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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.

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  • $\begingroup$ A bit off topic, but this integral has also has an elegant geometric (without using calculus) solution. $\endgroup$ – user185498 May 15 '16 at 15:48
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$$I=\int 1\cdot\sqrt{1-x^2}dx=\int dx \sqrt{1-x^2}-\int \left(\frac{\sqrt{1-x^2}}{dx}\int dx\right)dx$$

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

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

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

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

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    $\begingroup$ Except that last integral is proved using trig substitution, so I'm not sure what's to be gained by integration by parts! $\endgroup$ – GFauxPas Nov 4 '14 at 20:52
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    $\begingroup$ @GFauxPas: the derivative of arcsin can be computed with implicit differentiation and found to be $\frac{1}{\sqrt{1-x^2}}$. $\endgroup$ – Matthew Leingang Jan 1 '15 at 14:44
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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 $\sin 2t = 2\sin t\cos t = 2x\sqrt{1-x^2}$.

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

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  • $\begingroup$ This is tackling a different integral to the original poster's. $\endgroup$ – Lord Shark the Unknown Jul 31 '17 at 5:50
  • $\begingroup$ No, it is linking up to the first answer in terms of integration by parts. He asked how to do it without using trig substitutions. This completes that. $\endgroup$ – Peter Driscoll Jul 31 '17 at 6:30

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