trig substitution

**Nervous** · December 13th 2012, 04:54 PM

What have I done wrong

$\int \frac{\sqrt{x^2-9}}{x}dx$

$sec\theta = \frac{x}{3}$

$x = 3sec\theta$

$dx = 3sec\theta tan\theta d\theta$

$\sqrt{x^2-9} = 3tan\theta$

Therefore...

$\int \frac{3tan\theta}{3sec\theta} 3sec\theta tan\theta d\theta$

$\int 3tan^2\theta d\theta$

$3 \int sec^2\theta - 1 d\theta$

$3tan\theta - 3x + C$

**Nervous** · December 13th 2012, 05:00 PM

And I know that equates to

$\sqrt{x^2-9} - 3x + C$

But mine book sayth the answer my soul so desireth is:

$\sqrt{x^2-9} - arctan (\frac {\sqrt {x^2-9}}{3}) + C$