# Thread: Problems trying to solve an integral

1. ## Problems trying to solve an integral

Hi everyone,
I need to solve this integral, but I get stuck in some point and I can't go on. I explain what I have been trying:

$\displaystyle \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du$

Changing the variable as

$\displaystyle t^2=\displaystyle\frac{u^2-x^2}{u^2-a^2}$

So "u" would be:

$\displaystyle t^2(u^2-a^2)=u^2-x^2$ Desarrollando esta expresión,llegamos a que:

$\displaystyle u^2(t^2-1)=t^2a^2-x^2$ Por lo que $\displaystyle u=\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}$

The derivative of u with respect to: $\displaystyle \frac{du}{dt}$

$\displaystyle \displaystyle\frac{du}{dt}=\displaystyle\frac{1}{2 }\displaystyle\frac{1}{\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}}\displaystyle\frac{(t^2-1)(2ta^2)-(t^2a^2-x^2)(2t)}{(t^2-1)^2}$

$\displaystyle \displaystyle\frac{du}{dt}=\displaystyle\frac{1}{2 }\displaystyle\frac{1}{\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}}\displaystyle\frac{2t(x^2-a^2)}{(t^2-1)^2}$

Replacing these terms in the original integral we have:

$\displaystyle \displaystyle\int\sqrt{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du=\displaystyle\int\sqrt{t^2}\dis \frac{a}{\sqrt{\frac{t^2a^2-x^2}{t^2-1}}}\frac{1}{\sqrt{\frac{t^2a^2-x^2}{t^2-1}}}\frac{2t(x^2-a^2)}{(t^2-1)^2}dt$

$\displaystyle \displaystyle\int\frac{t^2(x^2-a^2)a}{\left(\dfrac{t^2a^2-x^2}{t^2-1}\right)(t^2-1)^2}dt$

When I get to this point, I don't know how to go on. The result of the integral would be like this:

$\displaystyle \displaystyle\int\frac{t^2(x^2-a^2)a}{\left(\dfrac{t^2a^2-x^2}{t^2-1}\right)(t^2-1)^2}dt=\frac{a}{2}\ln\left(\frac{t+1}{t-1}\right)-\frac{x}{2}\ln\left(\frac{t+(x/a)}{t-(x/a)}\right)$

I tried operating in other way, like this:

$\displaystyle \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du$

Doing the change of variable $\displaystyle t^2=\displaystyle\frac{u^2-x^2}{u^2-a^2}$ we have this:

$\displaystyle 2tdt=\displaystyle\frac{2u(u^2-a^2)-2u(u^2-x^2)}{(u^2-a^2)^2}du=u\displaystyle\frac{(x^2-a^2)}{(u^2-a^2)^2}du$

$\displaystyle \displaystyle\frac{u}{(u^2-a^2)^2}du=\displaystyle\frac{t}{(x^2-a^2)}dt$

But again, I don't know how to continue.

Could you tell me what could I do to go on with the integral?

2. ## Re: Problems trying to solve an integral

If you factor the denominator $\displaystyle u^2-a^2$ into $\displaystyle u^2\left(1-\left(\frac{a}{u}\right)^2\right)$, you can play with some change of variables, since you have a $\displaystyle a/u$ outside the root. But I don't know if this is going to work. Just my 2 cents.

3. ## Re: Problems trying to solve an integral

Originally Posted by nekovolta
Hi everyone,
I need to solve this integral, but I get stuck in some point and I can't go on. I explain what I have been trying:

$\displaystyle \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du$

Changing the variable as

$\displaystyle t^2=\displaystyle\frac{u^2-x^2}{u^2-a^2}$

So "u" would be:

$\displaystyle t^2(u^2-a^2)=u^2-x^2$ Desarrollando esta expresión,llegamos a que:

$\displaystyle u^2(t^2-1)=t^2a^2-x^2$ Por lo que $\displaystyle u=\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}$

The derivative of u with respect to: $\displaystyle \frac{du}{dt}$

$\displaystyle \displaystyle\frac{du}{dt}=\displaystyle\frac{1}{2 }\displaystyle\frac{1}{\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}}\displaystyle\frac{(t^2-1)(2ta^2)-(t^2a^2-x^2)(2t)}{(t^2-1)^2}$

$\displaystyle \displaystyle\frac{du}{dt}=\displaystyle\frac{1}{2 }\displaystyle\frac{1}{\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}}\displaystyle\frac{2t(x^2-a^2)}{(t^2-1)^2}$

Replacing these terms in the original integral we have:

$\displaystyle \displaystyle\int\sqrt{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du=\displaystyle\int\sqrt{t^2}\dis \frac{a}{\sqrt{\frac{t^2a^2-x^2}{t^2-1}}}\frac{1}{\sqrt{\frac{t^2a^2-x^2}{t^2-1}}}\frac{2t(x^2-a^2)}{(t^2-1)^2}dt$

$\displaystyle \displaystyle\int\frac{t^2(x^2-a^2)a}{\left(\dfrac{t^2a^2-x^2}{t^2-1}\right)(t^2-1)^2}dt$

When I get to this point, I don't know how to go on. The result of the integral would be like this:

$\displaystyle \displaystyle\int\frac{t^2(x^2-a^2)a}{\left(\dfrac{t^2a^2-x^2}{t^2-1}\right)(t^2-1)^2}dt=\frac{a}{2}\ln\left(\frac{t+1}{t-1}\right)-\frac{x}{2}\ln\left(\frac{t+(x/a)}{t-(x/a)}\right)$

I tried operating in other way, like this:

$\displaystyle \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du$

Doing the change of variable $\displaystyle t^2=\displaystyle\frac{u^2-x^2}{u^2-a^2}$ we have this:

$\displaystyle 2tdt=\displaystyle\frac{2u(u^2-a^2)-2u(u^2-x^2)}{(u^2-a^2)^2}du=u\displaystyle\frac{(x^2-a^2)}{(u^2-a^2)^2}du$

$\displaystyle \displaystyle\frac{u}{(u^2-a^2)^2}du=\displaystyle\frac{t}{(x^2-a^2)}dt$

But again, I don't know how to continue.

Could you tell me what could I do to go on with the integral?
Is this integral supposed to be w.r.t. u or w.r.t. x?

4. ## Re: Problems trying to solve an integral

Hi, the integral is with respect to "u".

5. ## Re: Problems trying to solve an integral

Can you tell us from where did you see this integral?

6. ## Re: Problems trying to solve an integral

Hi,

find attached 2 pictures scanned from the book where I saw that integral. As you can see, in the book appears $\displaystyle a+rho$ I put $\displaystyle u$ so, in the book the integral is:

$\displaystyle \displaystyle\int\sqrt[ ]{\frac{(a+rho)^2-x^2}{(a+rho)^2-a^2}}(\frac{a}{a+rho})d(a+rho)$

and I put this one:

$\displaystyle \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}(\frac{a}{u})du$

but the integral with respect to u should be the same, right?