Is this integral solvable? $$\int_0^{x_0} \frac{dx}{\sqrt{2g(k-\sqrt{l^2+x^2})}}$$
I came across this while doing some physics work, I couldn't solve it using trig. Substitutions or putting $l^2+x^2=t$, and neither could wolfram alpha solve it.
 A: The integral has not a well defined primitive, in terms of simple functions. It's clearly an Elliptic functions integral, and I think it arises by some pendulum problem, doesn't it?
Anyway, I'll try to show you the way to compute it, but keep in mind that in the very last part the need of Mathematica for the "exact" solution is required. Anyway, the first steps are really trivial.
Starting with taking out of the integral what is a constant, namely $g$ for the moment:
$$\frac{1}{\sqrt{2g}}\int_0^{x_0}\ \frac{\text{d}x}{\sqrt{k - \sqrt{\ell^2 + x^2}}}$$
now substitute
$$x = \ell\sinh\phi$$ so that $$\text{d}x = \ell\cosh\phi\ \text{d}\phi$$
Your extrema will now go from $\ell$ to $\ell\sinh x_0$. Then do the math and you'll get
$$\frac{\ell}{\sqrt{2g}}\int_{\ell}^{\operatorname{arcsinh}(x_0/\ell)}\ \frac{\cosh\phi\ \text{d}\phi}{\sqrt{k - \ell\cosh\phi}}$$
Collecting $k$ and renaming $\frac{\ell}{k} = \ell_0$ you get
$$\frac{\ell}{\sqrt{2gk}}\int_{\ell}^{\operatorname{arcsinh}(x_0/\ell)}\ \frac{\cosh\phi\ \text{d}\phi}{\sqrt{1 - \ell_0\cosh\phi}}$$
And here the simple part ends.
What comes next, is an application of the Mathematica software that will lead you to obtain the result in terms of Elliptic functions. I'll give you the unbounded integral result, then it's all about substitution:
$$\color{blue}{\int\frac{\cosh\phi\ \text{d}\phi}{\sqrt{1 - \ell_0\cosh\phi}} = }$$
$$\color{blue}{ = - \frac{2i \cdot\left((\ell_0 - 1)\cdot \mathsf{E}\left[\frac{i\phi}{2}, \frac{2\ell_0}{\ell_0 - 1}\right] + \mathsf{F}\left[\frac{i\phi}{2}, \frac{2\ell_0}{\ell_0 - 1}\right]\right)\sqrt{\frac{\ell_0\cosh\phi - 1}{\ell_0 - 1}}}{\ell_0\sqrt{1 - \ell_0\cosh\phi}}}$$
where we have the Elliptic integral of the Second Kind:
$\mathsf{E}(A, B) = \int_0^A\ \sqrt{1 - B^2\sin^2\theta}\ \text{d}\theta$
and the Elliptic integral of the Fist Kind:
$\mathsf{F}(A, B) = \int_0^A\ \frac{\text{d}\theta}{\sqrt{1 - B^2\sin^2\theta}}$
Now it's your turn to evaluate the result for the given extrema.
More information about Elliptic integrals here:
https://en.wikipedia.org/wiki/Elliptic_integral
More about Elliptic Integral of the First kind here:
http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html
More about Elliptic Integral of the Second Kind here:
http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html
More about the Pendulum:
https://en.wikipedia.org/wiki/Pendulum_(mathematics)#Arbitrary-amplitude_period
