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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.

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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 $\displaystyle \int_{\pi/4}^{\pi/6} \frac{\sec\theta \tan\theta}{\sec^{2}\theta \cdot \tan\theta} \ d\theta = \int_{\pi/4}^{\pi/6} \cos\theta \ d\theta$.

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  • $\begingroup$ 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
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$$\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$$
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
$$=\frac{\sqrt3 - \sqrt2}{2}$$

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  • $\begingroup$ 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
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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 $dt=\sinh(x)dx$

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

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

$$=\int{\text{sech}^2(x)dx}=\tanh(x)+C$$

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

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