# Thread: Can you check this Integration?

1. ## Can you check this Integration?

Hi I'm just practicing my trig integrals, can you tell me
if I got this right.

$\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}}$

The answer I got is

$\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}} = - \frac{1}{\sqrt{6}} arcsin( \frac{\sqrt{6}}{e^x}) + C$

But my book says the answer is;

$\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}} = \frac{1}{\sqrt{6}} arcsec( \frac{e^x}{\sqrt{6}}) + C$

I think they are the same because of the minus sign and
the fact that inside the brackets of arcsec things are
upside down but I'm not 100%.

2. Originally Posted by gerardhoyle
Hi I'm just practicing my trig integrals, can you tell me
if I got this right.

$\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}}$

The answer I got is

$\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}} = - \frac{1}{\sqrt{6}} arcsin( \frac{\sqrt{6}}{e^x}) + C$

But my book says the answer is;

$\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}} = \frac{1}{\sqrt{6}} arcsec( \frac{e^x}{\sqrt{6}}) + C$

I think they are the same because of the minus sign and
the fact that inside the brackets of arcsec things are
upside down but I'm not 100%.
$\displaystyle u = e^x$

$\displaystyle du = e^x \, dx$ ... $\displaystyle dx = \frac{du}{u}$

$\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}}$

$\displaystyle \int \frac{du}{u\sqrt{u^2 - 6}}$

looks like the integral form for an arcsecant to me ...

3. $\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}}$

by the picture,

$\displaystyle tan \theta = \frac{\sqrt{6}}{\sqrt{e^{2x} - 6}}$

and,

$\displaystyle \frac{1}{\sqrt{6}}tan \theta = \frac{1}{\sqrt{e^{2x} - 6}}$

so the integral becomes;

$\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}} =\int \frac{1}{\sqrt{6}}tan \theta dx$

Now to calculate dx, (this gets a bit nuts);

$\displaystyle u = e^x$

$\displaystyle du = e^x dx$

$\displaystyle dx = \frac{du}{e^x} = \frac{du}{u}$

so, the integral is NOW;

$\displaystyle \int \frac{1}{\sqrt{6}} \ tan \theta \ \frac{du}{u}$

We first get rid of the $\displaystyle u$ by using the fact that,

$\displaystyle sin \theta = \frac{\sqrt{6}}{u}$

$\displaystyle \frac{1}{\sqrt{6}} \ sin \theta = \frac{1}{u}$

and this gets our integral even closer;

$\displaystyle \int \frac{1}{\sqrt{6}} \ tan \theta \ \frac{1}{\sqrt{6}} \ sin \theta du$

cleaning it up;

$\displaystyle \int \frac{1}{6} \ tan \theta \ sin \theta du$

All thats left to get rid of is $\displaystyle du$,

Because

$\displaystyle sin \theta = \frac{\sqrt{6}}{e^x}$

and,

$\displaystyle e^x = u = \frac{\sqrt{6}}{sin \theta} = \sqrt{6} \ csc \theta$

we see that;

$\displaystyle u = \sqrt{6} \ csc \theta$

so,

$\displaystyle du = - \sqrt{6} \ csc \theta \ cot \theta \ d \theta$

and the original integral finally becomes:

$\displaystyle \int \frac{1}{6} \ tan \theta \ sin \theta \ du = \int - \frac{1}{6} \ tan \theta \ sin \theta \ \sqrt{6} \ csc \theta \ cot \theta \ d \theta$

Cleaning it up;

$\displaystyle \int - \frac{1}{\sqrt{6}} \ tan \theta \ sin \theta \ csc \theta \ cot \theta \ d \theta$

and this reduces to;

$\displaystyle - \frac{1}{\sqrt{6}} \int d \theta$

$\displaystyle - \frac{1}{\sqrt{6}} \ \theta + C$

to get theta I'll use;

$\displaystyle sin \theta = \frac{\sqrt{6}}{e^x}$

$\displaystyle \theta = arcsin ( \frac{\sqrt{6}}{e^x})$

so the final answer I get is;

$\displaystyle - \frac{1}{\sqrt{6}} \ arcsin ( \frac{\sqrt{6}}{e^x}) + C$

ergo;

$\displaystyle \int \frac{dx}{\sqrt{e^{2x} - 6}} dx = - \frac{1}{\sqrt{6}} arcsin( \frac{\sqrt{6}}{e^x}) + C$

There must of been some slip-up but I haven't seen it

Unless it's correct

4. maybe ... take the derivative of your solution and find out.