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

I am trying to integrate $$\int \frac{\sqrt{x^2-1}}{x^4}dx$$ via trig substitution. I decided to substitute $$x = \sec\theta$$ into the square root and $$dx = \sec\theta \tan\theta\,d\theta$$.

$$\int \frac{\sqrt{\sec^2 \theta-1^2}}{\sec^4\theta} \,dx = \int \frac{\sqrt{\tan^2\theta + 1 - 1}}{\sec^4\theta}\,dx = \int \frac{\tan\theta}{\sec^4\theta} \sec\theta \tan\theta\,d\theta = \int \dfrac{\tan^2\theta}{\sec^3\theta}\,d\theta$$

Here is where I am currently stuck. I attempted substitution with $$u = \sec\theta, du = \sec x \tan x dx$$ but that didn't seem to work out. I wasn't able to get an integration by parts strategy working either.

I think the answer lies in some sort of trigonometry regarding $$\int \frac{\tan^2\theta}{\sec^3\theta}\,d\theta$$ that I am overlooking to further simplify the problem, but no idea what it is

• why did you multiply the numerator with sec$\theta$ tan$\theta$ without the denominator? Mar 6 '19 at 15:49
• That was a substitution of $dx = \sec\theta tan\theta$. I will make an edit to make it more clear Mar 6 '19 at 15:51
• Have you considered other substitutions? (not trig) Mar 6 '19 at 16:10
• @GeorgSaliba no, the practice is specifically for trig substitution Mar 6 '19 at 16:12
• @EvanKim oh okay, then the answer i was preparing defeats the purpose of the exercise. Mar 6 '19 at 16:14

Note that$$\frac{\tan^2\theta}{\sec^3\theta}=\sin^2\theta\cos\theta.$$A primitive of this will be $$\frac13\sin^3\theta$$.

• ah that was what I was looking for, thanks. Now with $\frac{1}{3} \sin^3\theta$, I know I need to replace $\theta$ with $x$, but I am stuck again. I thought maybe it was $\theta = \sin^{-1} (\frac{x}{1})$, but it does not even look close to the answer Mar 6 '19 at 16:02
• Since $x=\sec\theta=\frac1{\cos\theta}$, $\theta=\arccos\left(\frac1x\right)$. Mar 6 '19 at 16:11
• i don't follow the step from $\frac{1}{\cos\theta} = \arccos (\frac{1}{x})$. My brain keeps telling me that it shouldn't work because it is a $\frac{1}{\cos\theta}$ and not a $\cos\theta$ Mar 6 '19 at 16:22
• $$x=\frac1{\cos\theta}\iff\frac1x=\cos\theta\iff\theta=\arccos\left(\frac1x\right)$$ Mar 6 '19 at 16:31

Alternative solution (without trigonometry). Note that by integration by parts \begin{align} \int \frac{\sqrt{x^2-1}}{x^4}dx&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{3}\int \frac{1}{x^2\sqrt{x^2-1}}dx\\ &=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{6}\int \frac{D(1-1/x^2)}{\sqrt{1-1/x^2}}dx\\ &=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{\sqrt{x^2-1}}{3x}+c=\frac{(x^2-1)^{3/2}}{3x^3}+c. \end{align}

For completeness, here is yet another way of solving it:

Set $$x^2-1=t^2x^2$$ then $$x=\frac{1}{\sqrt{1-t^2}} \qquad t=\frac{\sqrt{x^2-1}}{x}$$ and $$dx=\frac {tdt}{(1-t^2)^{3/2}}$$

The integral becomes: $$I=\int x^{-4}txdx=\int tx^{-3}dx=\int t(1-t^2)^{3/2}\frac {tdt}{(1-t^2)^{3/2}}=\int t^2dt=\frac{t^3}{3}+c$$

then replace $$t$$...

PS: The substitution I used may seem arbitrary, but whenever you have an integrand of the form $$x^m(a+bx^n)^{r/s}$$, where $$\frac{m+1}{n}+\frac rs \in \mathbb{Z}$$, the substitution $$a+bx^n=t^sx^n$$ makes the integral more tractable.

• What kind of substitution is this? I don't recognize it as the regular u-substitution that I usually use. I don't think I've come across this (learned it) yet Mar 7 '19 at 23:53
• @EvanKim It's more of a trick (maybe in certain institutions in certain countries it is taught as part of the curriculum, but not necessarily everywhere). You can use it anytime the conditions are met. Mar 8 '19 at 9:42

\begin{aligned} & \int \frac{\sqrt{x^{2}-1}}{x^{4}} d x \\\stackrel{y=\frac{1}{x}}{=} &\int\frac{\sqrt{\frac{1}{y^{2}}-1}}{\frac{1}{y^{4}}}\left(-\frac{1}{y^{2}} d y\right)\\ =& \int \frac{y \sqrt{1-y^{2}}}{3} y \\ =&-\frac{\left(1-y^{2}\right)^{2}}{3}+C \\=&-\frac{\left(x^{2}-1\right)^{\frac{3}{2}}}{3 x^{3}}+C \end{aligned}