What is going on with this integral $\int \frac{dx}{\sqrt{e^{2x} - 9}}$?

Question

A few days ago, I was tasked to solve this integral:

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

The way taught was to recongize the integral as an arcsecant integral. I just can't wrap my head around how it can be arcsecant? The way I did it, which apparently was marked wrong, was to first u-sub:

$$ u := -2x \\ dx = -\frac{du}{2} $$

which makes the integral then

$$ -\frac{1}{2} \int \frac{du}{\sqrt{e^{-u} - 9}} $$

I subsitute again (this time using v-sub):

$$ v := e^{-u} - 9 \\ du = -e^u dv $$

which shifting the terms around makes this integral:

$$ -\int \frac{dv}{(v+9)\sqrt v} $$

I then subsitute for the final time (all to try and get arctangent):

$$ t := \frac{\sqrt{v}}{3} \\ dv = 6 \sqrt v \,dt $$ which results in

$$ \int \frac{6}{9t^2 + 9} dt \\ = \frac{2}{3} \int \frac{dt}{t^2 + 1}\\ = \frac{2}{3} \arctan{(t)} $$

which at this point, I see it is the arctangent integral. Following through and subsituting the things back in:

$$ \frac{2}{3} \arctan{(t)} \\ = - \frac{2}{3} \arctan{(\frac{\sqrt v}{3})} \\ = -\frac{2}{3} \arctan{(\frac{\sqrt{e^{-u} - 9}}{3})} \\ = \frac{1}{3} \arctan{(\frac{\sqrt{e^{2x} - 9}}{3})} + C $$

Now, I am clearly lost on whether this is right or wrong, AFAIK, I see nothing wrong with my method so I boil down to 3 questions:

1. Is the above method valid and the answer listed is correct?
2. How would one solve it to be arcsecant?
3. Are the functions shifts of each other or is there still something wrong?

Answer 1

Rewrite the integrand by multiplying and dividing by $e^x$

$$\int\frac{e^x dx}{e^x\sqrt{e^x-9}} = \int \frac{d(e^x)}{e^x\sqrt{e^x-9}} = \frac{1}{3}\sec^{-1}\left(\frac{e^x}{3}\right)+C$$

By drawing a triangle we can see that

$$\sec 3\theta = \frac{e^x}{3} \implies \tan 3\theta = \frac{\sqrt{e^{2x}-9}}{3}$$

Thus we also obtain your arctan solution, which is equivalent to the arcsec solution

$$\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{e^{2x}-9}}{3}\right)+C$$

and similarly, we can obtain the arcsin solution

$$\sin 3\theta = \frac{e^x}{\sqrt{e^{2x}-9}}\implies \frac{1}{3}\arcsin\left(\frac{1}{\sqrt{1-9e^{-x}}}\right)+C$$

Answer 2

To get the arcsecant part, let $u = e^x$. Then, $du = u\,dx \Leftrightarrow dx = \frac{du}{u}$. Also, $e^{2x} - 9$ becomes $u^2 - 9$. The integral becomes $$\int\frac{du}{u\sqrt{u^2-9}}.$$ This can be seen as the derivative of an arcsecant because $$\frac{d}{dx}\left(\frac{1}{a}\sec^{-1}\frac{x}{a}\right) = \frac{1}{x\sqrt{x^2 - a^2}}.$$

This means that $$\int\frac{du}{u\sqrt{u^2-9}} = \frac{1}{3}\sec^{-1}\left(\frac{u}{3}\right).$$ Substituting it back gives us $$\frac{1}{3}\sec^{-1}\left(\frac{e^x}{3}\right).$$ Therefore, $$\int \frac{dx}{\sqrt{e^{2x} - 9}} = \frac{1}{3}\sec^{-1}\left(\frac{e^x}{3}\right).$$

Answer 3

The thing to do is to check by differentiation. Since you used integration by substitution repeatedly, you should expect to use differentiation by substitution (i.e. the chain rule) repeatedly:

\begin{align} & \frac 1 3 \arctan\left(\frac{\sqrt{e^{2x} - 9}} 3 \right) \\[8pt] = {} & \frac 1 3\cdot \frac 1 {1 + \left( \frac{\sqrt{e^{2x}-9}} 3 \right)^2} \cdot \frac 1 3 \cdot\frac 1 {2\sqrt{e^{2x}-9}}\cdot e^{2x} \cdot 2 \\ & \text{(In the line above we simply repeatedly used the chain rule.)} \\[8pt] = {} & \frac 1 9 \cdot \frac 9 {9 + (e^{2x}-9)} \cdot \frac 1 {\sqrt{e^{2x}-9}} \cdot e^{2x} \\[8pt] = {} & \frac 1 {\sqrt{e^{2x}-9}}. \end{align}

Answer 4

$$ \begin{aligned} \int\frac{d x}{\sqrt{e^{2x}-9}} &=\int \frac{e^{-x}}{\sqrt{1-9 e^{-2 x}} }d x\\ &=-\frac{1}{3} \int \frac{d(3e^{-x})}{\sqrt{1-\left(3 e^{-x}\right)^{2}}} \\ &=-\frac{1}{3} \sin ^{-1}\left(3 e^{-x}\right)+C \end{aligned} $$