# Thread: Integrate Sqrt[x^2-a]?

1. ## 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

2. 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 $\displaystyle x$

3. 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.

4. try letting $\displaystyle x=\sqrt{a}\cosh(\theta)$

5. mathstud what if a is negative - then we can't consider sqrt(a) ?

6. 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 $\displaystyle x=\sqrt{-a}\sinh(\theta)$

7. 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

8. 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

9. 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 ?

10. 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

11. Originally Posted by coverband
you're a very stupid man jhevon
And you're banned for 2 weeks to cool off. None of that.