How to integrate $\int\frac{du}{u\sqrt{c-2u}} $ I was looking through these notes, and at the top of the second page it says you can integrate
$$\mathrm{d}\zeta = \frac{\mathrm{d}u}{u\sqrt{c-2u}} $$
to get
$$u(\zeta) = \frac{c}{2}\mathrm{sech}^2(\frac{1}{2}\sqrt{c}(\zeta - \zeta_0)).$$
I can't see where that comes from. Is there any trig identity I should use?

Take the integral:
$$\int \frac{1}{u \sqrt{c-2 u}} \ du$$
For the $$\int \frac{1}{u \sqrt{c-2 u}}$$, substitute $u = c-2 u$ and $du = -2 du$. Then we have : 
 $$\int \frac{1}{\sqrt{u} (u-c)} \ du$$
For the integrand $\frac{1}{\sqrt{u} (u-c)}$, substitute $s = \sqrt{u}$ and $ds = \frac{1}{2 \sqrt{u}} du$, then we have :
$$2 \int \frac{1}{s^{2}-c} ds$$
Factor $-c$ from the denominator, then :
 $$ 2 \int \frac{-1}{c (1-\frac{s^2}{c})} ds$$
Factor out constants, then :
 $$\frac{-2}{c} \int \frac{1}{(1-\frac{s^2}{c})} ds$$
For the integrand $\frac{1}{(1-\frac{s^2}{c})}$, substitute $p = \frac{s}{\sqrt{c}}$ and $dp = \frac{1}{\sqrt{c}} ds$:
$$=\frac{-2}{\sqrt{c}} \int \frac{1}{(1-p^{2})} dp$$
The integral of $\frac{1}{(1-p^2)}$ is $tanh^{-1}(p)$, then :
 = $$-\frac{2 tanh^{-1}(p))}{\sqrt{c}}+constant$$
Substitute back for $p = \frac{ s}{\sqrt{c}}$:
= $$-\frac{2tanh^{-1}(\frac{s}{\sqrt{c}})}{\sqrt{c}}$$
Substitute back for $s = \sqrt{u}$:
= = $$-\frac{2tanh^{-1}(\frac{\sqrt{u}}{\sqrt{c}})}{\sqrt{c}}$$
Substitute back for $u = c-2 u$:
$$-\frac{2tanh^{-1}(\frac{\sqrt{c-2u}}{\sqrt{c}})}{\sqrt{c}}$$
