<s>[INST] Trig Subs issues Problem:
Use a trigonometric substitution to find
$$I = \int_\sqrt{3}^2{\frac{\sqrt{x^2-3}}{x}}\;dx$$
Give both an exact answer (involving $\pi$ and a square root) and a decimal estimate to 3 significant digits.

Here is the work I have so far.
$$I = \int_\sqrt{3}^2{\frac{\sqrt{x^2-3}}{x}}\;dx$$
$$1 + \tan^2\Theta = \sec^2\Theta$$
$$\sec^2\Theta - 1 =\ tan^2\Theta$$
$$x = \sqrt{3}\sec\Theta$$

$$\sqrt{(\sqrt{3}\sec\Theta)^2 - 3}$$
$$\sqrt{3(1+\tan^2\Theta)-3}$$
$$\sqrt{3\tan^2\Theta}$$
$$\sqrt{3} \ tan\Theta$$
$$I = \int_\sqrt{3}^2{\frac{\sqrt{3}\tan\Theta}{\sqrt{3}\sec\Theta}}\;dx$$
$$\frac{dx}{d\Theta} = (\sqrt{3}\sec\Theta)'$$
$$dx = \sqrt{3}\tan\Theta\sec \Theta d\Theta$$
I feel like around here I may have messed up and I'm not sure if I should switch the limits of integration to $\Theta$ numbers of $0$ to $\frac{\pi}{6}$ yet or not. Any help is appreciated =) Thanks.
 [/INST] You did well. 
Swith the limit when you replace $dx$ by $d\Theta$.
\begin{align}
I&= \int_0^{\frac{\pi}6} \frac{\tan \Theta}{\sec \Theta} \sqrt{3} \tan \Theta \sec \Theta \, d\Theta \\
&= \sqrt{3} \int_0^{\frac{\pi}{6}} \tan^2\Theta\, d\Theta\\
&= \sqrt{3} \int_0^{\frac{\pi}6} \sec^2 \Theta -1 \,d\Theta\\
&= \sqrt{3} \left( \tan \Theta - \Theta\right) \mid_0^\frac{\pi}{6} \\
&= \sqrt{3} \left( \frac1{\sqrt3}-\frac{\pi}6\right)\\
&= 1 - \frac{\pi\sqrt{3}}{6}
\end{align}
</s>