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Evaluating $$\int\sqrt{\frac{1-x^2}{1+x^2}}\mathrm dx$$

I had read the similar problem, but it doesn't work.

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mja
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  • No matter what you do, this is not going to be a pretty integral. – Shailesh Aug 26 '15 at 15:47
  • Do a trigonometric substitution. –  Aug 26 '15 at 15:49
  • $1 + tan^{2}{\theta} = sec^{2}{\theta}$ – Taylor Aug 26 '15 at 15:50
  • $1 - sin^{2}{\theta} = cos^{2}{\theta}$ – Taylor Aug 26 '15 at 15:51
  • It seems that this is non-elementary, Mathematica replies $-i E(i,\text{arsinh},(x),-1)$, where $E$ is some elliptic function. – mickep Aug 26 '15 at 16:36
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    Byrd, P. F. and Friedman, M. D., Handbook Of Elliptic Integrals For Engineers And Physicists, page 50. There is a section devoted to integrals involving $\sqrt{a^2-t^2}$ and $\sqrt{b^2+t^2}$. 214.03 is what you are looking for. The substitution that's being used is : $$ \sin^2u = \frac{t^2(a^2+b^2)}{a^2(b^2+t^2)}$$ – Ali Aug 26 '15 at 16:54
  • @Ali I had tried by your way. We obtain $\sin^2u=\dfrac{2x^2}{1+x^2},\quad \cos^2u=\dfrac{1-x^2}{1+x^2}$. But when we calculus the differentiation, it follows the complexity. – mja Aug 27 '15 at 01:06
  • @DDK: I honesty didn't try the substitution myself. I'll be glad to copy the corresponding pages from the book for you. – Ali Aug 28 '15 at 15:18

1 Answers1

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I had read the similar problem, but it doesn't work.

Of course it doesn't ! The integral you posted is nothing else than the formula for the arc length

of the $($co$)$sine function, which is rather famous for giving rise historically to the study of elliptic

integrals ! In particular, $\displaystyle\int_0^\tfrac\pi2\sqrt{1+\sin^2x}~dx~=~\int_0^\tfrac\pi2\sqrt{1+\cos^2x}~dx~=~\int_0^1\sqrt{\frac{1~{\color{red}+}~x^2}{1~{\color{red}-}~x^2}}~dx$

$=~\dfrac{\Gamma^2\bigg(\dfrac14\bigg)}{4\sqrt{2\pi}}~{\color{red}+}~\dfrac{\pi\sqrt{2\pi}}{\Gamma^2\bigg(\dfrac14\bigg)}$ . Of course, you will immediately object that this is not the integral you
posted; but this is only half-true, since we have $\displaystyle\int_0^1\sqrt{\frac{1~{\color{red}-}~x^2}{1~{\color{red}+}~x^2}}~dx~=~\dfrac{\Gamma^2\bigg(\dfrac14\bigg)}{4\sqrt{2\pi}}~{\color{red}-}~\dfrac{\pi\sqrt{2\pi}}{\Gamma^2\bigg(\dfrac14\bigg)}$ .

The reason for this lies in the fact that, in general, $\displaystyle\int_0^1\sqrt{\frac{1+x^n}{1-x^n}}~dx~=~a\cdot2^{a-1}~\bigg[\frac12~B\bigg(\frac a2,\frac a2\bigg)$

${\color{red}+}~B\bigg(\dfrac{a+1}2,\dfrac {a+1}2\bigg)\bigg]$, where $a={\color{red}+}~\dfrac1n$ , and $\displaystyle\int_0^1\sqrt[n]{\frac{1+x^2}{1-x^2}}~dx~=~a\cdot2^{a-1}~\bigg[\frac12~B\bigg(\frac a2,\frac a2\bigg)~{\color{red}-}$

${\color{red}-}~B\bigg(\dfrac{a+1}2,\dfrac {a+1}2\bigg)\bigg]$, where $a={\color{red}-}~\dfrac1n$ . See the beta and $\Gamma$ functions for more information.

Lucian
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