Trigonometric substitution and triangles I'm learning trigonometric substitutions and am having a bit of trouble understanding the intuition behind the conversions (why do most use secant?). If you could explain the conversions geometrically using a triangle, that would be very helpful. For example, if we have $$\int \frac{\sqrt{x^2-4}}{x}\,dx$$ I tried to construct a triangle like so:

To get $$\sin(\theta)=\frac{\sqrt{x^2-4}}{x}$$
But this is incorrect for use as a substitution. Why?

It's not that it's wrong to take on the integral the way you propose -- it's just not awfully convenient.  If we implicitly differentiate your substitution equation, we obtain
$$ \cos \theta \ d\theta \ = \ \frac{x \ \cdot \frac{1}{2} \ (x^2 - 4)^{-1/2} \ \cdot 2x \ - \ (x^2 - 4)^{1/2} \ \cdot \ 1}{x^2} \ dx $$
$$ = \ \frac{4}{x^2 \ (x^2 - 4)^{1/2}} \ dx \ = \ \left(\frac{2}{x} \right)^2  \cdot \  \frac{2}{  \ (x^2 - 4)^{1/2}} \ \cdot \ \frac{1}{2} \ \ dx $$
$$ = \frac{1}{2} \ \cdot \ \cos^2 \theta \ \cdot \ \frac{1}{\tan \theta} \ \ dx $$
[using your right triangle] 
$$ \Rightarrow \ \ dx \ = \ 2 \ \tan \theta \ \cdot \ \frac{\cos \theta}{\cos^2 \theta} \ \ d\theta \ = \ 2 \ \tan \theta \ \sec \theta \ \ d\theta \ \ , $$
which is just the result Mike shows.  Following your lead, the integral becomes
$$ \int \ \sin \theta \ ( \ 2 \   \tan \theta \ \sec \theta \ \ d\theta  \ ) \ = \ 2 \int \ \sin \theta \ \tan \theta \ \frac{1}{\cos \theta} \ \ d\theta \ = \ 2 \int \   \tan^2 \theta   \ \ d\theta \ \ . $$
There are easier ways to get there, though.
