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

J

johnk

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Jun 2, 2007
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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...
 
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 I looked at the site and it appears to be an overly complicated, outdated method.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.
 
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...
 
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.
 
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 :|...
 
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