# Trig substitution

• Feb 26th 2009, 05:14 PM
silencecloak
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
• Feb 26th 2009, 05:33 PM
mollymcf2009
Quote:

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 $\displaystyle 2\pi$
• Feb 26th 2009, 05:39 PM
silencecloak
Quote:

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 (Nerd)
• Feb 26th 2009, 05:50 PM
mollymcf2009
Quote:

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 (Nerd)

$\displaystyle arcsec (1) = \Theta = 0$