1. ## general integration formula

Hi all,

Is there any general formula for calculating integrals of the type $\displaystyle \int{\frac{dx}{x\sqrt{x^2-a^2}}}$ other than $\displaystyle \frac{1}{a}arc\sec{\frac{|x|}{a}}+C$ (guess the function is not even included in LaTeX)? I was trying to figure it out and got something like $\displaystyle -\frac{1}{a}\arcsin{\frac{a}{|x|}}+C$ but I'm not sure if it's (always) correct and where absolute values should be put.

Thank you.

EDIT:
Guess I got it: $\displaystyle -\frac{1}{|a|}\arcsin{\frac{|a|}{x}}+C$. Can someone confirm if it's correct? Thanks.

2. I get the arcsec.

$\displaystyle \int\frac{1}{x\sqrt{x^{2}-a^{2}}}dx$

Let $\displaystyle x=asec(u), \;\ dx=asec(u)tan(u)du$

$\displaystyle \frac{asec(u)tan(u)}{asec(u)\sqrt{a^{2}(sec^{2}(u)-1)}}du$

This all whittles down to:

$\displaystyle \int\frac{1}{a}du=\frac{u}{a}+C$

But, $\displaystyle u=sec^{-1}(\frac{x}{a})$

And we get $\displaystyle \frac{sec^{-1}(\frac{x}{a})}{a}+C$

3. Thanks. It's just that using $\displaystyle arc\sec$ or $\displaystyle \sec$ is very uncommon in my neck of the woods. So I wanted to get something connected with $\displaystyle \sin$. Here's my reasoning:

$\displaystyle \int{\frac{dx}{x\sqrt{x^2-a^2}}}$
Let $\displaystyle \frac{1}{x}=t$
So $\displaystyle x=\frac{1}{t}\Longrightarrow{dx=-\frac{1}{t^2}dt}$
That way we get $\displaystyle \int{\frac{-t}{t^2\sqrt{\frac{1}{t^2}-a^2}}}dt$ which yields $\displaystyle -\int{\frac{dt}{\sqrt{1-a^2t^2}}}$
Now we can write it as $\displaystyle -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}$ and use the formula $\displaystyle \int{\frac{dx}{\sqrt{a^2-x^2}}}=\arcsin{\frac{x}{|a|}}+C$
That said, $\displaystyle -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}=-\frac{1}{a}\arcsin{at}+C$
And, finally coming back to our substitution, we get:
$\displaystyle \boxed{\int{\frac{dx}{x\sqrt{x^2-a^2}}}=-\frac{1}{a}\arcsin{\frac{a}{x}}+C}$
What do you think about it? I'm not sure about the absolute values, to my mind it should be $\displaystyle -\frac{1}{|a|}\arcsin{\frac{|a|}{x}}+C$.

4. Originally Posted by disclaimer
Thanks. It's just that using $\displaystyle arc\sec$ or $\displaystyle \sec$ is very uncommon in my neck of the woods. So I wanted to get something connected with $\displaystyle \sin$. Here's my reasoning:

$\displaystyle \int{\frac{dx}{x\sqrt{x^2-a^2}}}$
Let $\displaystyle \frac{1}{x}=t$
So $\displaystyle x=\frac{1}{t}\Longrightarrow{dx=-\frac{1}{t^2}dt}$
That way we get $\displaystyle \int{\frac{-t}{t^2\sqrt{\frac{1}{t^2}-a^2}}}dt$ which yields $\displaystyle -\int{\frac{dt}{\sqrt{1-a^2t^2}}}$
Now we can write it as $\displaystyle -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}$ and use the formula $\displaystyle \int{\frac{dx}{\sqrt{a^2-x^2}}}=\arcsin{\frac{x}{|a|}}+C$
That said, $\displaystyle -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}=-\frac{1}{a}\arcsin{at}+C$
And, finally coming back to our substitution, we get:
$\displaystyle \boxed{\int{\frac{dx}{x\sqrt{x^2-a^2}}}=-\frac{1}{a}\arcsin{\frac{a}{x}}+C}$
What do you think about it? I'm not sure about the absolute values, to my mind it should be $\displaystyle -\frac{1}{|a|}\arcsin{\frac{|a|}{x}}+C$.
I think you mean to say $\displaystyle arcos\bigg(\frac{a}{x}\bigg)$ using the identity taht $\displaystyle arcos(u(x))=arcsec\bigg(\frac{1}{u(x)}\bigg)$...but I think the $\displaystyle |a|$ is incorrect the only absolute values I have seen in relation to this is the derivative of $\displaystyle arsec(x)$ or $\displaystyle arccsc(x)$

5. That looks better. Since $\displaystyle \left(\arcsin{x}\right)'=\frac{1}{\sqrt{1-x^2}}$ and $\displaystyle \left(\arccos{x}\right)'=-\frac{1}{\sqrt{1-x^2}}$, the integral $\displaystyle \int{\frac{dx}{x\sqrt{x^2-a^2}}}$ would be $\displaystyle \frac{1}{a}\arccos{\frac{a}{x}}+C$ regardless of the sign of $\displaystyle a$. Thanks.

6. Substitution $\displaystyle z^2=x^2-a^2$ kills the problem.

7. Originally Posted by Krizalid
Substitution $\displaystyle z^2=x^2-a^2$ kills the problem.
Your substitution leads to $\displaystyle \frac{1}{a}\arctan{\frac{\sqrt{x^2-a^2}}{a}}+C$, am I right?

8. That's what you need to find by yourself. Differentiate your result, does yield the integrand?