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?
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.
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
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.
Since the constant (C or C') is arbitary, the ln(1/3) term makes no meaningful difference
of course. That makes perfect sense. Thanks!
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
You have made a mistake in substituting the values of sec t and tan t when converting the integral back to x.
Here's the correct substitution process:
Start with the integral ∫dx/√(x^2-9).
Substitute x = 3sec(t).
Then we have dx = 3sec(t)tan(t)dt, and √(x^2-9) = √(9sec^2(t) - 9) = 3tan(t).
Now rewrite the integral in terms of t: ∫(3sec(t)tan(t))/(3tan(t)) dt.
This simplifies to ∫ sec(t) dt.
Integrating sec(t) gives us ln|sec(t) + tan(t)| + C.
To convert the answer back to x, we need to substitute the values of sec(t) and tan(t) back in terms of x:
sec(t) = x/3,
tan(t) = √(x^2 - 9)/3.
So the final answer should be ln|x/3 + (√(x^2 - 9)/3)| + C.
Make sure to substitute the correct values for sec(t) and tan(t) to get the right final answer.