calculus

Integrate: dx/sqrt(x^2-9)
Answer: ln(x + sqrt(x^2 - 9)) + C

I'm getting the wrong answer. Where am I going wrong:

Substitute: x = 3 * sec t

sqrt(x^2 - 9) = sqrt(3) * tan t
dx = sqrt(3) * sec t * tan t

Integral simplifies to: sec t dt
Integrates to: ln|sec t + tan t| + C

t = sec^-1 x/3
sec t = x/3
tan t = sqrt(x^2 - 9) / 3

Converting answer to x is
ln|x/3 + sqrt(x^2 - 9) / 3| + C

My answer doesn't match solution. Where am I going wrong?

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asked by mathstudent
  1. I don't think you made the substitution correctly. If x = 3 sec t, then
    sqrt (x^2 -9) = sqrt (9 sec^2t - 9)
    = 3 sqrt(sec^2t-1) = 3 tan t

    and
    dx = 3 sec t tan t dt

    You seem to be getting sqrt 3 instead of 3.

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    posted by drwls
  2. Ack! Actually, I just typed that up wrong. I didn't make that mistake on paper. My answer is still coming up wrong. Thanks for helping drwls.

    sqrt(x^2 - 9) = 3 * tan t
    dx = 3 * sec t * tan t * dt

    The rest is the same:

    Integral simplifies to: sec t dt
    Integrates to: ln|sec t + tan t| + C

    t = sec^-1 x/3
    sec t = x/3
    tan t = sqrt(x^2 - 9) / 3

    Converting answer to x is
    ln|x/3 + sqrt(x^2 - 9) / 3| + C

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    posted by mathstudent
  3. Both answers are correct- the book's and yours. Here's why: Factor out the 1/3 in your answer and it can be written
    ln{1/3)[x+ sqrt(x^2 - 9)]} + C
    = ln (1/3) + ln[x+ sqrt(x^2 - 9)] + C
    = ln[x+ sqrt(x^2 - 9)] + C'
    where C' is a different constant.

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    posted by drwls
  4. Since the constant (C or C') is arbitary, the ln(1/3) term makes no meaningful difference

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    posted by drwls
  5. of course. That makes perfect sense. Thanks!

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    posted by mathstudent
  6. hmm there seems to be a problem...

    sec=opp/hyp, while tan=opp/adj

    therefore, sec=x/3 AND tan=sqrt(x^2-9)/3 cannot happen. If so can you please explain this. It may be simple algebra from this point, but it's 3:45 too! Thanks

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    posted by calchelp

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