# general integration formula

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• April 19th 2008, 06:18 AM
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general integration formula
Hi all,

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

Thank you. (Smile)

EDIT:
Guess I got it: $-\frac{1}{|a|}\arcsin{\frac{|a|}{x}}+C$. Can someone confirm if it's correct? Thanks.
• April 19th 2008, 08:05 AM
galactus
I get the arcsec.

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

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

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

This all whittles down to:

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

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

And we get $\frac{sec^{-1}(\frac{x}{a})}{a}+C$
• April 19th 2008, 08:58 AM
disclaimer
Thanks. It's just that using $arc\sec$ or $\sec$ is very uncommon in my neck of the woods. So I wanted to get something connected with $\sin$. Here's my reasoning:

$\int{\frac{dx}{x\sqrt{x^2-a^2}}}$
Let $\frac{1}{x}=t$
So $x=\frac{1}{t}\Longrightarrow{dx=-\frac{1}{t^2}dt}$
That way we get $\int{\frac{-t}{t^2\sqrt{\frac{1}{t^2}-a^2}}}dt$ which yields $-\int{\frac{dt}{\sqrt{1-a^2t^2}}}$
Now we can write it as $-\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}$ and use the formula $\int{\frac{dx}{\sqrt{a^2-x^2}}}=\arcsin{\frac{x}{|a|}}+C$
That said, $-\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}=-\frac{1}{a}\arcsin{at}+C$
And, finally coming back to our substitution, we get:
$\boxed{\int{\frac{dx}{x\sqrt{x^2-a^2}}}=-\frac{1}{a}\arcsin{\frac{a}{x}}+C}$
What do you think about it? I'm not sure about the absolute values, to my mind it should be $
-\frac{1}{|a|}\arcsin{\frac{|a|}{x}}+C
$
.
• April 19th 2008, 09:03 AM
Mathstud28
Quote:

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

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

I think you mean to say $arcos\bigg(\frac{a}{x}\bigg)$ using the identity taht $arcos(u(x))=arcsec\bigg(\frac{1}{u(x)}\bigg)$...but I think the $|a|$ is incorrect the only absolute values I have seen in relation to this is the derivative of $arsec(x)$ or $arccsc(x)$
• April 19th 2008, 09:34 AM
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That looks better. Since $\left(\arcsin{x}\right)'=\frac{1}{\sqrt{1-x^2}}$ and $\left(\arccos{x}\right)'=-\frac{1}{\sqrt{1-x^2}}$, the integral $\int{\frac{dx}{x\sqrt{x^2-a^2}}}$ would be $\frac{1}{a}\arccos{\frac{a}{x}}+C$ regardless of the sign of $a$. Thanks. (Smile)
• April 19th 2008, 03:34 PM
Krizalid
Substitution $z^2=x^2-a^2$ kills the problem.
• April 20th 2008, 12:03 AM
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Quote:

Originally Posted by Krizalid
Substitution $z^2=x^2-a^2$ kills the problem.

Your substitution leads to $\frac{1}{a}\arctan{\frac{\sqrt{x^2-a^2}}{a}}+C$, am I right?
• April 20th 2008, 09:26 AM
Krizalid
That's what you need to find by yourself. Differentiate your result, does yield the integrand?