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. 
 A: $t=\sec(x)$
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
A: there is no need to go back to the original variable. you could change the limits of integration at the same time like:
change of variable: $t = 1/\cos x, dt = \frac{\sin x}{\cos^2 x} dx$  and at $t = \sqrt 2, x = \pi/4$ and $t = 2, x = \pi/3$
$$\int_{\sqrt 2}^2 \dfrac{1}{t^3\sqrt{t^2 - 1}} \ dt = 
\int_{\pi/4}^{\pi/3} \cos^2 x \ dx
 =  \frac{1}{2}\int_{\pi/4}^{\pi/3} (1 + \cos 2x) \ dx
$$
A: Let $x=\cosh t$, and use the fact that $\cosh^2t-\sinh^2t=1$, along with $\cosh't=\sinh t$ and $\sinh't=\cosh t$. You'll get $$I=\int\frac{\sinh t}{\cosh^3t\cdot\sinh t}dt=\int\frac1{\cosh^3t}dt=\int\frac{\cosh t}{\cosh^4t}dt=\int\frac{d(\sinh t)}{(1+\sinh^2t)^2}=\int\frac{du}{(1+u^2)^2}$$ which is trivial. $($Of course, you'll have to pay attention to evaluating the new limits of integration at each substitution$)$.
A: By substituting $u=\sqrt{t^2-1}$ you get using $\newcommand{\dd}{\; \mathrm{d}}\dd u = \frac{t}{\sqrt{t^2-1}} \dd t$ and $t^2=u^2+1$ that
$$\int \frac1{t^3\sqrt{t^2-1}} \dd t= \int \frac1{t^4} \cdot \frac{t}{\sqrt{t^2-1}} \dd t = \int \frac1{(u^2+1)^2} \dd u.$$
(So in the end this seems just as a variation of Lucian's answer.)
A: Probably the easiest thing to do at this point is to use that for $0<x<\frac{\pi}{2}$,
$\;\;\;\sec x=\sqrt{2}\implies \cos x=\frac{1}{\sqrt{2}}\implies x=\frac{\pi}{4}$ and 
$\;\;\;\sec x=2\implies \cos x=\frac{1}{2}\implies x=\frac{\pi}{3}$.

Alternatively, you can use $t=\sec x\implies\cos x=\frac{1}{t}$ and $\sin x=\frac{\sqrt{t^2-1}}{t}$ $\;\;$if $0<x<\frac{\pi}{2}$,
and $\sin 2x=2\sin x\cos x$.
