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Arc Length of the Sine Function

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Post Tue Nov 29, 2016 6:17 pm

Posts: 10
On the Wikipedia page for the sine function the arc length for a full period of $\sin(x)$ is given as

Image


I'm trying to derive this result and so far I have gotten to the integral

$$L=4\int_0^1\sqrt{\frac{1+t^2}{1-t^2}}dt.$$

I'm aware that this is a complete elliptic integral of the second kind and that $\frac{\Gamma^2\left(1/4\right)}{\sqrt{2\pi}}$ is the Lemniscate Constant, but that is really all I know. My guesses of how to continue so far are to try to transform the integral into lemniscatic integrals and apply the beta function. I'm stuck in a rut because I don't know any identities that can help me go on, so if you guys have any hints, substitutions, transformations, or identities to share that would be great.
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Post Tue Nov 29, 2016 10:39 pm

Posts: 27
The substution $u = t^4$ transforms the integral into:

\(\displaystyle \large \int_0^1 \frac{u^{-\frac{3}{4}} + u^{-\frac{1}{4}}}{\sqrt{1-u}} \, \text{d}u\)

The standard definition of the Beta Function can be applied immediately to yield the desired result.

Post Tue Nov 29, 2016 10:42 pm

Posts: 10
Paradoxica wrote:
The substution $u = t^4$ transforms the integral into:

\(\displaystyle \large \int_0^1 \frac{u^{-\frac{3}{4}} + u^{-\frac{1}{4}}}{\sqrt{1-u}} \, \text{d}u\)

The standard definition of the Beta Function can be applied immediately to yield the desired result.



Thankyou! I can't believe I didn't see that! I had tried $u=t^2$ and didn't get anywhere.
The internet is a $\sum^{\infty}_{n=0}$tube$_n$ filled with cats!




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