# Trig substitution

• Feb 26th 2009, 06:14 PM
silencecloak
Trig substitution
$\int_1^2 \frac{\sqrt{x^2-1}}{x}dx$

$x = sec\theta$

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

$dx = sec\theta*tan\theta$

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

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

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

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

Thanks
• Feb 26th 2009, 06:33 PM
mollymcf2009
Quote:

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

$x = sec\theta$

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

$dx = sec\theta*tan\theta$

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

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

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

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

Thanks

I think your new limits should be 0 & $\frac{\pi}{3}$ not $2\pi$
• Feb 26th 2009, 06:39 PM
silencecloak
Quote:

Originally Posted by mollymcf2009
I think your new limits should be 0 & $\frac{\pi}{3}$ not $2\pi$

$1 = sec\theta ; \theta = 0$ ?

Im probably gonna get slammed for this, but that's not what my calculator says (Nerd)
• Feb 26th 2009, 06:50 PM
mollymcf2009
Quote:

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

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

$arcsec (1) = \Theta = 0$