Integrating $\int_{\sqrt{2}}^2 \frac{1}{t^3\sqrt{t^2-1}}\,dt$. I am trying to compute
$$
\int_{\sqrt{2}}^2 \frac{1}{t^3\sqrt{t^2-1}}\,dt.
$$
This is what I got so far: 
$t=\sec(x)$ and $dt=\sec(x)\tan(x)x\,dx$ 
So plugging this in gives me 
$$
\int \frac{1}{\sec^3(x)}\cdot\sqrt{\sec^2(x)-1}\sec(x)\tan(x)\,dx.
$$
By my trig property $1+\tan^2(x)=\sec^2(x)$, I get
$$
\int \frac{1}{\sec^3(x)\tan(x)}\sec(x)\tan(x)\,dx.
$$
Then I simplify and get
$$
\int \frac{1}{\sec^2(x)}\,dx.
$$
By the trig property I get
$$
\int\cos^2(x)\,dx.
$$
Then I get $(x/2)+(1/4)\sin(2x).$ 
So now I need to get back to my original variable so I solve for $x$:
$t=\sec(x)$ and $x=\mathrm{arcsec}(t)$. 
It's this part I'm confused about. I don't know what step to take next. I know I am supposed to draw my triangle, I just don't know how to get there. 

$t=\sec(x)$
You can build a right triangle for this to translate your answer in terms of $t$ to in terms of $x$.
Also here's a hint: $\sin(2x)=2\sin(x)\cos(x)$
oops didn't realize it was a definite integral
