How to integrate $\int\frac{du}{u\sqrt{c-2u}} $

Question

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?

Answer 1

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 $v=\sqrt{c-x}, dv=\frac{-dx}{2\sqrt{c-x}}, x=c-v^2$. $$\int\frac{dx}{x\sqrt{c-x}}=\int\frac{-2dv}{c-v^2}=-\frac{2}{c}\int\frac{dv}{1-(\frac{v}{\sqrt c})^2}=-\frac{2}{c}\mathrm{arctanh}(\frac{v}{\sqrt c})+\zeta_0$$ $$\zeta=-\frac{2}{c}\mathrm{arctanh}(\frac{v}{\sqrt c})+\zeta_0=-\frac{2}{c}\mathrm{arctanh}(\frac{\sqrt{c-2u}}{\sqrt c})+\zeta_0$$ $$u=\frac{c}{2}(1-\tanh^2(-\frac{ c}{2}(\zeta - \zeta_0)))=\frac{c}{2}\mathrm{sech}^2(-\frac{ c}{2}(\zeta - \zeta_0))=\frac{c}{2}\mathrm{sech}^2(\frac{ c}{2}(\zeta - \zeta_0))$$

Answer 2

HINT

Change the variable $c-2u=t^2$

Answer 3

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.

Answer 4

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.

Answer 5

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 $u = c-2 u$:

$$-\frac{2tanh^{-1}(\frac{\sqrt{c-2u}}{\sqrt{c}})}{\sqrt{c}}$$