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I have that $l/r = 1+e.\cos(x)$, for $l = a(1-e^2)$ (constant). The question asks for the mean distance over angle of the planet from the sun, where the planet moves on an elliptical orbit with the sun at a focus. and gives the formula (and answer):
$$\frac{1}{2\pi} \int^{2\pi}_0 r \, dx = a(1-e^2)^{\frac{1}{2}}$$

I know I am meant to use the substitution $t = \tan(x/2)$, and then use a substitution again later, but my integral results in a function arctan, and I don't know how to get a constant from that.

Any help much appreciated, thank you.

AMath
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    Can you please clarify? Total distance of what? That's not the correct expression for the arclength of an ellipse, which involves elliptic integrals; see here, for example: https://math.stackexchange.com/questions/433094/how-to-determine-the-arc-length-of-ellipse – Hans Lundmark Nov 13 '18 at 08:16
  • Sorry - the question I have (undergraduate Physics) this (I missed a 2pi): – AMath Nov 13 '18 at 08:34
  • \int^{2\pi}_0 r dx = 2\pi a(1-e^2)^{\frac{1}{2}} – AMath Nov 13 '18 at 08:34
  • Where x is the angle. Sorry - can't work out how to add a picture, hence the latex type! – AMath Nov 13 '18 at 08:34
  • But $\int r , dx$ is not the correct formula for arc length in polar coordinates: https://math.stackexchange.com/questions/760757/arc-length-in-polar-coordinates-why-isnt-ds-r%C3%97d-theta. So what is it that you are trying to compute, really...? – Hans Lundmark Nov 13 '18 at 09:35
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    Oh! The original question asks for the mean distance over angle of the planet from the sun and gives the formula (and answer): – AMath Nov 13 '18 at 11:08
  • \frac{1}{2\pi} \int^{2\pi}_0 r dx = a(1-e^2)^{\frac{1}{2}} – AMath Nov 13 '18 at 11:09
  • Planet moves on an elliptical orbit with the sun at a focus. – AMath Nov 13 '18 at 11:10
  • Here are some answers that might help (or did you get that far already?): https://math.stackexchange.com/questions/1740458/finding-int-fracdxab-cos-x-without-weierstrass-substitution – Hans Lundmark Nov 13 '18 at 12:09
  • Thank you for your help. I did read all of these. I found a lot on using expansions etc, but the instruction was to use the tan double angle formulae to do a substitution. I thought I did all of that correctly, but ended up with a(1+e)^{\frac{1}{2}}(1-e)^{\frac{1}{2}}arctan(\frac{2\pi\sqrt{1-e}}{1+e}}. – AMath Nov 13 '18 at 12:16

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