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I am getting stuck on this trig substitution problem.

$$\int\frac1{x^2\sqrt{x^2 - 9}}~\mathrm dx.$$

$$x = 3 \sec\theta,\qquad\theta = \sec^{-1} \sqrt{\frac{x^2}{9}},\qquad\mathrm dx = \sec\theta\ \tan\theta\ \mathrm d\theta$$

I can get to here, but I don't know how to finish it (perhaps I have made a mistake before this point?)

$$\int\frac{3\sec\theta\ \tan\theta}{9\sec^2\theta(3\sec\theta -3)}~\mathrm d\theta.$$

If anyone could help from here, I'd appreciate it.

Thanks.

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+1 For showing some work. –  JavaMan Aug 28 '11 at 18:43
    
If $x=3\sec\theta$, then $\theta = \sec^{-1}(x/3)$. Note that $\sqrt{x^2/9} = |x|/3$, not $x/3$. –  Arturo Magidin Aug 28 '11 at 18:46
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4 Answers

up vote 3 down vote accepted

Notice first that you dropped a square root in the denominator. Also, $dx = 3 \sec \theta \tan \theta d \theta$. Otherwise, everything looked fine so far:

$$ \int \frac{1}{x^2\sqrt{x^2 - 9}}dx = \int \frac{3 \sec \theta \tan \theta}{9 \sec^2 \theta \sqrt{9 \sec^2 \theta - 9}} d \theta. $$

Now, what is a simpler way to write $\sqrt{9 \sec^2 \theta - 9}$?

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Additional hint: you'll want to use $1+\tan^2\theta=\sec^2\theta$ at some point... –  J. M. Aug 28 '11 at 18:52
    
A simpler way would be: 3 $\sqrt{sec^2(\theta)-1}$ or 3 $\sqrt{tan^2(\theta)}$ ? –  rkMathUser Aug 28 '11 at 19:02
    
@rkMathUser: the second looks more helpful, then clear the square root sign (remembering tor worry about signs) –  Ross Millikan Aug 29 '11 at 12:39
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I'd suggest to use the substitution $x:=3\cosh t$ instead. This leads to $$\eqalign{I&=\int{1\over 9\cosh^2 t\ 3\sinh t}\ 3\sinh t\ dt=\int{1\over 9\cosh^2 t}\ dt\cr &={1\over9}\tanh t+C={\sqrt{\cosh^2 t -1}\over9\cosh t}+C={\sqrt{x^2-9}\over 9 x}+C\ .\cr}$$

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Sadly, hyperbolic functions are apparently not so familiar to the kids these days... but I upvoted anyway. –  J. M. Aug 29 '11 at 11:48
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I think you made a mistake in substituting.

Added. Or rather, you made an algebra mistake. You seem to have gone from $$\sqrt{9\sec^2\theta - 9}$$ to $$3\sec\theta - 3.$$ That's incorrect. The square root does not distribute over sums and differences; that is, the square root of a difference is not the difference of the square roots (for example, $\sqrt{5} = \sqrt{9-4}$ is not equal to $\sqrt{9}-\sqrt{4} = 3-2=1$).

If $x=3\sec\theta$, then $x^2 - 9 = 9\sec^2\theta - 9 = 9(\sec^2\theta-1) = 9\tan^2\theta$, so that $\sqrt{x^2-9} = \sqrt{9\tan^2\theta} = 3|\tan\theta|$. For your substitution to work, though, you want to restrict $\theta$ to a nice interval where tangent is positive, so you can drop the absolute value bars.

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Using your substitution $x=3\sec \theta $ and cancelling $\sec \theta$ in the numerator and denominator, I got

$$I=\int \frac{1}{x^{2}\sqrt{x^{2}-9}}\ \textrm{d}x=\int \frac{ \tan \theta }{3\ \sec \theta \ \sqrt{9\sec ^{2}\theta -9}}\ \textrm{d}\theta=\int \frac{\tan \theta }{9\ \sec \theta \ \sqrt{\sec ^{2}\theta -1}}\ \textrm{d}\theta.$$

It is easy to see that

$$\frac{\tan \theta }{ \sec \theta \sqrt{\sec ^{2}\theta -1}}= \cos \theta .$$

So

$$I=\int \frac{1}{9}\cos \theta \ \textrm{d}\theta = \dots .$$

Added. Just to confirm Christian Blatter's evaluation. $$\begin{eqnarray*} I &=&\int \frac{1}{9}\cos \theta \,d\theta =\frac{1}{9}\sin \theta +C \\ &=&\frac{1}{9}\sqrt{1-\cos ^{2}\theta }+C=\frac{1}{9}\sqrt{1-\frac{1}{\sec ^{2}\theta }} +C\\ &=&\frac{1}{9}\sqrt{1-\frac{9}{x^{2}}}+C=\frac{\sqrt{x^{2}-9}}{9x}+C. \end{eqnarray*}$$

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