# Thread: Integration Help (using trig identities)

1. ## Integration Help (using trig identities)

please help! any tips or help would be awesome. ive gotten through part of the problem, but get lost.

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

$\displaystyle x=3\sec\Theta\\$

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

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

-clay

2. You're looking good, except you did not square your 3 in the radical after you made the sub.

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

Then, it simplifies to:

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

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

3. Originally Posted by galactus
Then, it simplifies to:

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

2$\displaystyle \int 3tan^{2}(t) dt$
One question, how did you get from 1 to 2, I understand that $\displaystyle sec^2(t)-1=tan^2(t)$, but what happens to the radical and the denominator?

4. Obviously, the square root of tan^2(t) gives tan(t).

The denominator cancels with the 3sec(t) in the dt

5. Well, you don't actually need any sort of trig. identity, so put $\displaystyle u^2=x^2-9$ and you'll turn your integral into an arctangent.

6. Yes, of course, but perhaps the poster is required to use trig sub. That was my impression anyway.