# Thread: trig sub [trig question]

1. ## 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 $-3tan^{-1} \frac{\sqrt{x^2-9}}{3}$?

2. ## Re: trig sub [trig question]

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 $-3tan^{-1} \frac{\sqrt{x^2-9}}{3}$?
please post the original integral

3. ## 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 $\sqrt{x^2-9}-3 arctan\sqrt{x^2-9}/3$

4. ## Re: trig sub [trig question]

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

Simple Trig substitution; find my error

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