Elliptic integral into usual form 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.
 A: 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
$$\int_0^\varphi(2-\cos t)^{-3/2}\,dt=\frac2{\sqrt3}E\left(\sin^{-1}\sqrt{\frac{3(1-\cos\varphi)}{2(2-\cos\varphi)}},\frac23\right)+K$$
