# 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, 2019 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, 2019 at 15:51
• Have you considered other substitutions? (not trig) Mar 6, 2019 at 16:10
• @GeorgSaliba no, the practice is specifically for trig substitution Mar 6, 2019 at 16:12
• @EvanKim oh okay, then the answer i was preparing defeats the purpose of the exercise. Mar 6, 2019 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, 2019 at 16:02
• Since $x=\sec\theta=\frac1{\cos\theta}$, $\theta=\arccos\left(\frac1x\right)$. Mar 6, 2019 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, 2019 at 16:22
• $$x=\frac1{\cos\theta}\iff\frac1x=\cos\theta\iff\theta=\arccos\left(\frac1x\right)$$ Mar 6, 2019 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, 2019 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, 2019 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 y \sqrt{1-y^{2}}d y \\ =&\frac{\left(1-y^{2}\right)^{\frac{3}{2}}}{3}+C \\=&\frac{\left(x^{2}-1\right)^{\frac{3}{2}}}{3 x^{3}}+C \end{aligned}