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Homework Help: Trig substitution integration problem, test in 1hr 30min

  1. May 6, 2008 #1
    1. The problem statement, all variables and given/known data

    √(9-x²) / (x²)

    2. Relevant equations

    Just trig substitution

    3. The attempt at a solution

    Ok, for trig sub I did

    so putting it into the equation


    where do I go from here, I tried getting help at Math forum, but they were complete tools. Plus I have had a good experience with this site in the past
  2. jcsd
  3. May 6, 2008 #2


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    Science Advisor

    You do understand, don't you that many of the people who post here, also post at other forums?

    Since you say this is an integral, √(9-x²) / (x²) is NOT the problem statement: it has to be something like [itex]\int \sqrt{9- x^2}/x^2 dx[/itex].

    Yes, with [itex]x= 3sin(\theta)[/itex] [itex]\sqrt{9- x^2}= 3\sqrt{1- sin^2(\theta)}= 3cos(\theta)[/itex]. Now, what is x2 in terms of [itex]\theta[/itex]. You need to replace that too. And what is dx? You need to write that in terms of [itex]\theta[/itex].
  4. May 6, 2008 #3
    I still think I am making a mistake somewhere

    I did

    so------- 3cosΘ/9sin²Θ= (1/3)cotΘcscΘ


    using the triangle i found
  5. May 6, 2008 #4
  6. May 6, 2008 #5
    I forgot to take into acount the dx, but when I do i get

    (3cosΘ/9sin²Θ)*(3cosΘ)=9cos²Θ/9sin²Θ= cot²

    but the integral of cot is ln(sinU) or it can be -ln(cscu)

    I am stuck
  7. May 6, 2008 #6
    x = 3siny

    dx = 3cosy dy

    (9-9siny^2)^1/2. 3cosy dy/9siny^2

    (9-9siny^2) = 9cosy^2 because cosy^2 + sin ^2 = 1, multiply both sides by 9.

    it becomes

    3cosy . 3 cos y dy / 9 siny^2

    9cosy^2 dy / 9 siny^2

    it becomes cosy^2 dy / sin^2 which is coty^2 dy

    but 1 + coty^2 = cscy^2. coty^2 = cscy^2 - 1

    now integrate this cscy^2 - 1

    you get -cot(y) - y.

    I am not so sure about my solution.
  8. May 6, 2008 #7
    Integrals can have many forms of answers and this one is one of them but what they share is that they have common output.
    Last edited by a moderator: May 3, 2017
  9. May 6, 2008 #8
    What Racer has done/said is correct; in this case, his solution is right, and integrals can be of different form. I agree after having made an error the first time :S
    Last edited: May 6, 2008
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