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

*

*Is the above method valid and the answer listed is correct?

*How would one solve it to be arcsecant?

*Are the functions shifts of each other or is there still something wrong?

 A: 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
$$\sin 3\theta = \frac{e^x}{\sqrt{e^{2x}-9}}\implies \frac{1}{3}\arcsin\left(\frac{1}{\sqrt{1-9e^{-x}}}\right)+C$$
A: 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, $$\int \frac{dx}{\sqrt{e^{2x} - 9}} = \frac{1}{3}\sec^{-1}\left(\frac{e^x}{3}\right).$$
A: 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}
A: $$
\begin{aligned}
\int\frac{d x}{\sqrt{e^{2x}-9}} &=\int \frac{e^{-x}}{\sqrt{1-9 e^{-2 x}} }d x\\
&=-\frac{1}{3} \int \frac{d(3e^{-x})}{\sqrt{1-\left(3 e^{-x}\right)^{2}}} \\
&=-\frac{1}{3} \sin ^{-1}\left(3 e^{-x}\right)+C
\end{aligned}
$$
