Trig substitution

**silencecloak** · February 26th 2009, 05:14 PM

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

$x = sec\theta$

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

$dx = sec\theta*tan\theta$

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

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

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

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

Thanks

**mollymcf2009** · February 26th 2009, 05:33 PM

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 $2\pi$

**silencecloak** · February 26th 2009, 05:39 PM

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

**mollymcf2009** · February 26th 2009, 05:50 PM

Originally Posted by silencecloak

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

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

$arcsec (1) = \Theta = 0$

Thread: Trig substitution

LinkBack

Thread Tools

Display