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Trig sub. integrals

  1. Feb 11, 2005 #1
    [tex]\int \frac{sqrt(9x^2-4)}{x}dx[/tex]

    I'm stuck at this one part, but i'll show you my steps so you guys can see if i'm doing it right or not.

    factor out the 9...
    [tex]\int \frac{sqrt(9(x^2-4/9)}{x}dx[/tex]

    x = 2/3*sec(theta)
    dx = 2/3*sec(theta)tan(theta) d(theta)

    then i just subbed 2/3*sec(theta) for one of the X first.

    [tex]sqrt(9(4/9*sec(theta)^2 - 4/9))[/tex] => [tex]sqrt(4*sec(theta)^2-4[/tex]

    => [tex]sqrt(4(sec(theta)^2-1))[/tex]

    using trig. id.....

    => 2tan(theta)

    now i will sub in for the other x and dx.

    [tex]\int \frac{2*tan(theta)}{2/3*sec(theta)}*2/3*sec(theta)*tan(theta)[/tex]

    2/3*sec(theta) canceles out each other so...

    [tex]2\int tan(theta)^2[/tex]

    => trig id... [tex]2\int sec(theta)^2-1[/tex]

    well i know the integral of sec(theta)^2 is just tan(theta). but what is the integral of 1? it would be x in other cases, but what would the integral of 1 be in this case?
  2. jcsd
  3. Feb 11, 2005 #2
    Assuming you did the rest of the work right, you're integrating with respect to theta so [tex] \int {1 d\theta}[/tex] is [tex]\theta[/tex]
  4. Feb 11, 2005 #3
    yea i knew that, but i didnt know how to sub in for theta. i can sub in for tan(theta) by drawing a right triangle.

    tan(theta) = [tex]\frac{sqrt(9x^2-4)}{2}[/tex]

    so the answer should be [tex]2*\frac{9x^2-4}{2} - 2(theta)[/tex]

    what should i sub in for theta?
    Last edited: Feb 12, 2005
  5. Feb 11, 2005 #4
    [tex]\int [a(x)+b(x)]dx = \int a(x) dx +\int b(x) dx[/tex]

    Intergration of 1 will be theta (not x)

    [tex] \theta = cos^{-1}\frac{2}{3x} [/tex]
    Last edited: Feb 11, 2005
  6. Feb 11, 2005 #5


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    How about
    [tex]\frac{3x}{2}=\cosh u [/tex] ?

    I think it comes out nicely...

  7. Feb 12, 2005 #6
    the answer is:
    [tex]sqrt((9*x^2-4))+2*arctan\frac{2}{sqrt(9*x^2-4)}[/tex] (which looks extactly like my answer if i subbed in [tex]arctan\frac{2}{sqrt(9*x^2-4)}[/tex] for theta)
    i used a computer to find the answer, but i would like to know how to find theta. can someone help?

    how did you get [tex] \theta = cos^{-1}\frac{2}{3x} [/tex]

    i tried to sub in what you got for theta, but it told me that i was wrong (i submit my hw online and it checks my stuff)
    Last edited: Feb 12, 2005
  8. Feb 12, 2005 #7
    if you can draw the triangle, write something like [tex] \cos \theta = (something) [/tex] and take the inverse cosine of both sides.
  9. Feb 12, 2005 #8


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    Just solve for theta from the above equation.
  10. Feb 12, 2005 #9
    sec^(-1)*3/2x = theta

    i got that, am i doing something wrong?
  11. Feb 12, 2005 #10


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    So was your final answer:
    [tex]\sqrt{9x^2-4} - 2(arccos(2/3x))[/tex]
  12. Feb 12, 2005 #11


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    It should be 3x/2 = sec(theta)
  13. Feb 12, 2005 #12
    the correct answer is [tex]sqrt((9*x^2-4))+2*arctan\frac{2}{sqrt(9*x^2-4)}[/tex]

    that means that if theta = [tex]arctan\frac{2}{sqrt(9*x^2-4)}[/tex]

    i would get the correct answer

    on the other hand, if i subbed in theta = [tex] \theta = cos^{-1}\frac{2}{3x} [/tex]

    i would get the wrong answer
  14. Feb 12, 2005 #13


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    It seems like the way you solved it, the answer comes out to:


    but the correct answer is:


    But I can't find the source of the discrepancy. Can somebody see the reason? I can't find any errors in your work. All I can think of is tan(theta)=1/tan(Pi/2-theta). Maybe somehow this caused a problem somewhere?? :yuck:
    Last edited: Feb 12, 2005
  15. Feb 12, 2005 #14


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    How about another substitution
    [tex] \frac{2}{3x}=\sin t [/tex] ?

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