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How can I show that $$\int {(2 -\cos t)^{-3/2}} dt= \frac23 \left( (E(t/2|-2) + \frac{\sin t}{\sqrt{2 -\cos t}}\right)+C$$ ?

I just want to know how to kill the $\frac32$-th power and put it into usual elliptic integral. Having known the formula, I tried to use integral by part. Let $${\sqrt{2 -\cos t}}dt=-vdu,\quad \frac{\sin t}{\sqrt{2 -\cos t}}=uv.$$

Thus $du/u=-(2-\cos t)/\sin t$. But it does not work well. Any help is appreciated.

Kirby Lee
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    Elliptic integrals satisfy certain recurrence relations. If $y^2 = (1 - x^2) (1 - m x^2)$, we have $$\frac d {dx} \frac {x y} {1 - m x^2} = \frac 1 m \frac {1 - m x^2} y - \frac {1 - m} m \frac 1 {(1 - m x^2) y},$$ so $\int dx/((1 - m x^2) y)$ gives $E = \int(1 - m x^2)/y , dx$ plus an elementary function. Or, once you know the formula, you can just differentiate both sides. – Maxim Nov 23 '20 at 00:43

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To get the given form (which is returned by Wolfram Alpha), rewrite $2-\cos t=1+2\sin^2\frac t2$ and then substitute $u=\sin\frac t2,\frac{du}{dt}=\frac12\sqrt{1-u^2}$: $$\int_0^\varphi(2-\cos t)^{-3/2}\,dt=2\int_0^{\sin\varphi/2}\frac1{(1+2u^2)\sqrt{(1+2u^2)(1-u^2)}}\,du$$ This is an elliptic integral of the third kind (argument convention as in Mathematica/mpmath – arguments $n,\varphi,m$ from left to right): $$=2\Pi(-2,\varphi/2,-2)+K$$ Since $m=n=-2$, though, this can be simplified into an elliptic integral of the second kind plus another term (Byrd and Friedman 111.06, DLMF 19.6.13): $$=\frac2{1-(-2)}\left(E(\varphi/2,-2)-\frac{-2\sin\varphi/2\cos\varphi/2}{\sqrt{1-(-2)\sin^2\varphi/2}}\right)+K$$ $$=\frac23\left(E(\varphi/2,-2)+\frac{\sin\varphi}{\sqrt{2-\cos\varphi}}\right)+K$$

However, I would not be inclined to use this form since it has a negative parameter. B&F 291.01 directly gives the result as $$\int_0^\varphi(2-\cos t)^{-3/2}\,dt=\frac2{\sqrt3}E\left(\sin^{-1}\sqrt{\frac{3(1-\cos\varphi)}{2(2-\cos\varphi)}},\frac23\right)+K$$

Parcly Taxel
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  • Thanks for your answer and the reference. Would you mind giving some hint to B&F 291.01? Is it easily derived? – Kirby Lee Nov 23 '20 at 23:17
  • @KirbyLee 291.01 is just a handy formula. It can be derived from the WA one by an imaginary modulus (negative parameter) transformation. – Parcly Taxel Nov 24 '20 at 07:21