Integrate: dx/sqrt(x^2-9)

Question

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?

Answer 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.

Answer 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

Answer 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.

Answer 4

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

Answer 5

of course. That makes perfect sense. Thanks!

Answer 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