8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ If $x=3\sec\theta$, then $\theta = \sec^{-1}(x/3)$. Note that $\sqrt{x^2/9} = |x|/3$, not $x/3$. $\endgroup$ – Arturo Magidin Aug 28 '11 at 18:46
3
$\begingroup$

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 $\sqrt{9 \sec^2 \theta - 9}$?

$\endgroup$
  • $\begingroup$ Additional hint: you'll want to use $1+\tan^2\theta=\sec^2\theta$ at some point... $\endgroup$ – J. M. is a poor mathematician Aug 28 '11 at 18:52
  • $\begingroup$ A simpler way would be: 3 $\sqrt{sec^2(\theta)-1}$ or 3 $\sqrt{tan^2(\theta)}$ ? $\endgroup$ – ranonk Aug 28 '11 at 19:02
  • $\begingroup$ @rkMathUser: the second looks more helpful, then clear the square root sign (remembering tor worry about signs) $\endgroup$ – Ross Millikan Aug 29 '11 at 12:39
4
$\begingroup$

I'd suggest to use the substitution $x:=3\cosh t$ instead. This leads to $$\eqalign{I&=\int{1\over 9\cosh^2 t\ 3\sinh t}\ 3\sinh t\ dt=\int{1\over 9\cosh^2 t}\ dt\cr &={1\over9}\tanh t+C={\sqrt{\cosh^2 t -1}\over9\cosh t}+C={\sqrt{x^2-9}\over 9 x}+C\ .\cr}$$

$\endgroup$
  • $\begingroup$ Sadly, hyperbolic functions are apparently not so familiar to the kids these days... but I upvoted anyway. $\endgroup$ – J. M. is a poor mathematician Aug 29 '11 at 11:48
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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. $$\begin{eqnarray*} I &=&\int \frac{1}{9}\cos \theta \,d\theta =\frac{1}{9}\sin \theta +C \\ &=&\frac{1}{9}\sqrt{1-\cos ^{2}\theta }+C=\frac{1}{9}\sqrt{1-\frac{1}{\sec ^{2}\theta }} +C\\ &=&\frac{1}{9}\sqrt{1-\frac{9}{x^{2}}}+C=\frac{\sqrt{x^{2}-9}}{9x}+C. \end{eqnarray*}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.