I would appreciate any help finding a possible closed form solution of this integral.
$$\int\sqrt{\cosh(u)-\cos(v)}\cdot e^\frac{u}{2}~du$$
Any help would be greatly appreciated!
The solution for $\cos(v)=1$ is simple...
$\int\sqrt{\cosh(u)-\cos(v)}\cdot e^\frac{u}{2}~du$$
$$\frac1{\sqrt2}\int\sqrt{e^{2u}-2e^{u}+1}~du$$
$$\frac{e^u-u}{\sqrt2}=\int(e^{u}-1})~du$$ $$ etc.
Any insight for non-trivial solutions?
Substitute $z = e^{u}, e^{u/2}du = \frac{dz}{\sqrt z}$. $$ \int \sqrt{z\frac{z + z^{-1}}{2} - z\cos v} \frac{dz}{z} = \frac{1}{\sqrt{2}}\int \frac{\sqrt{z^2 + 1 - 2z\cos v}}{z} dz = (9) = \\ = \frac{1}{\sqrt{2}}\sqrt{z^2 + 1 - 2z\cos v} - \frac{\cos v}{\sqrt{2}} \int\frac{dz}{\sqrt{z^2 + 1 - 2z\cos v}} + \frac{1}{\sqrt{2}} \int \frac{dz}{z\sqrt{z^2 + 1 - 2z\cos v}} = (1, 4) = \\ = \frac{1}{\sqrt{2}}\sqrt{z^2 + 1 - 2z\cos v} - \frac{\cos v}{\sqrt{2}} \log(\sqrt{z^2 + 1 - 2z\cos v} + z - \cos v) + {} \\ {} + \frac{1}{\sqrt{2}} \log \frac{z}{\sqrt{z^2 + 1 - 2z\cos v} - z \cos v + 1} + C $$ where (9),(1) and (4) are from here.