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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.

3 Answers 3

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

  • 0
    That is the part I do not get I can not do the algebra to make sec equal to square root of 2.2012-06-04
1

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

  • 0
    I do not understand how these calculations are done.2012-06-04