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