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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

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  • $\begingroup$ why did you multiply the numerator with sec$\theta$ tan$\theta$ without the denominator? $\endgroup$ – Anonymous Mar 6 at 15:49
  • $\begingroup$ That was a substitution of $dx = \sec\theta tan\theta$. I will make an edit to make it more clear $\endgroup$ – Evan Kim Mar 6 at 15:51
  • $\begingroup$ Have you considered other substitutions? (not trig) $\endgroup$ – GeorgSaliba Mar 6 at 16:10
  • $\begingroup$ @GeorgSaliba no, the practice is specifically for trig substitution $\endgroup$ – Evan Kim Mar 6 at 16:12
  • $\begingroup$ @EvanKim oh okay, then the answer i was preparing defeats the purpose of the exercise. $\endgroup$ – GeorgSaliba Mar 6 at 16:14
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Note that$$\frac{\tan^2\theta}{\sec^3\theta}=\sin^2\theta\cos\theta.$$A primitive of this will be $\frac13\sin^3\theta$.

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  • $\begingroup$ 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 $\endgroup$ – Evan Kim Mar 6 at 16:02
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    $\begingroup$ Since $x=\sec\theta=\frac1{\cos\theta}$, $\theta=\arccos\left(\frac1x\right)$. $\endgroup$ – José Carlos Santos Mar 6 at 16:11
  • $\begingroup$ 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 $\cos\theta$ $\endgroup$ – Evan Kim Mar 6 at 16:22
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    $\begingroup$ $$x=\frac1{\cos\theta}\iff\frac1x=\cos\theta\iff\theta=\arccos\left(\frac1x\right)$$ $\endgroup$ – José Carlos Santos Mar 6 at 16:31
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Alternative solution (without trigonometry). Note that by integration by parts $$\begin{align} \int \frac{\sqrt{x^2-1}}{x^4}dx&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{3}\int \frac{1}{x^2\sqrt{x^2-1}}dx\\ &=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{6}\int \frac{D(1-1/x^2)}{\sqrt{1-1/x^2}}dx\\ &=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{\sqrt{x^2-1}}{3x}+c=\frac{(x^2-1)^{3/2}}{3x^3}+c. \end{align}$$

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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.

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  • $\begingroup$ 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 $\endgroup$ – Evan Kim Mar 7 at 23:53
  • $\begingroup$ @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. $\endgroup$ – GeorgSaliba Mar 8 at 9:42

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