# Showing that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left (\sqrt{a^2+1}-1\right)$.

How can I show that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?

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The indefinite integral is possible but is very complex (both that is uses $i$ and it is complicated). Wolfram alpha or other software can solve this directly with enough time. – kaine Jan 18 '13 at 20:16
Nice question (+1) – I'm an artist Jan 18 '13 at 21:38

Putting $x=a\sin\theta,dx=a\cos\theta d\theta$ and $x=\pm a,\theta=\pm\frac\pi2$

$$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx =\int _{-\frac\pi2}^{\frac\pi2}\frac{a^2\cos^2\theta}{1+a^2\sin^2\theta}d\theta$$ $$=\int _{-\frac\pi2}^{\frac\pi2}\frac{a^2\sec^2\theta}{(1+\tan^2\theta)(1+(a^2+1)\tan^2\theta)}d\theta$$ (Diving the numerator & the denominator by $\sec^4\theta$)

$$=\int _{-\infty}^{\infty}\frac{a^2}{(1+t^2)(1+(a^2+1)t^2)}dt$$ (Putting $\tan\theta = t$ as $\tan\theta=\pm\infty, t=\pm\frac\pi2$)

$$=\frac{a^2}{(a^2+1)}\int _{-\infty}^{\infty}\frac{1}{(1+t^2)(\frac1{(a^2+1)}+t^2)}dt$$

$$=\frac{\frac{a^2}{(a^2+1)}}{\left(1-\frac1{1+a^2}\right)}\left(\int _{-\infty}^{\infty}\frac1{(\frac1{(a^2+1)}+t^2)}dt-\int _{-\infty}^{\infty}\frac{1}{(1+t^2)}dt\right)$$ as $\frac1{(c+y)(b+y)}=\frac1{c-b}\frac{(c+y)-(b+y)}{(c+y)(b+y)}=\frac1{c-b}\left(\frac1{y+b}-\frac1{y+c}\right)$

$$=\left(\sqrt{1+a^2}\arctan (t\sqrt{1+a^2} )-\arctan t\right)_{-\infty}^{\infty}$$

So,$$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx=(\sqrt{1+a^2}-1)\frac\pi2-\left\{-(\sqrt{1+a^2}-1)\frac\pi2\right\}=(\sqrt{1+a^2}-1)\pi$$

Observe that we have put $x=a\sin\theta$ and $\tan\theta = t\implies \left(\frac ax\right)^2- \left(\frac 1t\right)^2=\csc^2\theta-\cot^2\theta=1$

$\implies t^2=\frac{x^2}{a^2-x^2}\iff x^2=\frac{a^2t^2}{1+t^2}, a^2-x^2=\frac{a^2}{1+t^2}$

So, if we straight away take $t=\frac x{\sqrt{a^2-x^2}}$ (assuming $a>0$)

$$\frac{dt}{dx}=\frac1{\sqrt{a^2-x^2}}+x\left(\frac{-1}2\right)\frac1{(a^2-x^2)^{\frac32}}(-2x)=\frac{a^2}{(a^2-x^2)^{\frac32}}$$

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

and if $x=\pm a,t=\frac{\pm a}0=\pm\infty$

So, $$\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx$$ becomes $$\int _{-\infty}^{\infty}\frac{a^2}{(1+t^2)(1+(a^2+1)t^2)}dt$$

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i thought it was simple integral ... – Santosh Linkha Jan 18 '13 at 20:31
Great answer, thank you! – Guest 86 Jan 18 '13 at 20:50
@Guest86, my pleasure. Hope I could make approach clear. – lab bhattacharjee Jan 19 '13 at 8:06

A slight variant of lab bhattacharjee's method provides a simpler solution:

Let

$$I(a) = \int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx = 2\int_{0}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx.$$

Then by a simple application of multivariable calculus,

\begin{align*} I'(a) &= 2 \left. \frac{\sqrt{a^2-x^2}}{1+x^2} \right|_{x=a} + 2 \int_{0}^{a} \frac{d}{da} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx \\ &= 2 \int_{0}^{a} \frac{a}{(1+x^2)\sqrt{a^2-x^2}} \, dx. \end{align*}

Then with the change of variable $x = a \sin\theta$, we have

\begin{align*} I'(a) &= 2 \int_{0}^{\frac{\pi}{2}} \frac{a}{1+a^2\sin^2\theta} \, d\theta \\ &= 2 \int_{0}^{\frac{\pi}{2}} \frac{a \sec^2\theta}{1+(a^2+1)\tan^2\theta} \, d\theta \\ &= 2 \int_{0}^{\infty} \frac{a}{1+(a^2+1)t^2} \, dt \qquad (t = \tan\theta) \\ &= \frac{\pi a}{\sqrt{a^2+1}}. \end{align*}

Thus by integrating, we must have

$$I(a) = \pi\sqrt{a^2+1} + C$$

for some constant $C$. But

$$I(0+) = \lim_{a\to0} \int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx = \lim_{a\to0} \int_{-1}^{1} \frac{\sqrt{1-x^2}}{(1/a)^2+x^2} \, dx = 0$$

and we must have $C = -\pi$. This proves the identity.

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$\newcommand{\Res}{\operatorname{Res}}$ Since the question was tagged complex-analysis, and nobody has given a solutions with purely complex methods, here goes.

Let $$f(z) = \frac{(a^2-z^2)^{1/2}}{1+z^2},$$ where $(a^2-z^2)^{1/2}$ denotes branch that is holomorphic on $\mathbb{C} \setminus [-a,a]$. (See this question for details.)

Next, let $\Gamma$ be a "dog bone" contour together with a large circle:

and integrate $f$ along $\Gamma$. On the "top" part of $[-a,a]$ we get the integral that we want. On the "bottom" part, the square root will pick up a minus sign from the branch cut and another minus sign from the orientation. It's straight forward to check that the integrals over the small circles tend to $0$ as their radii tend to $0$, and the integral over the large circle is basically the residue of $f$ at $\infty$. More precisely, by the residue theorem

\begin{align} 2\int_{-a}^a \frac{\sqrt{a^2-x^2}}{x^2+1}\,dx &= 2\pi i( \Res(f;i) + \Res(f;-i) - \Res(f;\infty)) \\ &= 2\pi i \bigg( \frac{\sqrt{a^2+1}}{2i} + \frac{-\sqrt{a^2+1}}{-2i} + i\bigg) \end{align}

which simplifies to the stated equality. (Note that $\Res(f;\infty) = \Res(-\dfrac1{z^2}f(\dfrac1z);0)$.)

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What a clean solution! I'm reading up on some of your previous answers, really cool, thanks! =) – Guest 86 Jan 21 '13 at 16:23
@Guest86 happy to help, it's not often that I get a chance to integrate along the cute dog bone contour. Btw, don't forget to upvote answers you find helpful. (Not just mine...) – mrf Jan 21 '13 at 16:46
I've ended up up-voting every single answer here (Just that reading them took me a while) – Guest 86 Jan 21 '13 at 21:43

Let $x = a \sin(y)$. Then we have $$\dfrac{\sqrt{a^2-x^2}}{1+x^2} dx = \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy$$ Hence, $$I = \int_{-a}^{a}\dfrac{\sqrt{a^2-x^2}}{1+x^2} dx = \int_{-\pi/2}^{\pi/2} \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy$$ Hence, $$I + \pi = \int_{-\pi/2}^{\pi/2} \dfrac{a^2 \cos^2(y)}{1+a^2 \sin^2(y)} dy + \int_{-\pi/2}^{\pi/2} dy = \int_{-\pi/2}^{\pi/2} \dfrac{1+a^2}{1+a^2 \sin^2(y)} dy\\ = \dfrac{1+a^2}2 \int_0^{2 \pi} \dfrac{dy}{1+a^2 \sin^2(y)}$$ Now $$\int_0^{2 \pi} \dfrac{dy}{1+a^2 \sin^2(y)} = \oint_{|z| = 1} \dfrac{dz}{iz \left(1 + a^2 \left(\dfrac{z-\dfrac1z}{2i}\right)^2 \right)} = \oint_{|z| = 1} \dfrac{4z^2 dz}{iz \left(4z^2 - a^2 \left(z^2-1\right)^2 \right)}$$ $$\oint_{|z| = 1} \dfrac{4z^2 dz}{iz \left(4z^2 - a^2 \left(z^2-1\right)^2 \right)} = \oint_{|z| = 1} \dfrac{4z dz}{i(2z + a(z^2-1))(2z - a(z^2-1))}$$ Now $$\dfrac{4z}{(2z + a(z^2-1))(2z - a(z^2-1))} = \dfrac1{az^2 - a + 2z} - \dfrac1{az^2 - a - 2z}$$ $$\oint_{\vert z \vert = 1} \dfrac{dz}{az^2 - a + 2z} = \oint_{\vert z \vert = 1} \dfrac{dz}{a \left(z + \dfrac{1 + \sqrt{1+a^2}}a\right) \left(z + \dfrac{1 - \sqrt{1+a^2}}a\right)} = \dfrac{2 \pi i}{2 \sqrt{1+a^2}}$$ $$\oint_{\vert z \vert = 1} \dfrac{dz}{az^2 - a - 2z} = \oint_{\vert z \vert = 1} \dfrac{dz}{a \left(z - \dfrac{1 + \sqrt{1+a^2}}a\right) \left(z - \dfrac{1 - \sqrt{1+a^2}}a\right)} = -\dfrac{2 \pi i}{2 \sqrt{1+a^2}}$$ Hence, $$\oint_{|z| = 1} \dfrac{4z dz}{i(2z + a(z^2-1))(2z - a(z^2-1))} = \dfrac{2 \pi i}i \dfrac1{\sqrt{1+a^2}} = \dfrac{2 \pi}{\sqrt{1+a^2}}$$ Hence, we get that $$I + \pi = \left(\dfrac{1+a^2}2\right) \dfrac{2 \pi}{\sqrt{1+a^2}} = \pi \sqrt{1+a^2}$$ Hence, we get that $$I = \pi \left(\sqrt{1+a^2} - 1 \right)$$

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Funny how the y'=2y+pi took me a while to get... The +pi idea was very cool! :D I wish I could up-vote this solution... – Guest 86 Jan 18 '13 at 21:07
@Marvis: here is an alternative to the integral where you employed complex analysis: $\frac{4}{(a^2+1)\cos^2 y}\int_0^{\pi/2} \frac{\mathrm{dy}}{\tan^2(y)+1/(a^2+1)} =\frac{4}{(a^2+1) }\int_0^{\infty} \frac{\mathrm{du}}{u^2+(\sqrt{1/(a^2+1)})^2}= \left[\frac{4\sqrt{a^2+1}}{a^2+1} \arctan(\sqrt{a^2+1} u)\right]_0^{\infty}=\frac{2\pi}{\sqrt{a^2+1}}$ – I'm an artist Jan 18 '13 at 21:27
@Chris'ssister Ah, yes. I realized there must be some nice trick like this but used the work-horse for such problems(complex analysis). Thanks. – user17762 Jan 18 '13 at 21:31

You can also convert it into an integral over the upper half-disk $D_a$ of radius $a$: $$\int_{D_a} {1 \over 1 + x^2}\,dy\,dx$$ (Do the $y$ integral first to convert it into the original form.) Since the integrand is even, it is twice the integral over the portion where $x > 0$. Changing to polar coordinates, this becomes $$2\int_0^a \int_0^{\pi \over 2} {r \over 1 + r^2\cos^2(\theta)}\,d\theta\,dr$$ $$= 2\int_0^a r \int_0^{\pi\over 2} {\sec^2(\theta) \over \sec^2(\theta) + r^2}\,d\theta\,dr$$ $$= 2\int_0^a r \int_0^{\pi\over 2} {\sec^2(\theta) \over \tan^2(\theta) + (1 + r^2)}\,d\theta\,dr$$ Doing a $u$ substitution this becomes $$= 2\int_0^a r \int_0^{\infty} {1 \over u^2 + (1 + r^2)}\,du\,dr$$ $$= 2\pi\int_0^a r {1 \over \sqrt{1 + r^2}}\,dr$$ $$=\pi \sqrt{1 + r^2}\,\,\,\bigg|_{r = 0}^{r = a}$$ $$=\pi\sqrt{1 + a^2} - \pi$$

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Interesting idea, Thanks! Several answers used the 1/cos^2 trick, is it a well known thing? (There's an unneeded 2 in the last integral expression) – Guest 86 Jan 21 '13 at 21:39