# Inverse Trig Function Integrals

• Dec 1st 2006, 11:00 AM
ron007
Inverse Trig Function Integrals
I did this exercises, but I have a doubt in the process, thanks for your help.
• Dec 1st 2006, 11:23 AM
ThePerfectHacker
The integral,
$\int \frac{1}{x\sqrt{x^2-4}}dx$
Is almost the inverse cosecant integral.
There are some adjustments. First,
$x^2-4=x^2-2^2$
Thus, it has form,
$\csc^{-1} (x/2)$
But note the following, that if the problem asked for,
$\int \frac{1}{|x|\sqrt{x^2-4}}dx$
Then the answer would be,
$\csc^{-1} (x/2)+C$
But it does not ask that.
In that case the integral is,
$\csc^{-1} |x/2|+C$
• Dec 1st 2006, 12:28 PM
ron007
Thanks but that is the integral of the secant, the cosecant is the same as the secant, but with negative sign. I did this:
• Dec 1st 2006, 12:44 PM
Soroban
Hello, ron007!

If you're not familiar with the arcsecant formula,
. . there's always Trig Substitution.

Quote:

$\int\frac{dx}{x\sqrt{9x^2-16}}$

Let $3x \,=\,4\sec\theta\quad\Rightarrow\quad x \,=\,\frac{4}{3}\sec\theta\quad\Rightarrow\quad dx\,=\,\frac{4}{3}\sec\theta\tan\theta\,d\theta$

Then: . $\sqrt{9x^2 - 16} \:=\:\sqrt{(3x)^2 - 16} \:=\:\sqrt{(4\sec\theta)^2 - 16} \:=\:\sqrt{16\sec^2\theta-16}$

. . $=\:\sqrt{16(\sec^2\theta - 1)} \:=\:\sqrt{16\tan^2\theta} \:=\:4\tan\theta$

Substitute: . $\int\frac{\frac{4}{3}\sec\theta\tan\theta\,d\theta }{\frac{4}{3}\sec\theta\!\cdot\!4\tan\theta} \;=\;\frac{1}{4}\int d\theta \:=\:\frac{1}{4}\theta + C$

Back-substitute
We have: . $3x \,=\,4\sec\theta\quad\Rightarrow\quad \sec\theta \,=\,\frac{3x}{4}\quad\Rightarrow\quad \theta\,=\,\text{arcsec}\frac{3x}{4}
$

Therefore, the answer is: . $\frac{1}{4}\text{arcsec}\frac{3x}{4} + C$

• Dec 1st 2006, 12:47 PM
ron007
Quote:

Originally Posted by Soroban
Hello, ron007!

If you're not familiar with the arcsecant formula,
. . there's always Trig Substitution.

Let $3x \,=\,4\sec\theta\quad\Rightarrow\quad x \,=\,\frac{4}{3}\sec\theta\quad\Rightarrow\quad dx\,=\,\frac{4}{3}\sec\theta\tan\theta\,d\theta$

Then: . $\sqrt{9x^2 - 16} \:=\:\sqrt{(3x)^2 - 16} \:=\:\sqrt{(4\sec\theta)^2 - 16} \:=\:\sqrt{16\sec^2\theta-16}$

. . $=\:\sqrt{16(\sec^2\theta - 1)} \:=\:\sqrt{16\tan^2\theta} \:=\:4\tan\theta$

Substitute: . $\int\frac{\frac{4}{3}\sec\theta\tan\theta\,d\theta }{\frac{4}{3}\sec\theta\!\cdot\!4\tan\theta} \;=\;\frac{1}{4}\int d\theta \:=\:\frac{1}{4}\theta + C$

Back-substitute
We have: . $3x \,=\,4\sec\theta\quad\Rightarrow\quad \sec\theta \,=\,\frac{3x}{4}\quad\Rightarrow\quad \theta\,=\,\text{arcsec}\frac{3x}{4}
$

Therefore, the answer is: . $\frac{1}{4}\text{arcsec}\frac{3x}{4} + C$

I'm confussed :( .
• Dec 1st 2006, 09:06 PM
Jameson
Quote:

Originally Posted by ron007
I'm confussed :( .

His explanation is very straightfoward. Have you done this kind of substitution? I hope your teacher didn't just ask you to memorize the formulas and not be able to derive them. What part are you confused on so he can help you more? Help us help you.
• Dec 2nd 2006, 05:54 AM
ron007
Quote:

Originally Posted by Jameson
His explanation is very straightfoward. Have you done this kind of substitution? I hope your teacher didn't just ask you to memorize the formulas and not be able to derive them. What part are you confused on so he can help you more? Help us help you.

I get the problem. I was doing an error in the substitution. Thank you guys for ur help.