# trig sub [trig question]

• Feb 6th 2014, 09:34 PM
Jonroberts74
trig sub [trig question]
I solved an integral using trig sub

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

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

I understand the 3 and sqrt{x^2-9}. My problems is, shouldn't it be $-3tan^{-1} \frac{\sqrt{x^2-9}}{3}$?
• Feb 7th 2014, 07:05 AM
romsek
Re: trig sub [trig question]
Quote:

Originally Posted by Jonroberts74
I solved an integral using trig sub

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

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

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

• Feb 7th 2014, 09:26 AM
Jonroberts74
Re: trig sub [trig question]
$\int \frac{\sqrt{x^2-9}}{x}dx$

I know my answer is correct, I just want to know how wolfram got $\sqrt{x^2-9}+ 3 arctan 3/\sqrt{x^2-9}$ instead of $\sqrt{x^2-9}-3 arctan\sqrt{x^2-9}/3$
• Feb 7th 2014, 10:14 AM
Tclack
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.