Integral of $\sqrt{1-x^2}$ using integration by parts

Question

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.

Answer 1

$$I=\int 1\cdot\sqrt{1-x^2}dx=\int dx \sqrt{1-x^2}-\int \left(\frac{\sqrt{1-x^2}}{dx}\int dx\right)dx$$

$$=x\sqrt{1-x^2}-\int\frac{-2x}{2\sqrt{1-x^2}}xdx$$

$$=x\sqrt{1-x^2}+\int\frac{1-(1-x^2)}{\sqrt{1-x^2}}dx$$

$$=x\sqrt{1-x^2}+\int\frac1{\sqrt{1-x^2}}dx-I$$

Now, $\int\frac1{\sqrt{1-x^2}}dx=\arcsin x+C$

Answer 2

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, $$\arcsin(z) = \int\frac1{\sqrt{1-x^2}}dx$$