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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?

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  • $\begingroup$ Is $v$ a parameter or a typo? $\endgroup$ – uranix Jun 13 '15 at 20:18
  • $\begingroup$ If i did not mention it (u,v) are real variables $\endgroup$ – T. Poindexter Jun 13 '15 at 20:18
  • $\begingroup$ No "v" is not a typo, treating cos(v) as a constant in this calculation, not sure y there is a down vote, really need help on this one. If it is not possible to be evaluated , pls let me know and ill use numerical approximations instead of wasting time. $\endgroup$ – T. Poindexter Jun 13 '15 at 20:19
  • $\begingroup$ Have you tried Wolfram Alpha? $\endgroup$ – uranix Jun 13 '15 at 20:20
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    $\begingroup$ $$\int e^{u/2} \sqrt{\cosh (u)-a} \, du=\frac{e^{u/2} \sqrt{-2 a+e^{-u}+e^u} \left(\sqrt{-2 a e^u+e^{2 u}+1}-a \log \left(\sqrt{-2 a e^u+e^{2 u}+1}-a+e^u\right)-\log \left(a \left(-e^u\right)+\sqrt{-2 a e^u+e^{2 u}+1}+1\right)+u\right)}{\sqrt{2} \sqrt{-2 a e^u+e^{2 u}+1}}$$ is what mathematica gives $\endgroup$ – grdgfgr Jun 13 '15 at 20:32
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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.

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