# How to prove that $\int_0^1 \exp\left(\frac{4x\sqrt{1-x^2}}{\sqrt{8x^2+1}}\right)\sqrt{\frac{1-8x^2+16x^4}{1+7x^2-8x^4}}dx=e-1$

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

• The right hand side is $\int_0^1 e^udu$. Have you tried substituting the exponent? ie. $u = \frac{4x\sqrt{1-x^2}}{\sqrt{8x^2+1}}$ Sep 18, 2013 at 8:37
• @PrahladVaidyanathan,I have try it,But not any usefull.. Sep 18, 2013 at 9:34
• 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. Sep 18, 2013 at 9:47
• some people say this integral is not easy.and I think must use Integration of other methods Sep 18, 2013 at 9:58
• 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 $x=1/2$. Sep 18, 2013 at 17:33

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, $$\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx=\frac{3(4y^2-1)dy}{(1+8y^2)\sqrt{(1-y^2)(1+8y^2)}}$$