# 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.

• Welcome to MSE! Please use LaTeX or Mathjax to format your question. – graydad Jan 23 '15 at 19:42
• @graydad Lighten up a bit when it's a first time user, please. At the very least, provide a link to the tutorial. Some folks have never heard of mathjax. – Namaste Jan 23 '15 at 19:43
• Are LaTeX and Mathjax websites – Jessica Garcia Tejeda Jan 23 '15 at 19:44
• It's a way to typeset equations and mathematics to display nicely. See mathjax tutorial – Namaste Jan 23 '15 at 19:44
• In the future I will use Mathjax. I'm currently on a tablet device and am a new user of MSE. Thank you for sharing that Aaron. – Jessica Garcia Tejeda Jan 23 '15 at 19:52

$t=\sec(x)$

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

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$)$.

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 $t = 2, x = \pi/3$

$$\int_{\sqrt 2}^2 \dfrac{1}{t^3\sqrt{t^2 - 1}} \ dt = \int_{\pi/4}^{\pi/3} \cos^2 x \ dx = \frac{1}{2}\int_{\pi/4}^{\pi/3} (1 + \cos 2x) \ dx$$

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.)

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 $\sin 2x=2\sin x\cos x$.

• I though that I couldn't just do that because I changed the variable. So because I changed the variable, I can't use the original bounds. – Jessica Garcia Tejeda Jan 23 '15 at 19:58
• You can either change the limits and then use the antiderivative you have (in terms of x), or you can write the antiderivative in terms of t and then use the original limits. – user84413 Jan 23 '15 at 20:03