# Integration Help (using trig identities)

• Jan 26th 2009, 09:33 AM
claymation14
Integration Help (using trig identities)
please help! any tips or help would be awesome. ive gotten through part of the problem, but get lost.

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

$x=3\sec\Theta\\$

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

$\int\frac{\sqrt{3\sec^2\Theta-9}}{3\sec\Theta}\cdot3\sec\Theta\tan\Theta\ d\Theta$

-clay (Rock)
• Jan 26th 2009, 09:49 AM
galactus
You're looking good, except you did not square your 3 in the radical after you made the sub.

$\frac{\sqrt{9sec^{2}(t)-9}}{3sec(t)}\cdot 3sec(t)tan(t)dt$

Then, it simplifies to:

$\frac{\sqrt{9(sec^{2}(t)-1)}}{3sec(t)}\cdot 3sec(t)tan(t)dt$

$\int 3tan^{2}(t) dt$
• Jan 26th 2009, 10:02 AM
claymation14
Quote:

Originally Posted by galactus
Then, it simplifies to:

1 $\frac{\sqrt{9(sec^{2}(t)-1)}}{3sec(t)}\cdot 3sec(t)tan(t)dt$

2 $\int 3tan^{2}(t) dt$

One question, how did you get from 1 to 2, I understand that $sec^2(t)-1=tan^2(t)$, but what happens to the radical and the denominator?
• Jan 26th 2009, 10:14 AM
galactus
Obviously, the square root of tan^2(t) gives tan(t).

The denominator cancels with the 3sec(t) in the dt
• Jan 26th 2009, 10:33 AM
Krizalid
Well, you don't actually need any sort of trig. identity, so put $u^2=x^2-9$ and you'll turn your integral into an arctangent.
• Jan 26th 2009, 10:41 AM
galactus
Yes, of course, but perhaps the poster is required to use trig sub. That was my impression anyway.