show that

$$\int\limits_0^1 {\exp \left( {\frac{{4x\sqrt {1 - {x^2}} }}{{\sqrt {8{x^2} + 1} }}} \right)} \sqrt {\frac{{1 - 8{x^2} + 16{x^4}}}{{1 + 7{x^2} - 8{x^4}}}} dx = e - 1$$

I think this is nice integral,This problem is my china frend give me do it at yesterday, But I can't prove it. Thank you

my try:let $$u=\dfrac{4x\sqrt{1-x^2}}{\sqrt{8x^2+1}}$$

then $$du=-\dfrac{4(8x^4+2x^2-1)}{\sqrt{1-x^2}(8x^2+1)^{\frac{3}{2}}}$$

  • 2
    $\begingroup$ The right hand side is $\int_0^1 e^udu$. Have you tried substituting the exponent? ie. $u = \frac{4x\sqrt{1-x^2}}{\sqrt{8x^2+1}}$ $\endgroup$ Sep 18, 2013 at 8:37
  • $\begingroup$ @PrahladVaidyanathan,I have try it,But not any usefull.. $\endgroup$
    – math110
    Sep 18, 2013 at 9:34
  • $\begingroup$ Plotting the integrand is enough to convince me that you'll want to either split the integral into parts at $1/2$, and maybe use some sort of symmetry around that point. $\endgroup$
    – awwalker
    Sep 18, 2013 at 9:47
  • $\begingroup$ some people say this integral is not easy.and I think must use Integration of other methods $\endgroup$
    – math110
    Sep 18, 2013 at 9:58
  • $\begingroup$ If this function had an elementary anti-derivative, you'd have your answer by now. As it stands, it means that there's some sort of nice cancellation: perhaps the only non-elementary "part" of your integral has symmetry about $x=1/2$. $\endgroup$
    – awwalker
    Sep 18, 2013 at 17:33

1 Answer 1


Let us note that under the change of variables $y=\sqrt{\frac{1-x^2}{1+8x^2}}$ we have $x=\sqrt{\frac{1-y^2}{1+8y^2}}$ and the interval $(0,\frac12)$ is mapped to $(\frac12,1)$ and vice versa. Also, $$\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx=\frac{3(4y^2-1)dy}{(1+8y^2)\sqrt{(1-y^2)(1+8y^2)}}$$

Now our integral can be written as \begin{align} &\int_0^1e^{4xy}\frac{|1-4x^2|}{\sqrt{(1-x^2)(1+8x^2)}}dx=\\ =&\int_0^{\frac12}e^{4xy}\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx+\int_{\frac12}^1e^{4xy}\frac{4x^2-1}{\sqrt{(1-x^2)(1+8x^2)}}dx=\\ =&\int_0^{\frac12}e^{4xy}\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx+\int_0^{\frac12}e^{4xy}\frac{3(1-4y^2)}{(1+8y^2)\sqrt{(1-y^2)(1+8y^2)}}dy=\\ =&\int_0^{\frac12}e^{4xy}\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx+\int_0^{\frac12}e^{4xy}\frac{3(1-4x^2)}{(1+8x^2)\sqrt{(1-x^2)(1+8x^2)}}dx=\\ =&\int_0^{\frac12}e^{4xy}\frac{4(1-4x^2)(1+2x^2)}{(1+8x^2)\sqrt{(1-x^2)(1+8x^2)}}dx=\\ =&\int_0^{\frac12}e^{4xy}(4xy)'_xdx=\\ =&\left[\exp 4x\sqrt{\frac{1-x^2}{1+8x^2}}\,\right]_{0}^{\frac12}=\\ =&e-1. \end{align}

  • $\begingroup$ I take off my hat to you!! $\endgroup$ Sep 23, 2013 at 3:57
  • $\begingroup$ It's nice! Thank you +1 $\endgroup$
    – math110
    Sep 24, 2013 at 4:04

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