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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

Bernard
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Evan Kim
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5 Answers5

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

José Carlos Santos
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  • 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 – Evan Kim Mar 06 '19 at 16:02
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    Since $x=\sec\theta=\frac1{\cos\theta}$, $\theta=\arccos\left(\frac1x\right)$. – José Carlos Santos Mar 06 '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 $\cos\theta$ – Evan Kim Mar 06 '19 at 16:22
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    $$x=\frac1{\cos\theta}\iff\frac1x=\cos\theta\iff\theta=\arccos\left(\frac1x\right)$$ – José Carlos Santos Mar 06 '19 at 16:31
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Alternative solution (without trigonometry). Note that by integration by parts $$\begin{align} \int \frac{\sqrt{x^2-1}}{x^4}dx&=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{3}\int \frac{1}{x^2\sqrt{x^2-1}}dx\\ &=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{1}{6}\int \frac{D(1-1/x^2)}{\sqrt{1-1/x^2}}dx\\ &=-\frac{\sqrt{x^2-1}}{3x^3}+\frac{\sqrt{x^2-1}}{3x}+c=\frac{(x^2-1)^{3/2}}{3x^3}+c. \end{align}$$

Robert Z
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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.

entrelac
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  • 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 – Evan Kim Mar 07 '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. – entrelac Mar 08 '19 at 09:42
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$$ \begin{aligned} & \int \frac{\sqrt{x^{2}-1}}{x^{4}} d x \\\stackrel{y=\frac{1}{x}}{=} &\int\frac{\sqrt{\frac{1}{y^{2}}-1}}{\frac{1}{y^{4}}}\left(-\frac{1}{y^{2}} d y\right)\\ =& -\int y \sqrt{1-y^{2}}d y \\ =&\frac{\left(1-y^{2}\right)^{\frac{3}{2}}}{3}+C \\=&\frac{\left(x^{2}-1\right)^{\frac{3}{2}}}{3 x^{3}}+C \end{aligned} $$

Lai
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Your question is a very good question and also it is a very easy question if and only if you understand the trick to solve this question. Let me tell you how to solve this without any trigonometric substitution and in a very easy way. First of all take $x^{2}$ common from the numerator. Now the expression will become $(x×\sqrt{1-\frac{1}{x^{2}}})/x^{4}$. Now cancel $x$ and $x^{4}$. So, the final expression will become $\frac{\sqrt{1-\frac{1}{x^{2}}}}{x^{3}}$. Now I think you can solve it easily. Substitute $\frac{1}{x^{2}}$ $=$ $u$. So, $\frac{-2}{x^{3}}dx$ $=$ $du$. So, $\frac{1}{x^{3}}dx$ $=$ $\frac{-1}{2}du$. Now try to solve it on your own. I have given you enough hints.

Sohini
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