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Calculus II - Trigonometric Substitutions - Evaluate integral sqrt(x^2-9)/x dx

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Evaluate integral sqrt(x^2-9)/x dx

    2. Relevant equations



    3. The attempt at a solution

    I get this for my answer sqrt(x^2-9)+3sin^(-1)(3/x)+c which is apparently wrong because when I take the derivative of it I don't get the integrand
    my work is a attachment in this post
    wolfram alpha claims the answer is
    sqrt(x^2-9)+3tan^(-1)(3/sqrt(x^2-9))
    http://www.wolframalpha.com/input/?i=integral+sqrt%28x^2-9%29%2Fx+dx

    I'm not sure if I'm doing something wrong, i think i am doing something wrong becasue when I take the derivative I don't get the integrand but I don't see what I'm doing wrong

    thanks for any help
     

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  3. Aug 3, 2011 #2
    First of all, why are you torturing yourself? I get the feeling you LIKE trig substitution.
    Okay, now, I don't follow your work. Why do you have the integral initial and then it's suddenly equal to integral cos? Since you seem to use csc again (glutton for punishment...) the only two functions I woulod expect to see are csc and cot, as we learned we could do yesterday. So, I don't get the first step.
     
  4. Aug 3, 2011 #3
    Also, why not start to learn the Latex?
    First type [ tex ] with no spaces between the brackets and the word tex. Then, \sqrt{everything in the sqrt} for squirts, and \frac{top half}{bottom half} notice no slashy bar in between top and bottom. Go on and try it :) and then [ /tex ] no space to end the code. and \int for integral I think. Might want to look that one up. (I'm new too!)
     
  5. Aug 3, 2011 #4

    SammyS

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    Notice from your triangle that θ may be expressed in many different ways.

    Besides θ = sin-1(3/x), you could have any of the following & more:
    θ = csc-1(x/3)

    θ = tan-1(3/√(x2-9))

    ...​
     
  6. Aug 3, 2011 #5
    Oh geez. That's what you get for using unconventional trig subs :P
    My calculator has the BEST feature, I can put in two things and ask if they are equal. Saves me a lot of this kind of situation because the first thing I do when I don't get what the book has is ask if it's equal. :)
     
  7. Aug 3, 2011 #6
    Lol I don't like learning new things and don't want to use Latex I guess idk lol to lazy much rather just scan my paper, much less work...

    I don't like trig substitution but I'm trying to learn it, it's the section I'm working on at the moment in my textbook

    My first step... notice my triangle on the left

    cosine(theta)=adjacent/hypotenuse=sqrt(x^2-9)/x
    hence i can change the integrand to cosine(theta) dx... of course it's in terms of dx so I have to change that which I did by using the fact that
    sine(theta)=opposite/hypotenuse=3/x

    I then solved for x and got

    x= 3csc(theta)

    I then took the derrivative of this and got
    dx/dtheta = -3csc(theta)cot(theta)
    hence
    dx = -3csc(theta)cot(theta) dtheta

    I then plugged this back into my integral

    integral cosine(theta) dx = -3*integral[ cos(theta)csc(theta)cot(theta) dtheta]

    i then evaluated this integral

    -3[-theta -cot(theta)]
    if you look at my triangle cot(theta)=adjacent/opposite=sqrt(x^2-9)/3
    for theta I again used sin(theta)=3/x
    hence
    theta=sin^(-1)(3/x)

    hence i got this for my answer
    -3[-theta -cot(theta)] = 3sin^(-1)(3/x) + sqrt(x^2-9) + c

    note that the 3 canceled out from the second term

    like this makes perfect logic that would not to seem to be flawed and the intermediate steps in between are correct, at least appear to be, unless I'm doing something wrong like mixing up stupid trig identities because it's been so many years sense I originally learned them but I think I got everything correct and that my answer should be right... the only problem is that when i take the derivative of my answer I don't get the integrand =(
     
  8. Aug 3, 2011 #7
    ya wait so is my answer correct? or am i doing something wrong because I don't really see what, and ya i noticed that but

    d/dx(3 sin^(-1)(3/x)+sqrt(x^2-9)) = x/sqrt(x^2-9)-9/(sqrt(1-9/x^2) x^2)

    hence i think i may be doing something wrong
     
  9. Aug 3, 2011 #8

    SammyS

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    YES! It's correct !
     
  10. Aug 3, 2011 #9
    Yeah I see the first step now. Thanks for spelling it out for me :)

    Yes, your answer is correct. See if what you get when you differentiate might be equivalent to what you started with.
     
  11. Aug 3, 2011 #10
    x/sqrt(x^2-9)-9/(sqrt(1-9/x^2) x^2) = sqrt(x^2-9)/x
    ha i didn't know that
    guess my answer is right and yay we can use cosine substitutions ^_^
     
  12. Aug 3, 2011 #11
    Are you going to try hyperbolic substitution tomorrow? Because if you come back with sinh@ I swear I will not help you..... :P

    EDIT (You must forgive my sarcasm. I have an exam tomorrow and the prof lectured for two hours today and didn't take questions. I always tweak out before exams.)

    But one more sarcasm :)

    yay.
     
    Last edited: Aug 3, 2011
  13. Aug 3, 2011 #12
    hmmm... let me think about it... i don't think you can? Can you o_O??? I haven't studied the hyperbolic trigonometric functions yet =(, at least not enough detail to make any substitutions, but I thought there was no traditional way to define them with regards to sides of triangles like the rest of the trigonometric functions in terms of opposite, adjacent, hypotenuse? It would be interesting how that would turn out... or better yet :P I wounder for we can make a cis(theta) substitution >_<, i have done stuff like finding the exact answer of cos^(-1)(5/7) and stuff and had to learn how to use cis(theta)... hmm I don't think it would work... ha I bet you could ^_^ because cis(x)=cos(x)+isin(x) you would just have to divide by i if you had something like... hmm i don't know ^_^ but my thoughts are racing i want to see if you can... but ya so there's no convient way to define the hyperbolic functions in terms of sides of a triangle nor by using the cis function sense

    cosh(x)=(e^x+e^(-x))/2 as oppose to cos(x)=(e^(ix)+e^(-ix))/2

    hmmm i don't know?
     
  14. Aug 3, 2011 #13
    caffeine pills do the trick for me ^_^ take like two and you will be up all night and will get lots of studying done lol, one pill = two cups of coffee lolz
     
  15. Aug 3, 2011 #14

    SammyS

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    Yes, you can use cosine substitutions.

    Substitutions using hyperbolic trig functions is a very powerful integration technique.
     
  16. Aug 3, 2011 #15
    Could you give a example of a case in which one can use hyperbolic trig substitution?

    Also I imagine anything that can use a cis(theta) substitution can just be broken up into two seperate integrals and evaluated separately.
     
  17. Aug 4, 2011 #16
    could i just get like a example
     
  18. Aug 5, 2011 #17
    anyone?
     
  19. Aug 5, 2011 #18
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