1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Trig integrals Date
Integrating Trig Sep 7, 2005
Trig integral Aug 2, 2005
Integrals of Trig Functions May 20, 2005
Integration by trig. substitution May 16, 2005
Confusing Trig/rational Integral May 9, 2005