I'm really spinning my wheels on this one:

$\displaystyle \int^2_{\sqrt{2}} \frac{1}{t^3\sqrt{t^2-1}}dt$

Let's just start with the general antiderivative:

let $\displaystyle t = \sec \theta$ then $\displaystyle dt = \sec\theta \tan\theta$

I know that substituting $\displaystyle \sec\theta$ for $\displaystyle \sqrt{t^2-1}$ yields $\displaystyle \tan\theta$. This is what I get next (general antiderivative):

$\displaystyle \int \frac{\sec\theta \tan\theta d\theta}{\sec^3\theta \tan\theta}$

Assuming that this substitution is correct, I then simplify the equation by canceling common factors, and I get:

$\displaystyle \int\cos^2 \theta d\theta$

which integrates to $\displaystyle \frac{1}{2}\sin^2\theta + C$

So, two questions:

1) Am I right to this point (solving for the general antiderivative), and if not, where am I going wrong?

2) Where I am really stuck is in trying to get the definite integral. How do I convert my boundaries from t to $\displaystyle \theta$ or get my general antiderivative back into t? This is really confusing to me.

Can anybody help?

Thanks!