2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Welcome to MSE! Please use LaTeX or Mathjax to format your question. $\endgroup$ – graydad Jan 23 '15 at 19:42
  • $\begingroup$ @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. $\endgroup$ – Namaste Jan 23 '15 at 19:43
  • $\begingroup$ Are LaTeX and Mathjax websites $\endgroup$ – Jessica Garcia Tejeda Jan 23 '15 at 19:44
  • $\begingroup$ It's a way to typeset equations and mathematics to display nicely. See mathjax tutorial $\endgroup$ – Namaste Jan 23 '15 at 19:44
  • 3
    $\begingroup$ 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. $\endgroup$ – Jessica Garcia Tejeda Jan 23 '15 at 19:52
2
$\begingroup$

$t=\sec(x)$

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

$\endgroup$
2
$\begingroup$

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$)$.

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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.)

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jessica Garcia Tejeda Jan 23 '15 at 19:58
  • $\begingroup$ 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. $\endgroup$ – user84413 Jan 23 '15 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.