Re: trig sub [trig question]

Quote:

Originally Posted by

**Jonroberts74** I solved an integral using trig sub

the answer I got was $\displaystyle \sqrt{x^2-9} - 3 sec^{-1} \frac{x}{3} + C$

I checked my answer on wolfram it gave $\displaystyle \sqrt{x^2-9} + 3tan^{-1} \frac{3}{\sqrt{x^2-9} }$

I understand the 3 and sqrt{x^2-9}. My problems is, shouldn't it be $\displaystyle -3tan^{-1} \frac{\sqrt{x^2-9}}{3}$?

please post the original integral

Re: trig sub [trig question]

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

I know my answer is correct, I just want to know how wolfram got $\displaystyle \sqrt{x^2-9}+ 3 arctan 3/\sqrt{x^2-9}$ instead of $\displaystyle \sqrt{x^2-9}-3 arctan\sqrt{x^2-9}/3$

Re: trig sub [trig question]

Wow, I'm having the exact same problem. I posted this last night:

http://mathhelpforum.com/calculus/22...-my-error.html

I'll follow your post and you can follow mine. Maybe we'll double our chances of getting the correction.