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As title, how could one calculate $$\int_0^1 \frac{\sqrt{1-x^4}}{1+x^2}dx$$?

I thought of using $x=\sqrt{\sin u}$ substitution but to no avail.

  • 0
    It involves elliptic integrals. Or the gamma function. It is $$2K(-1)-E(-1)$$ – 2017-02-22
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    Does $\frac{\sqrt{1-x^4}}{1+x^2}=\sqrt{\frac{1-x^2}{1+x^2}}$ help? – 2017-02-22
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    this does not help since the integral leads to an elliptic function – 2017-02-22

2 Answers 2

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By substituting $x=\tan\theta$ we are left with $$ I=\int_{0}^{\pi/4}\sqrt{(1-\tan^2\theta)(1+\tan^2\theta)}\,d\theta =\int_{0}^{\pi/4}\frac{\sqrt{\cos(2\theta)}}{\cos^2\theta}\,d\theta=\int_{0}^{\pi/2}\frac{\sqrt{\cos\theta}}{1+\cos\theta}\,d\theta\tag{1}$$ that can be written in the following form: $$ I=\int_{0}^{1}\frac{\sqrt{x}}{(1+x)\sqrt{1-x^2}}\,dx = \int_{0}^{1}x^{1/2}(1-x^2)^{-3/2}\,dx-\int_{0}^{1}x^{5/2}(1-x^2)^{-3/2}\,dx\tag{2}$$ or in the following form: $$ \frac{1}{2}\left[\int_{0}^{1}x^{-1/4}(1-x)^{-3/2}\,dx-\int_{0}^{1}x^{3/4}(1-x)^{-3/2}\,dx\right]\tag{3}$$ that by Euler's Beta function equals:

$$ I = \color{red}{\frac{\sqrt{\pi}}{2}\left[\frac{\Gamma\left(\frac{1}{4}\right)^2}{\pi\sqrt{8}}-\frac{\pi\sqrt{8}}{\Gamma\left(\frac{1}{4}\right)^2}\right]}\tag{4}$$

that can be numerically evaluated very efficiently through the AGM :

$$ I = \color{red}{\frac{\sqrt{\pi}}{2}\left[\frac{\sqrt{\pi}}{\text{AGM}(1,\sqrt{2})}-\frac{\text{AGM}(1,\sqrt{2})}{\sqrt{\pi}}\right]}=\frac{\pi}{2\,\text{AGM}(1,\sqrt{2})}-\frac{\text{AGM}(1,\sqrt{2})}{2}. \tag{5}$$

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Let $z=x^{2}$ \begin{align} \int\limits_{0}^{1} \frac{\sqrt{1-x^{4}}}{1+x^{2}} dx &= \int\limits_{0}^{1} (1-x^{2})^{1/2}(1+x^{2})^{1/2} dx \\ &= \frac{1}{2} \int\limits_{0}^{1} (1-z)^{1/2} (1+z)^{-1/2} z^{-1/2} dz \\ &= \frac{\Gamma(1/2) \Gamma(3/2)}{2 \Gamma(2)} {}_{2}\mathrm{F}_{1}(1/2,1/2;2;-1) \\ &\approx 0.7119586 \end{align}

We used Gauss's hypergeometric function.