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?
 A: HINT
Change the variable $c-2u=t^2$
A: Take your integral and rewrite it as follow (I will use $x$ notation because I feel more comfortable lol):
$$\frac{1}{\sqrt{c}}\int \frac{\text{d}x}{x\sqrt{1 - bx}}$$
Where $b = 2/c$.
Now use the substitution
$$x = \frac{1}{b}\sin^2\theta$$
$$\text{d}x = \frac{2}{b}\sin\theta\cos\theta$$
To get
$$\frac{2}{\sqrt{c}}\int\frac{\sin\theta\cos\theta\ \text{d}\theta}{\sin^2\theta\cos\theta} = \frac{2}{\sqrt{c}}\int\frac{1}{\sin\theta}\ \text{d}\theta$$
This integralis trivial and its solution is
$$\int\frac{1}{\sin\theta}\ \text{d}\theta = \log \left(\sin \left(\frac{\theta}{2}\right)\right)-\log \left(\cos \left(\frac{\theta}{2}\right)\right)$$
From hereafter you can proceed by yourself I bet!
Note
If the final result doesn't match it's jut because you can express hyperbolic functions in terms of logarithms, no worries. 
A: Let $x=2u, dx=2du$.
$$\int\frac{du}{u\sqrt{c-2u}}=\int\frac{2du}{2u\sqrt{c-2u}}=\int\frac{dx}{x\sqrt{c-x}}$$
Let $v=\sqrt{c-x}, dv=\frac{-dx}{2\sqrt{c-x}}, x=c-v^2$.
$$\int\frac{dx}{x\sqrt{c-x}}=\int\frac{-2dv}{c-v^2}=-\frac{2}{c}\int\frac{dv}{1-(\frac{v}{\sqrt c})^2}=-\frac{2}{c}\mathrm{arctanh}(\frac{v}{\sqrt c})+\zeta_0$$
$$\zeta=-\frac{2}{c}\mathrm{arctanh}(\frac{v}{\sqrt c})+\zeta_0=-\frac{2}{c}\mathrm{arctanh}(\frac{\sqrt{c-2u}}{\sqrt c})+\zeta_0$$
$$u=\frac{c}{2}(1-\tanh^2(-\frac{ c}{2}(\zeta - \zeta_0)))=\frac{c}{2}\mathrm{sech}^2(-\frac{ c}{2}(\zeta - \zeta_0))=\frac{c}{2}\mathrm{sech}^2(\frac{ c}{2}(\zeta - \zeta_0))$$
A: WLOG, $c=2$. Then substitute $u=\cos^2t$ so that
$$\zeta=\int\frac{du}{u\sqrt{1-u}}=-\int\frac{2\cos t\sin t}{\cos^2t\sin t}dt=-2\int\frac{\cos t\,dt}{1-\sin^2t}=-2\text{ artanh}(\sin t)\\
=-2\text{ artanh}\sqrt{1-u}+C.$$
Now you can invert the relation.
A: 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 $u = c-2 u$:
$$-\frac{2tanh^{-1}(\frac{\sqrt{c-2u}}{\sqrt{c}})}{\sqrt{c}}$$
