Is there any way to compute the integral \begin{equation} \int_{-a}^{a} \sqrt{R^2-x^2} \,\mathrm{d}x, \qquad 0<a<R \end{equation} without using trig substitution or integration by parts?

I'm thinking to relate this to area of circle, but I couldn't find the relationship.

Thank you.

  • 1
    $\begingroup$ No that is not possible! $\endgroup$ – Jan Dec 7 '15 at 20:52
  • $\begingroup$ If you are allowed to cheat and look up in a table of integrals… $\endgroup$ – egreg Dec 7 '15 at 21:00
  • 2
    $\begingroup$ If you sketch the function under which you're integrating and the limits of integration you'll see that this is a part of the area of the circle with radius $R$. So if you wanted to do it geometrically, you could just use the formulas for the area of a (half) circle and for the area of a circular segment. $\endgroup$ – user137731 Dec 7 '15 at 21:01
  • 1
    $\begingroup$ The integral is 'half of the circle' - 'circular sector with $2\theta=2\cos^{-1}\frac aR$' + '2 triangles with base=$a$ and height=$\sqrt{R^2-a^2}$' $\endgroup$ – Kay K. Dec 7 '15 at 21:05
  • $\begingroup$ Maybe you can adapt this: en.wikipedia.org/wiki/… $\endgroup$ – egreg Dec 7 '15 at 21:51

Without trigonometric substitution, integration by parts, or appeal to the geometric interpretation of the integral, we need to devise some quite fortuitous manipulations. We proceed, therefore, and write

$$\begin{align} \sqrt{R^2-x^2}&=\frac12 \left(\sqrt{R^2-x^2}+\sqrt{R^2-x^2}\right)\\\\ &=\frac12 \left(\sqrt{R^2-x^2}+\sqrt{R^2-x^2}\right)+\frac{x^2}{\sqrt{R^2-x^2}}-\frac{x^2}{\sqrt{R^2-x^2}}\\\\ =&\frac12 \left(\sqrt{R^2-x^2}-\frac{x^2}{\sqrt{R^2-x^2}}\right)+\frac12\left(\sqrt{R^2-x^2}+\frac{x^2}{\sqrt{R^2-x^2}}\right)\\\\ &=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12\left(R^2-x^2\right)\left(\frac{1}{\sqrt{R^2-x^2}}+\frac{x^2}{(R^2-x^2)^{3/2}}\right)\\\\ &=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12\left(R^2-x^2\right)\frac{d\left(x/\sqrt{R^2-x^2}\right)}{dx}\\\\ &=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12R^2\left(\frac{1}{1+\left(\frac{x}{\sqrt{R^2-x^2}}\right)^2}\right)\frac{d\left(x/\sqrt{R^2-x^2}\right)}{dx}\\\\ &=\frac{d}{dx}\left(\frac12 x\sqrt{R^2-x^2}+\frac12 R^2 \arctan\left(\frac{x}{\sqrt{R^2-x^2}}\right) \right) \end{align}$$

Therefore, we arrive at

$$\begin{align} \int_{-a}^a\sqrt{R^2-x^2}\,dx&=2\int_0^a \sqrt{R^2-x^2}\,dx\\\\ &=\left.\left( x\sqrt{R^2-x^2}+ R^2 \arctan\left(\frac{x}{\sqrt{R^2-x^2}}\right) \right)\right|_0^a\\\\ &= a\sqrt{R^2-a^2}+ R^2 \arctan\left(\frac{a}{\sqrt{R^2-a^2}}\right) \end{align}$$

  • $\begingroup$ Wow, it's the first time I see such computation. Very nice work. Thank you. $\endgroup$ – dh16 Dec 8 '15 at 3:34
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Dec 8 '15 at 3:37

You can compute it by substitution: set $x=R\sin t,\enspace -\frac\pi2\le x\le\frac\pi2 $, whence $\;\mathrm d\mkern1mu x=\cos t\,\mathrm d\mkern1mu t$, so the integral becomes: \begin{align*} \int\sqrt{R^2-x^2}\,\mathrm d\mkern1mu x&=\int R\cos^2t\,\mathrm d\mkern1mu t=R\int \frac{1+\cos 2t}2\,\mathrm d\mkern1mu t\\&=\frac R2\Bigl(t+\frac12\sin2t\Bigr)=\frac R2\Bigl(t+\sin t\cos t\Bigr)\\ &=\frac R2\Bigl(\arcsin \frac xR+\frac xR\sqrt{1-\frac{x^2}{R^2}}\Bigr)\\ &=\frac R2\arcsin \frac xR+\frac {x\sqrt{R^2-x^2}}{2R}. \end{align*}

  • $\begingroup$ The OP specifically requested that no trigonometric substitution be permitted. $\endgroup$ – Mark Viola Dec 7 '15 at 21:41
  • $\begingroup$ Sorry§ I misread the question. I'll delete my answer. $\endgroup$ – Bernard Dec 7 '15 at 21:44

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