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
$\endgroup$ 64 Answers
$\begingroup$Note that$$\frac{\tan^2\theta}{\sec^3\theta}=\sin^2\theta\cos\theta.$$A primitive of this will be $\frac13\sin^3\theta$.
$\endgroup$ 4 $\begingroup$Alternative solution (without trigonometry). Note that by integration by parts$$\begin{align} \int \frac{\sqrt{x^2-1}}{x^4}dx&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{3}\int \frac{1}{x^2\sqrt{x^2-1}}dx\\ &=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{6}\int \frac{D(1-1/x^2)}{\sqrt{1-1/x^2}}dx\\ &=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{\sqrt{x^2-1}}{3x}+c=\frac{(x^2-1)^{3/2}}{3x^3}+c. \end{align}$$
$\endgroup$ $\begingroup$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.
$\endgroup$ 2 $\begingroup$$$ \begin{aligned} & \int \frac{\sqrt{x^{2}-1}}{x^{4}} d x \\\stackrel{y=\frac{1}{x}}{=} &\int\frac{\sqrt{\frac{1}{y^{2}}-1}}{\frac{1}{y^{4}}}\left(-\frac{1}{y^{2}} d y\right)\\ =& -\int y \sqrt{1-y^{2}}d y \\ =&\frac{\left(1-y^{2}\right)^{\frac{3}{2}}}{3}+C \\=&\frac{\left(x^{2}-1\right)^{\frac{3}{2}}}{3 x^{3}}+C \end{aligned} $$
$\endgroup$
