integral: x*sqrt(x^2-2*x+2)

[johnk]

Hello,

I need help with this integral:
$\displaystyle \int x\cdot\sqrt{x^2-2x+2}\ dx$

I need to use Euler substitution, but I don't know how to apply it properly. I tried to follow this site: http://planetmath.org/encyclopedia/Eule ... ation.html

So I started like this:
$\displaystyle \sqrt{x^2-2x+2} = -x+t$
(...)
$\displaystyle x = \frac{t^2-2}{2t-2}$

What to do now? I tried substituting it into $\displaystyle x\cdot(-x+t)$ from the original integral, but this doesn't seem to lead to the right solution. Just help me build an integral with no x's and I'll try to continue...

 

[galactus]

You MUST use Euler sub?. To tell you the truth, this is the first I have heard of Euler sub for integrals. I looked at the site and it appears to be an overly complicated, outdated method. Why would anyone require you use this?.
Are you in a history of math course or something?.


If you complete the square inside the radical:

$\displaystyle \int{x\sqrt{(x-1)^{2}+1}}dx$


You could make the substitution $\displaystyle u=x-1, \;\ x=u+1, \;\ du=dx$

Then you get:

$\displaystyle \int{(u+1)\sqrt{u^{2}+1}}du$

This isn't too bad. You can make another sub or use trig sub.

 

[johnk]

Thanks Galactus.

I did not have to use Euler substitution, but our instructor gave us a bunch of integrals and suggested it for almost all of them, including this one, so I thought that would be the best idea... Unfortunately, I've never seen it used yet so I had a hard time with it.

Your approach seems better, but I'm not sure if I'm getting anywhere... I chose substitution u = tan z to get rid of the sqrt:
$\displaystyle \int(u+1)\sqrt{u^2+1}\ du$

$\displaystyle u = \tan z;\ du = \frac{1}{\cos^2 z}\ dz$

$\displaystyle \int (\tan z + 1)\sqrt{\frac{1}{cos^2 z}}\frac{1}{cos^2 z}\ dz$

$\displaystyle \int \frac{(\tan z + 1)}{cos^3 z}\ dz$

What now? I'd think maybe substitution y = tan z/2 but then it becomes something horrible...

 

[galactus]

Use sec instead of cosine.

If you just sub $\displaystyle u=tan({\theta}), \;\ du=sec^{2}({\theta})d{\theta}$

Then you get:

$\displaystyle \int{(tan({\theta})+1)(\sqrt{tan^{2}({\theta})+1})sec^{2}({\theta})}d{\theta}$

This reduces down to:

$\displaystyle \int{sec^{3}({\theta})tan({\theta})+sec^{3}({\theta})}d{\theta}$

These aren't that bad.

The first, just let $\displaystyle u=sec({\theta}), \;\ du=sec({\theta})tan({\theta})d{\theta}$

Then we get $\displaystyle \int{u^{2}}du=\frac{1}{3}u^{3}=\frac{1}{3}sec^{3}({\theta})$

Just use parts or the reduction formula on the other.

 

[johnk]

The sec / csc functions are not very popular over here. In fact, no one even teaches those. But it works with $\displaystyle 1/\cos(\theta)$ just as well of course .

Integrating $\displaystyle 1/\cos^3(\theta)$ turns out to be the most labour-intensive bit, but it works in the end.

Thanks for your help. I'll probably have to ask about some more tomorrow when I try to do the rest :|...