I am trying to find $$\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$$

$t = \sec \theta$ $dt = \sec \theta \tan\theta $

$$\int_\sqrt{2}^2 \frac{dt}{\sec ^2 \theta \sqrt{\sec^2 \theta-1}}$$

$$\int_\sqrt{2}^2 \frac{dt}{\sec ^2 \theta \tan^2 \theta}$$

$$\int_\sqrt{2}^2 \frac{\sec \theta \tan\theta}{\sec ^2 \theta \tan^2 \theta}$$

$$\int_\sqrt{2}^2 \frac{1}{\sec \theta}$$

$$\int_\sqrt{2}^2 \cos \theta$$

$$\sin \theta$$

Then I need to make it in terms of t.

$t = \sec \theta$

So I just use the arcsec which is

$\theta =\operatorname{arcsec} t$

$$\sin (\operatorname{arcsec} t)$$

This is wrong but I am not sure why.


These are the mistakes you have made:

  • When you substitute $t = \sec\theta$ the limits will change accordingly.

  • And $\sqrt{\sec^{2}\theta -1} \neq \tan^{2}\theta$ , its $\tan\theta$.

  • When $t= \sec\theta$, $\theta$ will change from $\frac{\pi}{4}$ to $\frac{\pi}{3}$.

  • When you make the change there will be a $d\theta$ term.

  • Your integral will look like $\displaystyle \int_{\pi/4}^{\pi/6} \frac{\sec\theta \tan\theta}{\sec^{2}\theta \cdot \tan\theta} \ d\theta = \int_{\pi/4}^{\pi/6} \cos\theta \ d\theta$.

  • $\begingroup$ How do I change the limits? I do not understand that part. $\endgroup$ – toby yeats Jun 4 '12 at 14:45
  • $\begingroup$ I am not good enough with algebra to figure it out, but how do I find the limits? $\endgroup$ – toby yeats Jun 4 '12 at 17:00
  • $\begingroup$ @jordan: First when u had the t term the limit was from $\sqrt{2}$ to 2. when u put $t=sec\theta$ since $t=2$ therefore $\sec\theta = 2$ and when $t = \sqrt{2}$ you have to see when $\sec\theta = \sqrt{2}$. $\endgroup$ – user9413 Jun 4 '12 at 17:03
  • $\begingroup$ That is the part I do not get I can not do the algebra to make sec equal to square root of 2. $\endgroup$ – toby yeats Jun 4 '12 at 17:05

$$\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$$
Substitute :
$t = \sec \theta$
$dt = \sec \theta \tan\theta d\theta$
The limits will also change accordingly
When $t=2$ , $\theta = \ arc sec(2) = \frac{\pi}{3}$
When $t=\sqrt2$ , $\theta = \ arc sec(\sqrt2) = \frac{\pi}{4}$ $$=\int_\frac{\pi}{4}^\frac{\pi}{3} \frac{\sec \theta \tan\theta d\theta}{\sec ^2 \theta \tan \theta}$$
$$=\int_\frac{\pi}{4}^\frac{\pi}{3} \cos \theta d \theta$$
$$=\sin \frac{\pi}{3} - \sin \frac{\pi}{4}$$
$$=\frac{\sqrt3 - \sqrt2}{2}$$
I think the answer you got is also correct
$$\sin (\operatorname{arcsec} (t))$$ by applying simple trigonometric rules $$=\frac{\sqrt{t^2-1}}{t}$$
and then applying the limits we get the same answer
$$=\frac{\sqrt3 - \sqrt2}{2}$$

  • $\begingroup$ This isn't really any different from the other answer and it still doesn't address my confusion with how to actually find the new bounds. $\endgroup$ – toby yeats Jun 4 '12 at 17:17
  • $\begingroup$ @Jordan I have written there how to get the new limits immediately after substitution as at t=2 and our substitution is $t=\sec \theta$ put there t=2 and solve for $\theta$ same for lower limit $\endgroup$ – Saurabh Jun 4 '12 at 17:56
  • $\begingroup$ $t=\sec \theta$ so when t=2 so the equation becomes $ 2= \sec \theta$ implies $0.5 = \cos \theta$ implies $\theta = \frac{\pi}{3}$ do same for the lower limit $\endgroup$ – Saurabh Jun 4 '12 at 18:25
  • $\begingroup$ I do not understand how these calculations are done. $\endgroup$ – toby yeats Jun 4 '12 at 19:33

To compute $$\int \frac{dt}{t^2 \sqrt{t^2-1}}$$ you may note that $$\cosh^2(x)-\sinh^2(x)=1,$$ or $$\cosh^2(x)-1=\sinh^2(x),$$ so putting $t=\cosh(x),$ then $dt=\sinh(x)dx$

$$\int \frac{dt}{t^2 \sqrt{t^2-1}}=\int \frac{\sinh(x)dx}{\cosh^2(x)\sqrt{\cosh^2(x)-1}}$$

$$=\int \frac{\sinh(x)dx}{\cosh^2(x)\sqrt{\sinh^2(x)}}=\int\frac{dx}{\cosh^2(x)}$$


and I leave the change of the bounds $t=\sqrt{2}$ to $t=2$ to you.


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.