# trig substitution

• Dec 13th 2012, 04:54 PM
Nervous
trig substitution
What have I done wrong

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

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

$\displaystyle x = 3sec\theta$

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

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

Therefore...

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

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

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

$\displaystyle 3tan\theta - 3x + C$
• Dec 13th 2012, 05:00 PM
Nervous
Re: trig substitution
And I know that equates to

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

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

$\displaystyle \sqrt{x^2-9} - arctan (\frac {\sqrt {x^2-9}}{3}) + C$
• Dec 13th 2012, 05:15 PM
skeeter
Re: trig substitution
Quote:

Originally Posted by Nervous
What have I done wrong ...

...

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

$\displaystyle 3tan\theta - {\color{red}3x} + C$

last line should be

$\displaystyle 3\tan{\theta} - {\color{red}3\theta} + C$