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I have been trying to find the arc-length of $\sin^{-1}(x)$ over $[0,1]$.

Of course, it is given by the integral $$J=\int_0^1\sqrt{1+\frac1{1-x^2}}\ dx=\int_0^1 \sqrt{\frac{2-x^2}{1-x^2}}\, dx$$ To compute this, I used $x\mapsto \cos x$: $$J=\int_0^{\pi/2}\sqrt{\frac{1+1-\cos^2x}{1-\cos^2x}}\, \sin(x)dx=\int_0^{\pi/2}\sqrt{1+\sin^2x}\,dx=\mathrm{E}(-1)$$ where $$\mathrm{E}(k)=\int_0^{\pi/2}\sqrt{1-k\sin^2x}\,dx$$ is the complete elliptic integral of the second kind.

I am usually one to accept values of $\mathrm{E}$ as closed forms by themselves, but by chance, Wolfram kindly provided the explicit evaluation $$\mathrm{E}(-1)=\frac{\pi\sqrt{2\pi}}{\Gamma^2(1/4)}+\frac{\Gamma^2(1/4)}{4\sqrt{2\pi}}\tag{1}$$ Which I would like to know how to prove.

I was able to immediately recognize that $$\frac{\Gamma^2(1/4)}{4\sqrt{2\pi}}=\frac{1}{4\sqrt{2}}\int_0^1\frac{dx}{x^{3/4}(1-x)^{3/4}}$$ But I cannot seem to find an integral representation for the other chunk, namely $\frac{\pi\sqrt{2\pi}}{\Gamma^2(1/4)}$. I suspect that the solution involves the Jacobi elliptic functions and the Jacobi theta functions, but I have no idea how to use them. Could I have some help proving $(1)$? Thanks.

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  • $\begingroup$ This should be exactly the same as the length of $\sin$ over $[0,\pi/2]$, that is, $\int_0^{\pi/2}\sqrt{1+\cos^2x}dx$. Just to shorten your first computations. $\endgroup$ – ajotatxe Jun 6 at 16:52
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This is conventionally $\mathrm{E}(i)$, not $\mathrm{E}(-1)$. Its evaluation begins with $$\mathrm{E}(i)=\int_0^1\sqrt\frac{1+x^2}{1-x^2}\,dx=\int_0^1\frac{1+x^2}{\sqrt{1-x^4}}\,dx=\frac{1}{4}\mathrm{B}\Big(\frac{1}{4},\frac{1}{2}\Big)+\frac{1}{4}\mathrm{B}\Big(\frac{3}{4},\frac{1}{2}\Big)=\ldots$$

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  • $\begingroup$ okay that was way easier than I thought. $\endgroup$ – clathratus Jun 6 at 17:13

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