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Math Help - trig sub [trig question]

  1. #1
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    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}?
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  2. #2
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    Re: trig sub [trig question]

    Quote Originally Posted by Jonroberts74 View Post
    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
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  3. #3
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    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
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  4. #4
    Junior Member Tclack's Avatar
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    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.
    Thanks from Jonroberts74
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