TITLE: integrating $\int_{x_0}^{x}\arccos(a-\cos(x'))dx'$
QUESTION [5 upvotes]: I've been looking at an integral of the form 
$$\int_{x_0}^{x}\arccos(a-\cos(x'))dx'.$$ We can set $x_0$ as to give no contribution from the lower limit. I've tried to set $\arccos(x)=\pi/2-\arcsin(x)$, to simplify it.
I achieve the form
$$\int\frac{v\cos(v)}{\sqrt{1-(a-\sin(v))^2}}dv,$$ by setting $\sin(v)=a-\cos(x)$, I have also ignored the limits to focus on the integrand. I have looked at the table of integrals and series to see if there was a solution for this integral, but no luck. Does anybody possibly know a technique for solving it?

REPLY [2 votes]: I'll be more careful and explicitly write the limits every time.
$$I(a,b,c)=\int_b^c \arccos(a-\cos(y))dy$$
$$y=\arccos z$$
$$dy = -\frac{dz}{\sqrt{1-z^2}}$$
$$I=\int_{\cos b}^{\cos c} \arccos(a-z)\frac{dz}{\sqrt{1-z^2}}$$
$$z=a-u$$
$$u=a-z$$
$$I=\int_{a-\cos c}^{a-\cos b} \arccos(u)\frac{du}{\sqrt{1-(a-u)^2}}$$
$$u=\cos v$$
$$v=\arccos u$$
$$du=- \sin v$$
$$I=\int_{\arccos(a-\cos b)}^{\arccos(a-\cos c)} \frac{v \sin v dv}{\sqrt{1-(a-\cos v)^2}}$$
Yes, this is not fun.

Let's get a little back:
$$I=\int_{a-\cos c}^{a-\cos b} \arccos(u)\frac{du}{\sqrt{1-(a-u)^2}}$$
$$I=\frac{\pi}{2}\int_{a-\cos c}^{a-\cos b} \frac{du}{\sqrt{1-(a-u)^2}}-\int_0^1 \int_{a-\cos c}^{a-\cos b} \frac{u du dt}{\sqrt{1-u^2 t^2}\sqrt{1-(a-u)^2}}$$
First integral is elementary, second integral w.r.t. u is most likely incomplete elliptic integral. Some transformations are required, but this is what I would suggest. Integrating w.r.t. $u$ first, using special functions, then taking the $t$ integral.
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