt{1-x^2}} \,dx - \int \frac{1 }{\sqrt{1-x^2}} \,dx = I - \sin^{-1}x
$$
So, 
$$I =  x\sqrt{1-x^2} - (I-\sin^{-1}x)$$
and solving for $I$ yields 
$$\int \sqrt{1-x^2}\, dx = \frac12  x\sqrt{1-x^2} + \frac12 \sin^{-1}x + C$$

For completeness and comparison, I'll add the conventional solution using $x=\sin t$ substitution. Here $dx=\cos t\,dt$, so
$$
\int\sqrt{1-x^2}\,dx = \int \cos^2 t\,dt =\int \left(\frac12+\frac{\cos 2t}{2}\right)\,dt = \frac{t}{2}+\frac{\sin 2t}{4}+C $$ 
To return to $x$, note that $t=\sin^{-1}x$ and $\sin 2t = 2\sin t\cos t = 2x\sqrt{1-x^2}$. 
A: $$I=\int 1\cdot\sqrt{1-x^2}dx=\int dx \sqrt{1-x^2}-\int \left(\frac{\sqrt{1-x^2}}{dx}\int dx\right)dx$$
$$=x\sqrt{1-x^2}-\int\frac{-2x}{2\sqrt{1-x^2}}xdx$$
$$=x\sqrt{1-x^2}+\int\frac{1-(1-x^2)}{\sqrt{1-x^2}}dx$$
$$=x\sqrt{1-x^2}+\int\frac1{\sqrt{1-x^2}}dx-I$$
Now, $\int\frac1{\sqrt{1-x^2}}dx=\arcsin x+C$
