1. Trig substitution

$\displaystyle \int_1^2 \frac{\sqrt{x^2-1}}{x}dx$

$\displaystyle x = sec\theta$

$\displaystyle \sqrt{x^2-1} = tan\theta$

$\displaystyle dx = sec\theta*tan\theta$

$\displaystyle x = 2 ; \theta = \pi/3$

$\displaystyle x = 1 ; \theta = 2\pi$

$\displaystyle \int_{2\pi}^{\pi/3} \frac{tan\theta}{sec\theta}sec\theta*tan\theta$

This isnt yielding me the correct result, where am I going wrong?

Thanks

2. Originally Posted by silencecloak
$\displaystyle \int_1^2 \frac{\sqrt{x^2-1}}{x}dx$

$\displaystyle x = sec\theta$

$\displaystyle \sqrt{x^2-1} = tan\theta$

$\displaystyle dx = sec\theta*tan\theta$

$\displaystyle x = 2 ; \theta = \pi/3$

$\displaystyle x = 1 ; \theta = 2\pi$

$\displaystyle \int_{2\pi}^{\pi/3} \frac{tan\theta}{sec\theta}sec\theta*tan\theta$

This isnt yielding me the correct result, where am I going wrong?

Thanks

I think your new limits should be 0 & $\displaystyle \frac{\pi}{3}$ not $\displaystyle 2\pi$

3. Originally Posted by mollymcf2009
I think your new limits should be 0 & $\displaystyle \frac{\pi}{3}$ not $\displaystyle 2\pi$
$\displaystyle 1 = sec\theta ; \theta = 0$ ?

Im probably gonna get slammed for this, but that's not what my calculator says

4. Originally Posted by silencecloak
$\displaystyle 1 = sec\theta ; \theta = 0$ ?

Im probably gonna get slammed for this, but that's not what my calculator says
$\displaystyle arcsec (1) = \Theta = 0$