I did this exercises, but I have a doubt in the process, thanks for your help.

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- Dec 1st 2006, 11:00 AMron007Inverse Trig Function Integrals
I did this exercises, but I have a doubt in the process, thanks for your help.

- Dec 1st 2006, 11:23 AMThePerfectHacker
The integral,

$\displaystyle \int \frac{1}{x\sqrt{x^2-4}}dx$

Is almost the inverse cosecant integral.

There are some adjustments. First,

$\displaystyle x^2-4=x^2-2^2$

Thus, it has form,

$\displaystyle \csc^{-1} (x/2)$

But note the following, that if the problem asked for,

$\displaystyle \int \frac{1}{|x|\sqrt{x^2-4}}dx$

Then the answer would be,

$\displaystyle \csc^{-1} (x/2)+C$

But it does not ask that.

In that case the integral is,

$\displaystyle \csc^{-1} |x/2|+C$ - Dec 1st 2006, 12:28 PMron007
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 PMSoroban
Hello, ron007!

If you're not familiar with the arcsecant formula,

. . there's always Trig Substitution.

Quote:

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

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

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

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

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

Back-substitute

We have: .$\displaystyle 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: .$\displaystyle \frac{1}{4}\text{arcsec}\frac{3x}{4} + C$

- Dec 1st 2006, 12:47 PMron007
- Dec 1st 2006, 09:06 PMJameson
- Dec 2nd 2006, 05:54 AMron007