# Integrate Sqrt[x^2-a]?

• Dec 2nd 2008, 06:24 AM
coverband
Integrate Sqrt[x^2-a]?
I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

does anyone know how to get this/derive this i.e. not take from tables !!!!

Thanks
• Dec 2nd 2008, 07:05 AM
Jhevon
Quote:

Originally Posted by coverband
I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

does anyone know how to get this/derive this i.e. not take from tables !!!!

Thanks

that is correct.

to get the derivative of the answer, you need to use the product rule and the chain rule. are you familiar with these.

moreover, note that $\displaystyle \frac d{dx} \ln u = \frac {u'}u$ (by the chain rule)

where $\displaystyle u$ is some function of $\displaystyle x$
• Dec 2nd 2008, 07:39 AM
Chop Suey
Quote:

Originally Posted by Jhevon
that is correct.

to get the derivative of the answer, you need to use the product rule and the chain rule. are you familiar with these.

moreover, note that $\displaystyle \frac d{dx} \ln u = \frac {u'}u$ (by the chain rule)

where $\displaystyle u$ is some function of $\displaystyle x$

I could be wrong, but I think s/he was asking how the result was reached.
• Dec 4th 2008, 12:53 PM
Mathstud28
try letting $\displaystyle x=\sqrt{a}\cosh(\theta)$
• Dec 4th 2008, 12:56 PM
coverband
mathstud what if a is negative - then we can't consider sqrt(a) ?
• Dec 4th 2008, 12:57 PM
Mathstud28
Quote:

Originally Posted by coverband
mathstud what if a is negative - then we can't consider sqrt(a) ?

Ok then make it $\displaystyle \sqrt{x^2+(-a)}$ and make the sub $\displaystyle x=\sqrt{-a}\tan(\theta)$ or $\displaystyle x=\sqrt{-a}\sinh(\theta)$
• Dec 4th 2008, 12:59 PM
coverband
but we want to integrate Sqrt[x^2-a] and i forsee difficulty for if a<0 we can not use sqrt(a) in our substitutions
• Dec 4th 2008, 01:00 PM
Mathstud28
Quote:

Originally Posted by coverband
but we want to integrate Sqrt[x^2-a] and i forsee difficulty for if a<0 we can not use sqrt(a) in our substitutions

I edited my post
• Dec 4th 2008, 01:00 PM
coverband
Quote:

Originally Posted by Mathstud28
Ok then make it $\displaystyle \sqrt{x^2+(-a)}$ and make the sub $\displaystyle x=\sqrt{-a}\tan(\theta)$ or $\displaystyle x=\sqrt{-a}\sinh(\theta)$

but you took the square root of a neg number there. Is that allowed ?
• Dec 4th 2008, 01:00 PM
Mathstud28
Quote:

Originally Posted by coverband
but you took the square root of a neg number there. Is that allowed ?

But you said what if $\displaystyle a<0\implies{-a>0}$ so yes
• Dec 4th 2008, 01:01 PM
Jameson
Quote:

Originally Posted by coverband
you're a very stupid man jhevon

And you're banned for 2 weeks to cool off. None of that.