Trig substitution integration of $\int1/(x^2\sqrt{x^2 - 9}) dx$ - stuck on a problem 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.
 A: I'd suggest to use the substitution $x:=3\cosh t$ instead. This leads to
$$\eqalign{I&=\int{1\over 9\cosh^2 t\ 3\sinh t}\ 3\sinh t\ dt=\int{1\over 9\cosh^2 t}\ dt\cr &={1\over9}\tanh t+C={\sqrt{\cosh^2 t -1}\over9\cosh t}+C={\sqrt{x^2-9}\over 9 x}+C\ .\cr}$$
A: 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 $\sqrt{9 \sec^2 \theta - 9}$?
A: 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. 
A: 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. 
$$\begin{eqnarray*}
I &=&\int \frac{1}{9}\cos \theta \,d\theta =\frac{1}{9}\sin \theta +C \\
&=&\frac{1}{9}\sqrt{1-\cos ^{2}\theta }+C=\frac{1}{9}\sqrt{1-\frac{1}{\sec
^{2}\theta }} +C\\
&=&\frac{1}{9}\sqrt{1-\frac{9}{x^{2}}}+C=\frac{\sqrt{x^{2}-9}}{9x}+C.
\end{eqnarray*}$$
